{"id":16,"date":"2024-04-04T07:30:15","date_gmt":"2024-04-04T04:30:15","guid":{"rendered":"https:\/\/sisu.ut.ee\/huvitavkeemia\/28-arvutused-saagise-ja-kao-pohjal\/"},"modified":"2025-02-12T23:30:19","modified_gmt":"2025-02-12T21:30:19","slug":"28-arvutused-saagise-ja-kao-pohjal","status":"publish","type":"page","link":"https:\/\/sisu.ut.ee\/huvitavkeemia\/28-arvutused-saagise-ja-kao-pohjal\/","title":{"rendered":"2.8. Arvutused saagise ja kao p\u00f5hjal"},"content":{"rendered":"<p>\n\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">V\u00f5ib juhtuda, et tegelike saaduste hulk erineb oluliselt teoreetiliselt arvutatud kogustest. <\/span><\/span><\/span>Reaktsiooni saagis n\u00e4itab, mitu protsenti moodustab tegelikult saadud reaktsioonisaaduse hulk (v\u00f5i mass) reaktsiooniv\u00f5rrandi j\u00e4rgi teoreetiliselt arvutatud reaktsioonisaaduse hulgast (v\u00f5i massist). J\u00e4rgnevas videos saad vaadata, kuidas arvutada reaktsiooni saagist.\n<\/p>\n<p>\n\t<\/p><div class=\"ratio ratio-16x9 mb-3\"><div class=\"video-placeholder-wrapper video-placeholder-wrapper--16x9\">\n\t\t\t    <div class=\"video-placeholder d-flex justify-content-center align-items-center\">\n\t\t\t        <div class=\"overlay text-white p-2 w-100 text-center d-block justify-content-center align-items-center\">\n\t\t\t            <div>Kolmandate osapoolte sisu n\u00e4gemiseks palun n\u00f5ustu k\u00fcpsistega.<\/div>\n\t\t\t            <button class=\"btn btn-secondary btn-sm mt-1 consent-change\">Muuda n\u00f5usolekut<\/button>\n\t\t\t        <\/div>\n\t\t\t    <\/div>\n\t\t\t<\/div>\n<\/div>\n\n<h6>\n\tAllikas:\u00a0<a data-url=\"https:\/\/youtu.be\/0epwo6KCfbQ\" href=\"https:\/\/youtu.be\/0epwo6KCfbQ\" target=\"_blank\" title=\"\" rel=\"noopener\">https:\/\/youtu.be\/0epwo6KCfbQ<\/a><br>\n<\/h6>\n<p>\n\tAine massi v\u00f5i hulka, mis protsessi tulemusel saadakse, m\u00f5jutavad paljud tegurid, n\u00e4iteks gaasiliste ainete korral v\u00f5ib osa ainet lenduda.\n<\/p>\n<p>\n\tReaktsiooni saagise arvutamisel kasutatakse protsendi arvutamise reegleid. Saagise, tavaliselt saagise protsendi leidmiseks saab kasutada j\u00e4rgmist valemit:\n<\/p>\n<p style=\"text-align: center\">\n\t$P(saagis)=\\frac{m(protsessi \u00a0k\u00e4igus \u00a0saadud \u00a0aine \u00a0mass)}{m(arvutatud\u00a0 mass)})\\times 100$\n<\/p>\n<p>\n\tSarnaselt massidega saab saagist arvutada ka ainehulga kaudu:\n<\/p>\n<p style=\"text-align: center\">\n\t$P(saagis)=\\frac{n(protsessi \u00a0k\u00e4igus \u00a0saadud \u00a0a\\mspace{0mu}inehulk)}{n(arvutatud \u00a0a\\mspace{0mu}inehulk)}\\times 100$\n<\/p>\n<h2>\n\tN\u00e4idis\u00fclesanne 1. Reaktsiooni saagise arvutamine<br>\n<\/h2>\n<table class=\"table table-hover\" border=\"0\" cellpadding=\"1\" cellspacing=\"1\" style=\"width: 100%\">\n<tbody>\n<tr>\n<td>\n\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">1. Loe \u00fclesande tekst m\u00f5ttega l\u00e4bi.<\/span><\/span><\/span>\n\t\t\t<\/td>\n<td style=\"width: 70%\">\n<p>\n\t\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">Arvuta saagise protsent, kui 6,5 grammi\u00a0 fruktoosi lagunemisel allpool toodud reaktsiooni alusel tekkis 1,27 dm<sup>3<\/sup> s\u00fcsihappegaasi (nt).<\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<p>\n\t\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub><\/span><\/span><\/span><span lang=\"et\"><span style=\"line-height:115%\"><span> \u2192 2C<\/span><\/span><\/span><sub><span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">2<\/span><\/span><\/span><\/sub><span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">H<sub>5<\/sub>OH + 2CO<sub>2<\/sub><\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">2. Jooni tekstis alla k\u00f5ik arvud.<\/span><\/span><\/span>\n\t\t\t<\/td>\n<td>\n<p>\n\t\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">Arvuta saagise protsent, kui<u> 6,5<\/u> grammi fruktoosi lagunemisel allpool toodud reaktsiooni alusel tekkis <u>1,27<\/u> dm<sup>3<\/sup> s\u00fcsihappegaasi (nt).<\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<p>\n\t\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub><\/span><\/span><\/span><span lang=\"et\"><span style=\"line-height:115%\"><span> \u2192 2C<\/span><\/span><\/span><sub><span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">2<\/span><\/span><\/span><\/sub><span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">H<sub>5<\/sub>OH + 2CO<sub>2<\/sub><\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p style=\"margin-top:12.0pt;margin-right:0cm;margin-bottom:12.0pt;margin-left:0cm\">\n\t\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">3. Omista arvule kindel f\u00fc\u00fcsikaline suurus. Seda on lihtne teha n\u00e4iteks \u00fchiku alusel.<\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<\/td>\n<td>\n\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">6,5 g fruktoosi on l\u00e4hteaine kogus. 1,27 dm<sup>3<\/sup> on s\u00fcsihappegaasi ruumala, mis protsessi k\u00e4igus l\u00f5puks saadi.<\/span><\/span><\/span>\n\t\t\t<\/td>\n<\/tr>\n<tr>\n<td>\n\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">4. Kirjuta v\u00e4lja andmed ning otsitav suurus.<\/span><\/span><\/span>\n\t\t\t<\/td>\n<td>\n<p style=\"margin-top:12.0pt;margin-right:0cm;margin-bottom:12.0pt;margin-left:0cm\">\n\t\t\t\t\t<i><span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">m<\/span><\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">(C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub>) = 6,5 g<\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<p style=\"margin-top:12.0pt;margin-right:0cm;margin-bottom:12.0pt;margin-left:0cm\">\n\t\t\t\t\t<i><span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">V<\/span><\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">(CO<sub>2<\/sub>) = 1,27 dm<sup>3<\/sup><\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<p style=\"margin-top:12.0pt;margin-right:0cm;margin-bottom:12.0pt;margin-left:0cm\">\n\t\t\t\t\t<i><span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">P<\/span><\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">(saagis) = ?<\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">5. Kirjuta v\u00e4lja valemid ning kui vaja, avalda valemist otsitav suurus.<\/span><\/span><\/span>\n\t\t\t<\/td>\n<td>\n<p style=\"margin-top:12.0pt;margin-right:0cm;margin-bottom:12.0pt;margin-left:0cm\">\n\t\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">S\u00fcsihappegaasi saab leida valemiga<\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<p style=\"margin-top:12.0pt;margin-right:0cm;margin-bottom:12.0pt;margin-left:0cm\">\n\t\t\t\t\t$n=\\frac{V}{V_{m}}$\n\t\t\t\t<\/p>\n<p style=\"margin-top:12.0pt;margin-right:0cm;margin-bottom:12.0pt;margin-left:0cm\">\n\t\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub> massi saab leida\u00a0 <i>m<\/i>(C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub>) = <i>n\u00a0 <\/i><span style=\"background:white\">$\\times $ <i>M<\/i>.<\/span><\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<p style=\"margin-top:12.0pt;margin-right:0cm;margin-bottom:12.0pt;margin-left:0cm\">\n\t\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">Saagise protsendi leidmiseks saab kasutada j\u00e4rgmist valemit:<\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<p>\n\t\t\t\t\t$P(saagis)=\\frac{m(reaalse \u00a0protsessi \u00a0k\u00e4igus \u00a0saadud \u00a0aine \u00a0mass)}{m(teoreetiliselt \u00a0saadud \u00a0aine \u00a0mass)})\\times 100$\n\t\t\t\t<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p style=\"margin-top:12.0pt;margin-right:0cm;margin-bottom:12.0pt;margin-left:0cm\">\n\t\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">6. Asenda valemitesse arvud ning teosta arvutused.<\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<\/td>\n<td>\n<p style=\"margin-top:12.0pt;margin-right:0cm;margin-bottom:12.0pt;margin-left:0cm\">\n\t\t\t\t\tPraktiliselt tekkinud s\u00fcsihappegaasi hulk:\n\t\t\t\t<\/p>\n<p style=\"margin-top:12.0pt;margin-right:0cm;margin-bottom:12.0pt;margin-left:0cm\">\n\t\t\t\t\t$n(CO_{2})=\\frac{V}{V_{m}}=\\frac{1,27 dm^{3}}{22,4 dm^{3}\/mol}=0,057 mol$\n\t\t\t\t<\/p>\n<p style=\"margin-top:12.0pt;margin-right:0cm;margin-bottom:12.0pt;margin-left:0cm\">\n\t\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">S\u00fcsihappegaasi teoreetilise hulga arvutamiseks arvutame fruktoosi molaarmassi ja selle kaudu fruktoosi hulga moolides:<\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<p style=\"margin-top:12.0pt;margin-right:0cm;margin-bottom:12.0pt;margin-left:0cm\">\n\t\t\t\t\t<i><span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">M<\/span><\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">(C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub>) = 6 <span style=\"background:white\">$\\times $ 12 + 12 $\\times $ 1 + <\/span>6 <span style=\"background:white\">$\\times $ 16 =\u00a0 180 g\/mol<\/span><\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<p style=\"margin-top:12.0pt;margin-right:0cm;margin-bottom:12.0pt;margin-left:0cm\">\n\t\t\t\t\t$n(C_{6}H_{12}O_{6})=\\frac{m}{M}=\\frac{6,5 g}{180 g\/mol}=0,036\u00a0mol$\n\t\t\t\t<\/p>\n<p style=\"margin-top:12.0pt;margin-right:0cm;margin-bottom:12.0pt;margin-left:0cm\">\n\t\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">Reaktsiooniv\u00f5rrandi alusel tekib 1 mooli C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub> kohta 2 mooli CO<sub>2<\/sub>. <\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<p style=\"margin-top:12.0pt;margin-right:0cm;margin-bottom:12.0pt;margin-left:0cm\">\n\t\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">J\u00e4relikult peaks s\u00fcsihappegaasi teoreetiliselt tekkima<\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<p style=\"margin-top:12.0pt;margin-right:0cm;margin-bottom:12.0pt;margin-left:0cm\">\n\t\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\"><em>n<\/em>(CO<sub>2<\/sub>) = 2\u00a0<span style=\"background:white\">$\\times $ 0,036 mol = 0,072 mol<\/span><\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<p style=\"margin-top:12.0pt;margin-right:0cm;margin-bottom:12.0pt;margin-left:0cm\">\n\t\t\t\t\t<span lang=\"et\"><span style=\"background:white\"><span style=\"line-height:115%\"><span style=\",serif\">Reaktsiooni saagis:<\/span><\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<p>\n\t\t\t\t\t$P(saagis)=\\frac{0,057 mol}{0,072 mol}\\times 100=79,2%\\approx 79%$\n\t\t\t\t<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>\n\t<\/p><div class=\"accordion mb-3\">\n        <div class=\"accordion-item accordion-item--white\">\n        <h2 class=\"accordion-header\" id=\"accordion-69f5ba06520d4-heading\">\n            <button class=\"accordion-button collapsed\" type=\"button\" data-bs-toggle=\"collapse\" data-bs-target=\"#accordion-69f5ba06520d4-collapse\" aria-expanded=\"true\" aria-controls=\"accordion-69f5ba06520d4-collapse\">\u00dclesanne 1<\/button>\n        <\/h2>\n        <div id=\"accordion-69f5ba06520d4-collapse\" class=\"accordion-collapse collapse\" aria-labelledby=\"accordion-69f5ba06520d4-heading\">\n            <div class=\"accordion-body\">\n\n\n<div class=\"h5p-iframe-wrapper\"><div class=\"video-placeholder-wrapper video-placeholder-wrapper--fixed\" style=\"height: 366px;\">\n\t\t\t    <div class=\"video-placeholder d-flex justify-content-center align-items-center\">\n\t\t\t        <div class=\"overlay text-white p-2 w-100 text-center d-block justify-content-center align-items-center\">\n\t\t\t            <div>Kolmandate osapoolte sisu n\u00e4gemiseks palun n\u00f5ustu k\u00fcpsistega.<\/div>\n\t\t\t            <button class=\"btn btn-secondary btn-sm mt-1 consent-change\">Muuda n\u00f5usolekut<\/button>\n\t\t\t        <\/div>\n\t\t\t    <\/div>\n\t\t\t<\/div>\n<\/div>\n\n\n<p><\/p><\/div>\n        <\/div>\n        <\/div>\n    <\/div>\n<p><\/p><div class=\"accordion mb-3\">\n        <div class=\"accordion-item accordion-item--white\">\n        <h2 class=\"accordion-header\" id=\"accordion-69f5ba06520ed-heading\">\n            <button class=\"accordion-button collapsed\" type=\"button\" data-bs-toggle=\"collapse\" data-bs-target=\"#accordion-69f5ba06520ed-collapse\" aria-expanded=\"true\" aria-controls=\"accordion-69f5ba06520ed-collapse\">\u00dclesande 1 lahendused<\/button>\n        <\/h2>\n        <div id=\"accordion-69f5ba06520ed-collapse\" class=\"accordion-collapse collapse\" aria-labelledby=\"accordion-69f5ba06520ed-heading\">\n            <div class=\"accordion-body\">\n<p>1.1. 5 g KClO<sub>3<\/sub> kuumutati ja see lagunes kaaliumkloriidiks ja hapnikuks. Mis on saagise protsent, kui reaktisooni tulemusena tekkis 1,5 g hapnikku. Anna vastus t\u00e4isarvuna.<\/p>\n<p>2KClO<sub>3<\/sub> \u2192 2KCl + 3O<sub>2<\/sub><\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus<\/b>: Saagise protsent on 77%<\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus: <\/b>Esmalt arvutame, kui palju hapnikku oleks teoreetiliselt pidanud tekkima. Meil on teada KClO<sub>3<\/sub> mass <i>m<\/i>=5 g, seega saame arvutada selle moolide arvu:<\/p>\n<p style=\"margin-left: 80px;\">M(<span lang=\"ET\" style=\"color: black;\">K<\/span><span style=\"color: black;\">Cl<\/span><span lang=\"ET\" style=\"color: black;\">O<sub>3<\/sub><\/span>) = 122,5 g\/mol<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$n = {m \\over M}={5g \\over 122.5g\/mol}=0.0408mol$<\/span><\/p>\n<p style=\"margin-left: 40px;\">Reaktsiooniv\u00f5rrandi j\u00e4rgi on KClO<sub>3<\/sub> ja hapniku moolsuhe 2:3, seega oleks hapnikku pidanud tekkima <i><span lang=\"ET\">0,0408 \u00d7\u00a0<\/span><\/i><i><span lang=\"ET\">3 \/\u00a0<\/span><\/i><i><span lang=\"ET\">2\u00a0<\/span><\/i><i><span lang=\"ET\">= 0,061<\/span><\/i>\u00a0mooli. Leiame n\u00fc\u00fcd tekkinud hapniku massi kaudu, kui palju seda tegelikult tekkis:<\/p>\n<p style=\"margin-left: 80px;\">M(O<sub>2<\/sub>) = 32 g\/mol<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$n = {m \\over M}={1.5g \\over 32g\/mol}=0.047mol$<\/span><\/p>\n<p style=\"margin-left: 40px;\">Kuna n\u00fc\u00fcd on teada nii eeldatav hapniku ainehulk kui ka tegelik hapniku ainehulk, saame arvutada saagise:<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$P(Saagis) = {0.047mol \\over 0.061mol}*100=77$<\/span><\/p>\n<p>\u00a0<\/p>\n<p>1.2. 5 mooli l\u00e4mmastiku reageerimisel vesinikuga tekkis 2,5 mol ammoniaaki (NH<sub>3<\/sub>). Arvutage reaktasiooni saagise protsent. Anna vastus t\u00e4isarvuna.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus: <\/b>25%<\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus:<\/b> Kirjutame ja tasakaalustame reaktsiooniv\u00f5rrandi: N<sub>2<\/sub> + 3H<sub>2<\/sub> \u2192 2NH<sub>3<\/sub><\/p>\n<p style=\"margin-left: 40px;\">Reaktsiooniv\u00f5rrandist n\u00e4eme, et l\u00e4mmastiku ja ammoniaagi moolsuhe on 1:2. Seega oleks 5 mooli l\u00e4mmastiku reageerimisel pidanud teoreetiliselt tekkima 2\u00d75=10 mol ammoniaaki. Tekkis aga 2,5 mol, seega on reaktsiooni saagis:<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$P(Saagis) = {2.5mol \\over 10mol}*100=25$<\/span><\/p>\n<p>\u00a0<\/p>\n<p>1.3. 17,5 mooli l\u00e4mmastiku reageerimisel katal\u00fcsaatori juuresolekul piisava hulga vesinikuga tekkis 115 g ammoniaaki. Arvutage reaktsiooni saagise % ja eraldunud ammoniaagi ruumala (nt.). Anna vastus t\u00e4isarvuna.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus<\/b>: 19%, 152 dm<sup>3<\/sup><\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus:<\/b> Kirjutame ja tasakaalustame reaktsiooniv\u00f5rrandi: N<sub>2<\/sub> + 3H<sub>2<\/sub> \u2192 2NH<sub>3<\/sub><\/p>\n<p style=\"margin-left: 40px;\">Reaktsiooniv\u00f5rrandist n\u00e4eme, et l\u00e4mmastiku ja ammoniaagi moolsuhe on 1:2. Seega oleks 17,5 mooli l\u00e4mmastiku reageerimisel pidanud teoreetiliselt tekkima 2\u00d717,5=35 mol ammoniaaki. Leiame aga antud ammoniaagi massi kaudu, kui palju seda p\u00e4riselt tekkis:<\/p>\n<p style=\"margin-left: 80px;\">M(NH<sub>3<\/sub>)=17 g\/mol<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$n = {m \\over M}={115g \\over 17g\/mol}=6.77mol$<\/span><\/p>\n<p style=\"margin-left: 40px;\">Reaktsiooni saagis on seega:<\/p>\n<p style=\"padding-left: 80px;\"><span class=\"math-tex\">$P(Saagis) = {6.77mol \\over 35mol}*100=19$<\/span><\/p>\n<p style=\"margin-left: 40px;\">Ja eraldunud ammoniaagi ruumala:<i><\/i><\/p>\n<p style=\"margin-left: 80px;\"><i><span lang=\"ET\">V\u00a0=\u00a0n\u00a0\u00d7\u00a0<\/span><\/i><i><span lang=\"ET\">V<\/span><\/i><i><span lang=\"ET\"><sub>m<\/sub>\u00a0<\/span><\/i><i><span lang=\"ET\">=\u00a0<\/span><\/i><span lang=\"ET\">6,77 mol \u00d7 22,4 <\/span><span lang=\"ET\">dm<\/span><sup><span lang=\"ET\"><sup>3<\/sup><\/span><\/sup><span lang=\"ET\">\/mol = 151,5 \u2248 152 d<\/span><span lang=\"ET\">m<\/span><sup><span lang=\"ET\"><sup>3<\/sup><\/span><\/sup><\/p>\n<p style=\"margin-left: 40px;\">\u00a0<\/p>\n<p>1.4. 340 g ammoniaagi (NH<sub>3<\/sub>) s\u00fcnteesiks kulus 896 dm<sup>3<\/sup> vesinikku. Leida s\u00fcnteesi saagise protsent. Anna vastus t\u00e4isarvuna.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus<\/b>: 75%<\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus: <\/b>Kirjutame ja tasakaalustame reaktsiooniv\u00f5rrandi: N<sub>2<\/sub> + 3H<sub>2<\/sub> \u2192 2NH<sub>3<\/sub><\/p>\n<p style=\"margin-left: 40px;\">Esmalt arvutame, kui palju ammoniaaki oleks teoreetiliselt pidanud tekkima. Meil on teada vesiniku ruumala <i>V<\/i>=896 dm<sup>3<\/sup>, seega saame arvutada selle moolide arvu:<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$n = {V \\over V_m}={896dm^3 \\over 22.4dm^3\/mol}=40mol$<\/span><\/p>\n<p style=\"margin-left: 40px;\">Reaktsiooniv\u00f5rrandist n\u00e4eme, et vesiniku ja ammoniaagi moolsuhe on 3:2. Seega oleks 40 mooli l\u00e4mmastiku reageerimisel pidanud teoreetiliselt tekkima <i><span lang=\"ET\">40 \u00d7\u00a0<\/span><\/i><i><span lang=\"ET\">2 \/\u00a0<\/span><\/i><i><span lang=\"ET\">3\u00a0<\/span><\/i><i><span lang=\"ET\">= 26,67<\/span><\/i>\u00a0mooli ammoniaaki. Leiame aga antud ammoniaagi massi kaudu, kui palju seda p\u00e4riselt tekkis:<\/p>\n<p style=\"margin-left: 80px;\">M(NH<sub>3<\/sub>)=17 g\/mol<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$n = {m \\over M}={340g \\over 17g\/mol}=20mol$<\/span><\/p>\n<p style=\"margin-left: 40px;\">Reaktsiooni saagis on seega:<\/p>\n<p style=\"padding-left: 80px;\"><span class=\"math-tex\">$P(Saagis) = {20mol \\over 26.67mol}*100=75$<\/span><\/p>\n<p>\u00a0<\/p>\n<p>1.5. V\u00e4\u00e4velhappe tootmisel kasutati l\u00e4hteainena p\u00fcriiti (FeS<sub>2<\/sub>). 200 kg p\u00fcriidi s\u00e4rdamisel saadi 180 kg v\u00e4\u00e4veldioksiidi. Kui suur on protsessi saagis? Anna vastus t\u00e4isarvuna.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus<\/b>: 84%<\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus:<\/b> Kirjutame ja tasakaalustame reaktsiooniv\u00f5rrandi:<\/p>\n<p style=\"margin-left: 80px;\">4FeS<sub>2<\/sub> +11 O<sub>2<\/sub> \u2192 2Fe<sub>2<\/sub>O<sub>3<\/sub> + 8SO<sub>2<\/sub><\/p>\n<p style=\"margin-left: 40px;\">Esmalt arvutame, kui palju v\u00e4\u00e4veldioksiidi oleks teoreetiliselt pidanud tekkima. Meil on teada FeS<sub>2<\/sub> mass <i>m<\/i>=200 kg, seega saame arvutada selle moolide arvu:<\/p>\n<p style=\"margin-left: 80px;\">M(FeS<sub>2<\/sub>) = 120 g\/mol=120 kg\/kmol<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$n = {m \\over M}={200kg \\over 120kg\/kmol}=1.67kmol$<\/span><\/p>\n<p style=\"margin-left: 40px;\">Reaktsiooniv\u00f5rrandist n\u00e4eme, et FeS<sub>2<\/sub> ja SO<sub>2<\/sub> moolsuhe on 4:8=1:2. Seega oleks 1,67 kilomooli FeS<sub>2<\/sub> reageerimisel pidanud teoreetiliselt tekkima 2\u00d71,67=3,34 kmol v\u00e4\u00e4veldioksiidi. Leiame aga antud v\u00e4\u00e4veldioksiidi massi kaudu, kui palju seda p\u00e4riselt tekkis:<\/p>\n<p style=\"margin-left: 80px;\">M(SO<sub>2<\/sub>)=64 g\/mol=64 kg\/kmol<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$n = {m \\over M}={180kg \\over 64kg\/kmol}=2.81kmol$<\/span><\/p>\n<p style=\"margin-left: 40px;\">Reaktsiooni saagis on seega:<i><\/i><\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$P(Saagis) = {2.81mol \\over 3.34mol}*100=84$<\/span><\/p><\/div>\n        <\/div>\n        <\/div>\n    <\/div>\n<h2>N\u00e4idis\u00fclesanne 2. Saaduse koguse arvutamine, arvestades reaktsiooni kadu<\/h2>\n<table class=\"table table-hover\" style=\"width: 100%;\" border=\"0\" cellspacing=\"1\" cellpadding=\"1\">\n<tbody>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">1. Loe \u00fclesande tekst m\u00f5ttega l\u00e4bi.<\/span><\/span><\/td>\n<td style=\"width: 70%;\">\n<p><span lang=\"et\"><span style=\"line-height: 115%;\">P\u00fcriidi (FeS<sub>2<\/sub>) p\u00f5letamisel (s\u00e4rdamisel) saadakse gaasiline SO<sub>3<\/sub>. Reaktsiooni kohta v\u00f5ib v\u00e4lja kirjutada j\u00e4rmise suhte:<br>FeS<sub>2<\/sub> \u2192 2SO<\/span><\/span><sub><span lang=\"et\"><span style=\"line-height: 115%;\">3<\/span><\/span><\/sub><br><span lang=\"et\"><span style=\"line-height: 115%;\">Mitu grammi SO<sub>3 <\/sub>on v\u00f5imalik saada 250 g p\u00fcriidi p\u00f5letamisel, kui kadu\u00a0on 12%?<\/span><\/span><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">2. Jooni tekstis alla k\u00f5ik arvud.<\/span><\/span><\/td>\n<td>\n<p><span lang=\"et\"><span style=\"line-height: 115%;\">P\u00fcriidi (FeS<sub>2<\/sub>) p\u00f5letamisel (s\u00e4rdamisel) saadakse gaasiline SO<sub>3<\/sub>. Reaktsiooni kohta v\u00f5ib v\u00e4lja kirjutada j\u00e4rmise suhte:<br>FeS<sub>2<\/sub> \u2192 2SO<\/span><\/span><sub><span lang=\"et\"><span style=\"line-height: 115%;\">3<\/span><\/span><\/sub><br><span lang=\"et\"><span style=\"line-height: 115%;\">Mitu grammi SO<sub>3 <\/sub>on v\u00f5imalik saada <u>250<\/u> g p\u00fcriidi p\u00f5letamisel, kui kadu\u00a0on <u>12<\/u>%?<\/span><\/span><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">3. Omista arvule kindel f\u00fc\u00fcsikaline suurus. Seda on lihtne teha n\u00e4iteks \u00fchiku alusel.<\/span><\/span><\/td>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">250 g on p\u00fcriidi mass, 12% on kadu, mis t\u00e4hendab, et protsessi k\u00e4igus saadava aine kogus on teoreetilisest v\u00e4iksem:\u00a0 100% \u2013 12%= 88%.<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">4. Kirjuta v\u00e4lja andmed ning otsitav suurus.<\/span><\/span><\/td>\n<td>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">m<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(p\u00fcriiit) = 250 g<\/span><\/span><\/p>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">P<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(kadu) = 12%<\/span><\/span><\/p>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">m<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(SO<sub>3<\/sub>) = ?<\/span><\/span><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">5. Kirjuta v\u00e4lja valemid ning kui vaja, avalda valemist otsitav suurus.<\/span><\/span><\/td>\n<td>\n<p><span lang=\"et\"><span style=\"line-height: 115%;\">Arvuta mitu mooli esialgset ainet hakkas reageerima.<\/span><\/span><\/p>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">$n(p\u00fcriit)=\\frac{m}{M}$<\/span><\/span><\/p>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">Seej\u00e4rel v\u00f5ta saaduste tekke puhul arvesse reaktsiooniv\u00f5rrandist tulenevat moolsuhet. 1 mooli p\u00fcriidi lagunemisel tekib 2 mooli SO<sub>3<\/sub>, seega tekkinud SO<sub>3<\/sub> ainehulk on kaks korda suurem p\u00fcriidi omast.<\/span><\/span><\/p>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">SO<sub>3<\/sub> massi saad leida valemist <i>m<\/i>(SO<sub>3<\/sub>) = <i>n\u00a0 <\/i><span style=\"background: white;\">$\\times $ <i>M<\/i>.<\/span><\/span><\/span><\/p>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\"><span lang=\"et\"><span style=\"background: white;\"><span style=\"line-height: 115%;\">Kuna protsessi k\u00e4igus on kadu 12% siis tekkinud aine mass on 88% arvutatud <\/span><\/span><\/span><span lang=\"et\"><span style=\"line-height: 115%;\">SO<sub>3<\/sub> massist.<\/span><\/span><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">6. Asenda valemitesse arvud ning teosta arvutused.<\/span><\/span><\/td>\n<td>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">M(FeS<sub>2<\/sub>) = 56 + 2 <span style=\"background: white;\">$\\times $ 32= 120 g\/mol<\/span><\/span><\/span><\/p>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\">$n(p\u00fcriit)=\\frac{m}{M}=\\frac{250 g}{120 g\/mol}=2,08 mol$<\/p>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\"><i><span lang=\"et\"><span style=\"background: white;\"><span style=\"line-height: 115%;\">n<\/span><\/span><\/span><\/i><span lang=\"et\"><span style=\"background: white;\"><span style=\"line-height: 115%;\">(<\/span><\/span><span style=\"line-height: 115%;\">SO<sub>3<\/sub><span style=\"background: white;\">) = 2 $\\times $ 2,08 = 4,16 mol<\/span><\/span><\/span><\/p>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\"><span lang=\"et\"><span style=\"background: white;\"><span style=\"line-height: 115%;\">M(<\/span><\/span><span style=\"line-height: 115%;\">SO<sub>3<\/sub><span style=\"background: white;\">) = 32 + 3 $\\times $ 16= 80 g\/mol<\/span><\/span><\/span><\/p>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">SO<sub>3\u00a0<\/sub><span style=\"background: white;\">teoreetiline mass:<\/span><\/span><\/span><br><i><span lang=\"et\"><span style=\"line-height: 115%;\">m<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(SO<sub>3<\/sub>) = <i>n\u00a0 <\/i><span style=\"background: white;\">$\\times $ <i>M <\/i>= 4,16 mol $\\times $ 80 g\/mol = 332,8 g <\/span><\/span><\/span><\/p>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\">Tegelik mass on teoreetilisest v\u00e4iksem:<\/p>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">m<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(SO<sub>3<\/sub>) = 0,88 <span style=\"background: white;\">$\\times $ 332,8 g = 293 g<\/span><\/span><\/span><\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>\u00a0<\/p>\n<p><\/p><div class=\"accordion mb-3\">\n        <div class=\"accordion-item accordion-item--white\">\n        <h2 class=\"accordion-header\" id=\"accordion-69f5ba06520f5-heading\">\n            <button class=\"accordion-button collapsed\" type=\"button\" data-bs-toggle=\"collapse\" data-bs-target=\"#accordion-69f5ba06520f5-collapse\" aria-expanded=\"true\" aria-controls=\"accordion-69f5ba06520f5-collapse\">\u00dclesanne 2<\/button>\n        <\/h2>\n        <div id=\"accordion-69f5ba06520f5-collapse\" class=\"accordion-collapse collapse\" aria-labelledby=\"accordion-69f5ba06520f5-heading\">\n            <div class=\"accordion-body\">\n\n\n<div class=\"h5p-iframe-wrapper\"><div class=\"video-placeholder-wrapper video-placeholder-wrapper--fixed\" style=\"height: 366px;\">\n\t\t\t    <div class=\"video-placeholder d-flex justify-content-center align-items-center\">\n\t\t\t        <div class=\"overlay text-white p-2 w-100 text-center d-block justify-content-center align-items-center\">\n\t\t\t            <div>Kolmandate osapoolte sisu n\u00e4gemiseks palun n\u00f5ustu k\u00fcpsistega.<\/div>\n\t\t\t            <button class=\"btn btn-secondary btn-sm mt-1 consent-change\">Muuda n\u00f5usolekut<\/button>\n\t\t\t        <\/div>\n\t\t\t    <\/div>\n\t\t\t<\/div>\n<\/div>\n\n\n<p><\/p><\/div>\n        <\/div>\n        <\/div>\n    <\/div>\n<p><\/p><div class=\"accordion mb-3\">\n        <div class=\"accordion-item accordion-item--white\">\n        <h2 class=\"accordion-header\" id=\"accordion-69f5ba0652101-heading\">\n            <button class=\"accordion-button collapsed\" type=\"button\" data-bs-toggle=\"collapse\" data-bs-target=\"#accordion-69f5ba0652101-collapse\" aria-expanded=\"true\" aria-controls=\"accordion-69f5ba0652101-collapse\">\u00dclesande 2 lahendused<\/button>\n        <\/h2>\n        <div id=\"accordion-69f5ba0652101-collapse\" class=\"accordion-collapse collapse\" aria-labelledby=\"accordion-69f5ba0652101-heading\">\n            <div class=\"accordion-body\">\n<p>2.1 0.08 mooli gaasilist vesinikkloriidi juhiti naatriumsulfiidi lahusesse. Arvutage, mitu mooli ja mitu dm<sup>3<\/sup> gaasilist divesiniksulfiidi eraldus normaaltingimustel (V<sub>m<\/sub> = 22.4 dm<sup>3<\/sup>\/mol), kui reaktsiooni saagis on 75%. Anna vastus t\u00e4psusega kaks kohta p\u00e4rast koma.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus:<\/b> 0,03 mol; 0,67 dm<sup>3<\/sup><\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus: <\/b>Kirjutame ja tasakaalustame reaktsiooniv\u00f5rrandi:<\/p>\n<p style=\"margin-left: 80px;\">Na<sub>2<\/sub>S + 2HCl \u2192 2NaCl + H<sub>2<\/sub>S<\/p>\n<p style=\"margin-left: 40px;\">Reaktsiooniv\u00f5rrandist n\u00e4eme, et HCl ja H<sub>2<\/sub>S moolsuhe on 2:1. Seega peaks H<sub>2<\/sub>S teoreetiliselt tekkima 0,08\/2=0,04 mol. Kuna aga saagis oli 75%, tekkis tegelikult 0,04\u00d70,75=0,03 mol divesiniksulfiidi. Leiame selle vastava ruumala:<i><\/i><\/p>\n<p style=\"margin-left: 80px;\"><i><span lang=\"ET\">V\u00a0=\u00a0n\u00a0\u00d7\u00a0<\/span><\/i><i><span lang=\"ET\">V<\/span><\/i><i><span lang=\"ET\">m\u00a0<\/span><\/i><i><span lang=\"ET\">=\u00a0<\/span><\/i><span lang=\"ET\">0,03 mol \u00d7 22,4 <\/span><span lang=\"ET\">dm<\/span><sup><span lang=\"ET\"><sup>3<\/sup><\/span><\/sup><span lang=\"ET\">\/mol = 0,672 \u2248 0,67 d<\/span><span lang=\"ET\">m<\/span><sup><span lang=\"ET\"><sup>3<\/sup><\/span><\/sup><\/p>\n<p style=\"margin-left: 120px;\">\u00a0<\/p>\n<p>2.2. 3,1 tonni puidu tselluloosi h\u00fcdrol\u00fc\u00fcsil tekkis 0.81 tonni gl\u00fckoosi. Saadud gl\u00fckoos k\u00e4\u00e4ritati t\u00e4ielikult etanooliks:<\/p>\n<p>C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub>\u21922C<sub>2<\/sub>H<sub>5<\/sub>OH + 2CO<sub>2<\/sub>.<\/p>\n<p>Etanooli deh\u00fcdraatimise saada eteen. Mitu m<sup>3<\/sup> eteeni v\u00f5ib saada normaaltingimustel etanoolist, kui kadu protsessil on 40%? Anna vastus t\u00e4isarvuna.<\/p>\n<p>C<sub>2<\/sub>H<sub>5<\/sub>OH \u2192 C<sub>2<\/sub>H<sub>4<\/sub> + H<sub>2<\/sub>O<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus: <\/b>121 m<sup>3<\/sup><\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus:<\/b> \u00dclesande lahendamiseks tuleb esmalt leida etanooli kogus, millest eteeni valmistati. Selleks l\u00e4htume gl\u00fckoosi k\u00e4\u00e4ritamise reaktsioonist, mis kulges t\u00e4ielikult. Kuna on teada, et gl\u00fckoosi mass on <i>m<\/i>=0,81 t=810 kg, leiame gl\u00fckoosi moolide arvu:<\/p>\n<p style=\"margin-left: 80px;\">M(gl\u00fckoos)=180 g\/mol=180 kg\/kmol<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$n = {m \\over M}={810kg \\over 180kg\/kmol}=4.5kmol$<\/span><\/p>\n<p style=\"margin-left: 36.0pt;\">Gl\u00fckoosi ja etanooli moolsuhte 2:1 j\u00e4rgi leiame, et etanooli tekkis 2\u00d74,5=9 kmol. Etanooli deh\u00fcdraatimise reaktsiooniv\u00f5rrandist n\u00e4eme, et etanooli ja eteeni moolsuhe on 1:1, seega eteeni teoreetiline saagis on 9 kmol. Teame, et kadu on 40%, seega on reaktsiooni saagis 60% ja eteeni tekib tegelikult 0,6\u00d79=5,4 kmol. Leiame selle ruumala:<i><\/i><\/p>\n<p style=\"margin-left: 80px;\"><i><span lang=\"ET\">V\u00a0=\u00a0n\u00a0\u00d7\u00a0<\/span><\/i><i><span lang=\"ET\">V<\/span><\/i><i><span lang=\"ET\"><sub>m<\/sub>\u00a0<\/span><\/i><i><span lang=\"ET\">=\u00a0<\/span><\/i><span lang=\"ET\">5,4 kmol \u00d7 22,4 <\/span><span lang=\"ET\">m<\/span><sup><span lang=\"ET\"><sup>3<\/sup><\/span><\/sup><span lang=\"ET\">\/kmol =1 20,96 \u2248 121 <\/span><span lang=\"ET\">m<\/span><sup><span lang=\"ET\"><sup>3<\/sup><\/span><\/sup><\/p>\n<p>\u00a0<\/p>\n<p>2.3. Mitu kilogrammi etanooli (C<sub>2<\/sub>H<sub>5<\/sub>OH) saab mahla k\u00e4\u00e4ritamisel toota, kui mahl sisaldab 25 kg gl\u00fckoosi (C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub>) ja k\u00e4\u00e4ritamise saagis on 72%? Anna vastus t\u00e4psusega \u00fcks koht peale koma.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus:<\/b> 9,2 kg etanooli<\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus:<\/b> Kirjutame ja tasakaalustame reaktsiooniv\u00f5rrandi:<\/p>\n<p style=\"margin-left: 80px;\">C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub>\u21922C<sub>2<\/sub>H<sub>5<\/sub>OH + 2CO<sub>2<\/sub><\/p>\n<p style=\"margin-left: 40px;\">Leiame k\u00f5igepealt gl\u00fckoosi moolide arvu:<\/p>\n<p style=\"margin-left: 80px;\">M(gl\u00fckoos)=180 g\/mol=180 kg\/kmol<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$n = {m \\over M}={25kg \\over 180kg\/kmol}=0.14kmol$<\/span><\/p>\n<p style=\"margin-left: 40px;\">Reaktsiooniv\u00f5rrandist n\u00e4eme, et gl\u00fckoosi ja etanooli moolsuhe on 1:2. Seega tekib etanooli teoreetiliselt 2\u00d70,14=0,28 kmol. Kuna saagis on 72%, tekib seda tegelikult 0,72\u00d70,28=0,20 kmol. Leiame selle abil n\u00fc\u00fcd etanooli massi:<\/p>\n<p style=\"margin-left: 40px;\">M(C<sub>2<\/sub>H<sub>5<\/sub>OH) = 46 g\/mol=46 kg\/kmol<\/p>\n<p style=\"margin-left: 80px;\"><i><span lang=\"ET\">m<\/span><\/i><sub>C<\/sub><sub>2<\/sub><sub>H<\/sub><sub>5<\/sub><sub>OH\u00a0<\/sub><i><span lang=\"ET\">=\u00a0n\u00a0\u00d7\u00a0M\u00a0=\u00a0<\/span><\/i><span lang=\"ET\">0,2 kmol \u00d7 46 kg\/kmol = 9,2 kg<\/span><\/p>\n<p style=\"margin-left: 40px;\">\u00a0<\/p>\n<p>2.4. Sama \u00fclesanne mis N\u00e4idis\u00fclesandes 1.5.<\/p><\/div>\n        <\/div>\n        <\/div>\n    <\/div>\n<h2>N\u00e4idis\u00fclesanne 3. L\u00e4hteainete vajaliku koguse arvutamine, arvestades reaktsiooni saagist<\/h2>\n<table class=\"table table-hover\" style=\"width: 100%;\" border=\"0\" cellspacing=\"1\" cellpadding=\"1\">\n<tbody>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">1. Loe \u00fclesande tekst m\u00f5ttega l\u00e4bi.<\/span><\/span><\/td>\n<td style=\"width: 70%;\">\n<p><span lang=\"et\"><span style=\"line-height: 115%;\">KCl t\u00f6\u00f6stusliku s\u00fcnteesiprotsessi kohta on teada, et saagis on 70%. Mitu grammi l\u00e4hteainet on vaja v\u00f5tta selleks, et tekiks 500 g KCl? Reaktsiooniv\u00f5rrand on<\/span><\/span><\/p>\n<p><span lang=\"et\"><span style=\"line-height: 115%;\">2KClO<sub>3 <\/sub><\/span><\/span><span lang=\"et\"><span style=\"line-height: 115%;\">\u2192 2KCl + 3O<\/span><\/span><sub><span lang=\"et\"><span style=\"line-height: 115%;\">2<\/span><\/span><\/sub><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">2. Jooni tekstis alla k\u00f5ik arvud.<\/span><\/span><\/td>\n<td>\n<p><span lang=\"et\"><span style=\"line-height: 115%;\">KCl t\u00f6\u00f6stusliku s\u00fcnteesiprotsessi kohta on teada, et saagis on <u>70<\/u>%. Mitu grammi l\u00e4hteainet on vaja v\u00f5tta selleks, et tekiks <u>500<\/u> g KCl? Reaktsiooniv\u00f5rrand on<\/span><\/span><\/p>\n<p><span lang=\"et\"><span style=\"line-height: 115%;\">2KClO<sub>3 <\/sub><\/span><\/span><span lang=\"et\"><span style=\"line-height: 115%;\">\u2192 2KCl + 3O<\/span><\/span><sub><span lang=\"et\"><span style=\"line-height: 115%;\">2<\/span><\/span><\/sub><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">3. Omista arvule kindel f\u00fc\u00fcsikaline suurus. Seda on lihtne teha n\u00e4iteks \u00fchiku alusel.<\/span><\/span><\/td>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">Protsessi saagis on 70%,<br>tekkinud KCl mass on 500 g.<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">4. Kirjuta v\u00e4lja andmed ning otsitav suurus.<\/span><\/span><\/td>\n<td>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">P<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(saagis) = 70%<\/span><\/span><\/p>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">m<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(KCl) = 500 g<\/span><\/span><\/p>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">m<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(KClO<sub>3<\/sub>) = ?<\/span><\/span><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">5. Kirjuta v\u00e4lja valemid ning kui vaja, avalda valemist otsitav suurus.<\/span><\/span><\/td>\n<td>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">Esmalt arvutame l\u00e4hteaine moolide arvu<\/span><\/span><\/p>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\">$n=\\frac{m}{M}$<\/p>\n<p><span lang=\"et\"><span style=\"line-height: 115%;\">L\u00e4hteaine ja saaduse moolide suhe on reaktsiooniv\u00f5rrandis sama, j\u00e4relikult<\/span><\/span><\/p>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">n<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(KCl ) = <i>n<\/i>(KClO<sub>3<\/sub>)<\/span><\/span><\/p>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">Seej\u00e4rel saab arvutada KClO<sub>3<\/sub> massi<\/span><\/span><\/p>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">m<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(KClO<sub>3<\/sub>) =\u00a0<i>n\u00a0 <\/i><span style=\"background: white;\">$\\times $ <i>M<\/i><\/span><\/span><\/span><\/p>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\"><span lang=\"et\"><span style=\"background: white;\"><span style=\"line-height: 115%;\">Kuna saagis on 70%, siis tuleb v\u00f5tta l\u00e4hteainet rohkem.<\/span><\/span><\/span><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">6. Asenda valemitesse arvud ja arvuta.<\/span><\/span><\/td>\n<td>\n<p><span lang=\"et\"><span style=\"line-height: 115%;\">Leiame saaduse hulga moolides:<\/span><\/span><\/p>\n<p><span lang=\"et\"><span style=\"line-height: 115%;\">M(KCl) = 39 + 35,5 = 74,5 g\/mol<\/span><\/span><\/p>\n<p>$n(KCl)=\\frac{m}{M}=\\frac{500 g}{74,5 g\/mol}=6,7 mol$<\/p>\n<p>J\u00e4relikult ka\u00a0<i><span lang=\"et\"><span style=\"line-height: 115%;\">n<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(KClO<sub>3<\/sub>) = 6,7 mol<\/span><\/span><\/p>\n<p><span lang=\"et\"><span style=\"line-height: 115%;\">Arvutame KClO<sub>3 <\/sub>massi:<\/span><\/span><\/p>\n<p><span lang=\"et\"><span style=\"line-height: 115%;\">M(KClO<sub>3<\/sub>) = 39 + 35,5 + 3 <span style=\"background: white;\">$\\times $ 16 = 122,5 g\/mol<\/span><\/span><\/span><\/p>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">m<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(KClO<sub>3<\/sub>) =\u00a0<i>n\u00a0 <\/i><span style=\"background: white;\">$\\times $ M =6,7 mol $\\times $ 122,5 g\/mol = 820,75 g<\/span><\/span><\/span><\/p>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\">$m(aine \u00a0koos \u00a0kaoga)=\\frac{100}{70}\\times 820,75 g=1172,5 g$<\/p>\n<p style=\"margin: 12.0pt 0cm 12.0pt 0cm;\"><span lang=\"et\"><span style=\"background: white;\"><span style=\"line-height: 115%;\">K\u00f5ige olulisem on sellist t\u00fc\u00fcpi \u00fclesannete puhul m\u00f5ista, et kuna reaktsiooni k\u00e4igus tekib kadu ehk saagis on alla 100%, siis selleks, et saada oodatavat hulka saadust, on vaja esialgset l\u00e4hteainet v\u00f5i l\u00e4hteaineid v\u00f5tta rohkem. Sama kehtib ka lisandite korral.<\/span><\/span><\/span><\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><\/p><div class=\"accordion mb-3\">\n        <div class=\"accordion-item accordion-item--white\">\n        <h2 class=\"accordion-header\" id=\"accordion-69f5ba0652106-heading\">\n            <button class=\"accordion-button collapsed\" type=\"button\" data-bs-toggle=\"collapse\" data-bs-target=\"#accordion-69f5ba0652106-collapse\" aria-expanded=\"true\" aria-controls=\"accordion-69f5ba0652106-collapse\">\u00dclesanne 3<\/button>\n        <\/h2>\n        <div id=\"accordion-69f5ba0652106-collapse\" class=\"accordion-collapse collapse\" aria-labelledby=\"accordion-69f5ba0652106-heading\">\n            <div class=\"accordion-body\">\n\n\n<div class=\"h5p-iframe-wrapper\"><div class=\"video-placeholder-wrapper video-placeholder-wrapper--fixed\" style=\"height: 366px;\">\n\t\t\t    <div class=\"video-placeholder d-flex justify-content-center align-items-center\">\n\t\t\t        <div class=\"overlay text-white p-2 w-100 text-center d-block justify-content-center align-items-center\">\n\t\t\t            <div>Kolmandate osapoolte sisu n\u00e4gemiseks palun n\u00f5ustu k\u00fcpsistega.<\/div>\n\t\t\t            <button class=\"btn btn-secondary btn-sm mt-1 consent-change\">Muuda n\u00f5usolekut<\/button>\n\t\t\t        <\/div>\n\t\t\t    <\/div>\n\t\t\t<\/div>\n<\/div>\n\n\n<p><\/p><\/div>\n        <\/div>\n        <\/div>\n    <\/div>\n<p><\/p><div class=\"accordion mb-3\">\n        <div class=\"accordion-item accordion-item--white\">\n        <h2 class=\"accordion-header\" id=\"accordion-69f5ba0652116-heading\">\n            <button class=\"accordion-button collapsed\" type=\"button\" data-bs-toggle=\"collapse\" data-bs-target=\"#accordion-69f5ba0652116-collapse\" aria-expanded=\"true\" aria-controls=\"accordion-69f5ba0652116-collapse\">\u00dclesande 3 lahendused<\/button>\n        <\/h2>\n        <div id=\"accordion-69f5ba0652116-collapse\" class=\"accordion-collapse collapse\" aria-labelledby=\"accordion-69f5ba0652116-heading\">\n            <div class=\"accordion-body\">\n<p>3.1 Gl\u00fckoos k\u00e4\u00e4ritati t\u00e4ielikult etanooliks:<\/p>\n<p>C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub>\u21922C<sub>2<\/sub>H<sub>5<\/sub>OH + 2CO<sub>2<\/sub>.<\/p>\n<p>Saadud etanoolist saadi deh\u00fcdraatimise teel 150 dm<sup>3<\/sup> (nt) eteeni, protsessi kadu oli 45 %.<\/p>\n<p>C<sub>2<\/sub>H<sub>5<\/sub>OH \u2192 C<sub>2<\/sub>H<sub>4<\/sub> + H2O<\/p>\n<p>Mitu kg gl\u00fckoosi pidi protsessi jaoks v\u00f5tma? Anna vastus t\u00e4isarvuna.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus: <\/b>1 kg<\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus:<\/b> Gl\u00fckoosi massi leidmiseks on vaja teada etanooli ainehulka. Saame selle eteeni ruumala ja eteeni saamise reaktsiooni saagise abil arvutada. Esmalt tuleb leida eteeni moolide arv:<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$n = {V \\over V_m}={150dm^3 \\over 22.4dm^3\/mol}=6.7mol$<\/span><\/p>\n<p style=\"margin-left: 40px;\">Kuna protsessi kadu oli 45%, vastab 6,7 mol j\u00e4relikult 55%-le. Teoreetiliselt oleks seega pidanud tekkima 6,7\/0,55=12,2 mol.<\/p>\n<p style=\"margin-left: 40px;\">Etanooli deh\u00fcdraatimise reaktsiooni j\u00e4rgi on etanooli ja eteeni moolsuhe 1:1, seega oli reageerivat etanooli 12,2 mol. Gl\u00fckoosi k\u00e4\u00e4ritamise reaktsiooni j\u00e4rgi on gl\u00fckoosi ja etanooli moolsuhe 1:2, millest saab tuletada, et gl\u00fckoosi oli 12,2\/2=6,1 mol. N\u00fc\u00fcd saame juba leida gl\u00fckoosi massi:<\/p>\n<p style=\"margin-left: 80px;\">M(gl\u00fckoos)=180 g\/mol<\/p>\n<p style=\"margin-left: 80px;\"><i><span lang=\"ET\">m<\/span><\/i><sub>gl\u00fckoos<\/sub>\u00a0<i style=\"text-align: center;\"><span lang=\"ET\">=\u00a0n\u00a0\u00d7\u00a0M\u00a0=\u00a0<\/span><\/i><span lang=\"ET\" style=\"text-align: center;\">6,1 mol \u00d7 180 g\/mol = 1098 g = 1,098 \u2248 1 kg<\/span><\/p>\n<p style=\"margin-left: 80px;\">\u00a0<\/p>\n<p>3.2. Mahla k\u00e4\u00e4ritamise tulemusena saadi 50 kg etanooli (C<sub>2<\/sub>H<sub>5<\/sub>OH). Mitu grammi gl\u00fckoosi (C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub>) pidi mahl sisaldama, kui k\u00e4\u00e4ritamise protsessi saagis oli 72%?<\/p>\n<p>C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub>\u21922C<sub>2<\/sub>H<sub>5<\/sub>OH + 2CO<sub>2<\/sub>. Anna vastus t\u00e4isarvuna.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus<\/b>: Gl\u00fckoosi l\u00e4heb tarvis 136 kg.<\/p>\n<p style=\"margin-left: 40px;\"><b><span lang=\"ET\" style=\"color: black;\">Lahendus:<\/span><\/b> Kirjutame ja tasakaalustame reaktsiooniv\u00f5rrandi:<\/p>\n<p style=\"margin-left: 80px;\">C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub>\u21922C<sub>2<\/sub>H<sub>5<\/sub>OH + 2CO<sub>2<\/sub><\/p>\n<p style=\"margin-left: 40px;\">Esmalt saame leida etanooli moolide arvu:<\/p>\n<p style=\"margin-left: 80px;\">M(C<sub>2<\/sub>H<sub>5<\/sub>OH) = 46 g\/mol=46 kg\/kmol<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$n = {m \\over M}={50kg \\over 46kg\/kmol}=1.09kmol$<\/span><\/p>\n<p style=\"margin-left: 40px;\">Kuna aga protsessi saagis oli 72%, pidanuks etanooli teoreetiliselt tekkima 1,09\/0,72=1,51 kmol.<\/p>\n<p style=\"margin-left: 40px;\">Reaktsiooniv\u00f5rrandi j\u00e4rgi n\u00e4eme, et gl\u00fckoosi ja etanooli moolsuhe on 1:2, seega oli gl\u00fckoosi moolide arv 1,51\/2=0,755 kmol. N\u00fc\u00fcd saame leida gl\u00fckoosi massi:<\/p>\n<p style=\"margin-left: 80px;\">M(gl\u00fckoos)=180 g\/mol=180 kg\/kmol<\/p>\n<p style=\"margin-left: 80px;\"><i><span lang=\"ET\">m<\/span><\/i><sub>gl\u00fckoos\u00a0<\/sub><i><span lang=\"ET\">=\u00a0n\u00a0\u00d7\u00a0M\u00a0=\u00a0<\/span><\/i><span lang=\"ET\">0,755 kmol \u00d7 180 kg\/kmol = 135,9 \u2248 136 kg<\/span><\/p>\n<p style=\"margin-left: 80px;\">\u00a0<\/p>\n<p>3.3. V\u00e4\u00e4velhappe tootmisel kasutati l\u00e4hteainena p\u00fcriiti (FeS<sub>2<\/sub>). P\u00fcriidi s\u00e4rdamisel (p\u00f5letamisel) saadi 180 kg v\u00e4\u00e4veldioksiidi. Kui palju p\u00fcriiti pidi s\u00e4ristamiseks v\u00f5tma, kui protsessi saagise protsent on 84%? Anna vastus t\u00e4isarvuna.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus: <\/b>201 kg\u00a0<\/p>\n<p style=\"margin-left: 40px;\"><b><span lang=\"ET\" style=\"color: black;\">Lahendus:<\/span><\/b> Kirjutame ja tasakaalustame reaktsiooniv\u00f5rrandi:<\/p>\n<p style=\"margin-left: 80px;\">4FeS<sub>2<\/sub> +11 O<sub>2<\/sub> \u2192 2Fe<sub>2<\/sub>O<sub>3<\/sub> + 8SO<sub>2<\/sub><\/p>\n<p style=\"margin-left: 40px;\"><span lang=\"ET\" style=\"color: black;\">Arvutame esmalt v\u00e4\u00e4veldioksiidi moolide arvu:<\/span><\/p>\n<p style=\"margin-left: 80px;\">M(SO<sub>2<\/sub>)=64 g\/mol=64 kg\/kmol<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$n = {m \\over M}={180kg \\over 64kg\/kmol}=2.81kmol$<\/span><\/p>\n<p style=\"margin-left: 40px;\">Kuna aga protsessi saagis oli 84%, pidanuks v\u00e4\u00e4veldioksiidi teoreetiliselt tekkima 2,81\/0,84=3,345 kmol.<\/p>\n<p style=\"margin-left: 40px;\">Reaktsiooniv\u00f5rrandist n\u00e4eme, et FeS<sub>2<\/sub> ja SO<sub>2<\/sub> moolsuhe on 4:8=1:2. Seega oli FeS<sub>2<\/sub> moolide arv 3,345\/2=1,673 kmol. Leiame selle massi:<\/p>\n<p style=\"margin-left: 40px;\">M(FeS<sub>2<\/sub>) = 120 g\/mol=120 kg\/kmol<i><\/i><\/p>\n<p style=\"margin-left: 40px;\"><i><span lang=\"ET\">m\u00a0=\u00a0n\u00a0\u00d7\u00a0M\u00a0=\u00a0<\/span><\/i><span lang=\"ET\">1,673 kmol \u00d7 120 kg\/kmol = 200,76 \u2248 201 kg<\/span><\/p>\n<p style=\"margin-left: 40px;\">\u00a0<\/p>\n<p>3.4. Vesiniku ja l\u00e4mmastiku vahelisel reaktsioonil k\u00f5rgel temperatuuril saadi 115 g ammoniaaki saagisega 25%. Arvuta mitu liitrit vesinikku v\u00f5eti reaktsiooniks?<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus:<\/b> Vesinikku v\u00f5eti 909 liitrit.<\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus:<\/b> Kirjutame ja tasakaalustame reaktsiooniv\u00f5rrandi: N<sub>2<\/sub> + 3H<sub>2<\/sub> \u2192 2NH<sub>3<\/sub><\/p>\n<p style=\"margin-left: 40px;\">Esmalt leiame olemasolevate andmete abil ammoniaagi moolide arvu:<\/p>\n<p style=\"margin-left: 80px;\">M(NH<sub>3<\/sub>)=17 g\/mol<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$n = {m \\over M}={115g \\over 17g\/mol}=6.765mol$<\/span><\/p>\n<p style=\"margin-left: 40px;\">Kuna aga protsessi saagis oli 25%, pidanuks ammoniaaki teoreetiliselt tekkima 6,765\/0,25=27,06 mol.<\/p>\n<p style=\"margin-left: 40px;\">Reaktsiooniv\u00f5rrandist n\u00e4eme, et ammoniaagi ja vesiniku moolsuhe on 2:3, seega vesiniku moolide arv oli <i><span lang=\"ET\">27,06 \u00d7\u00a0<\/span><\/i><i><span lang=\"ET\">3 \/\u00a0<\/span><\/i><i><span lang=\"ET\">2<\/span><\/i><i><span lang=\"ET\">= 40,6<\/span><\/i>\u00a0mol. N\u00fc\u00fcd saame leida vesiniku ruumala:<i><\/i><\/p>\n<p style=\"margin-left: 40px;\"><i><span lang=\"ET\">V\u00a0=\u00a0n\u00a0\u00d7\u00a0<\/span><\/i><i><span lang=\"ET\">V<\/span><\/i><i><span lang=\"ET\"><sub>m<\/sub>\u00a0<\/span><\/i><i><span lang=\"ET\">=\u00a0<\/span><\/i><span lang=\"ET\">40,6 mol \u00d7 22,4 <\/span><span lang=\"ET\">dm<\/span><sup><sub><span lang=\"ET\"><sup><sub>3<\/sub><\/sup><\/span><\/sub><\/sup><span lang=\"ET\">\/mol = 909,22 \u2248 909 <\/span><span lang=\"ET\">dm<\/span><sup><span lang=\"ET\"><sup>3<\/sup><\/span><\/sup><\/p><\/div>\n        <\/div>\n        <\/div>\n    <\/div>","protected":false},"excerpt":{"rendered":"<p>V\u00f5ib juhtuda, et tegelike saaduste hulk erineb oluliselt teoreetiliselt arvutatud kogustest. Reaktsiooni saagis n\u00e4itab, mitu protsenti moodustab tegelikult saadud reaktsioonisaaduse hulk (v\u00f5i mass) reaktsiooniv\u00f5rrandi j\u00e4rgi teoreetiliselt arvutatud reaktsioonisaaduse hulgast (v\u00f5i massist). J\u00e4rgnevas videos saad vaadata, kuidas arvutada reaktsiooni saagist. Kolmandate &#8230;<\/p>\n","protected":false},"author":269,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_acf_changed":false,"inline_featured_image":false,"footnotes":""},"class_list":["post-16","page","type-page","status-publish","hentry"],"acf":[],"_links":{"self":[{"href":"https:\/\/sisu.ut.ee\/huvitavkeemia\/wp-json\/wp\/v2\/pages\/16","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sisu.ut.ee\/huvitavkeemia\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/sisu.ut.ee\/huvitavkeemia\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/sisu.ut.ee\/huvitavkeemia\/wp-json\/wp\/v2\/users\/269"}],"replies":[{"embeddable":true,"href":"https:\/\/sisu.ut.ee\/huvitavkeemia\/wp-json\/wp\/v2\/comments?post=16"}],"version-history":[{"count":11,"href":"https:\/\/sisu.ut.ee\/huvitavkeemia\/wp-json\/wp\/v2\/pages\/16\/revisions"}],"predecessor-version":[{"id":988,"href":"https:\/\/sisu.ut.ee\/huvitavkeemia\/wp-json\/wp\/v2\/pages\/16\/revisions\/988"}],"wp:attachment":[{"href":"https:\/\/sisu.ut.ee\/huvitavkeemia\/wp-json\/wp\/v2\/media?parent=16"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}