{"id":11,"date":"2024-04-04T07:30:14","date_gmt":"2024-04-04T04:30:14","guid":{"rendered":"https:\/\/sisu.ut.ee\/huvitavkeemia\/27-lisandite-arvestamine-arvutustes\/"},"modified":"2025-02-12T23:23:45","modified_gmt":"2025-02-12T21:23:45","slug":"27-lisandite-arvestamine-arvutustes","status":"publish","type":"page","link":"https:\/\/sisu.ut.ee\/huvitavkeemia\/27-lisandite-arvestamine-arvutustes\/","title":{"rendered":"2.7. Lisandite arvestamine arvutustes"},"content":{"rendered":"<p>Eri ainete tootmises kasutatakse tihti l\u00e4hteaineid, mis ei ole t\u00e4iesti puhtad\u00a0ehk sisaldavad mingeid lisandeid. Seet\u00f5ttu tekib protsessi k\u00e4igus ka v\u00e4hem saadusi, kui tekiks puhaste ainete korral. J\u00e4rgnevast videost saad vaadata, kuidas arvestada lisandite m\u00f5ju erinevates arvutus\u00fclesannetes.<\/p>\n<p><\/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<h6>Allikas:\u00a0<a title=\"\" href=\"https:\/\/youtu.be\/0epwo6KCfbQ\" target=\"_blank\" rel=\"noopener\" data-url=\"https:\/\/youtu.be\/0epwo6KCfbQ\">https:\/\/youtu.be\/0epwo6KCfbQ<\/a><\/h6>\n<p>Kui l\u00e4hteainetes on lisandid, siis \u00a0puhta aine protsenti l\u00e4hteainetes saad arvutada protsentarvutuse reegli alusel:<\/p>\n<p style=\"text-align: center;\">$P(puhas aine)=\\frac{m(puhas aine)}{m(kogu aine mass koos lisandiga)}\\times 100$<\/p>\n<p>m(puhas aine) on puhta aine mass (\u00fchik g) ja \u00a0m(kogu aine mass koos lisandiga) on aine mass, mis sisaldab lisandeid (\u00fchik g).\u00a0<\/p>\n<p>Sarnaselt puhta aine protsendilise sisaldusega on v\u00f5imalik arvutada ka lisandi sisaldus aines<\/p>\n<p style=\"text-align: center;\">$P(lisandid)=\\frac{m(lisand)}{m(kogu aine mass koos lisandiga)}\\times 100$<\/p>\n<p>m(lisand) on lisandi mass (\u00fchik g).<\/p>\n<p>Kokku on l\u00e4hteaine ja lisandi protsent 100%: P(puhas aine) + P(lisand) = 100%\u00a0<\/p>\n<h2>N\u00e4idis\u00fclesanne 1. Lisandi protsendilise sisalduse arvutamine<\/h2>\n<table class=\"table table-hover\" style=\"width: 100%;\" border=\"0\" cellspacing=\"1\" cellpadding=\"1\">\n<tbody>\n<tr>\n<td style=\"width: 30%;\">1. Loe \u00fclesande tekst m\u00f5ttega l\u00e4bi.<\/td>\n<td>150 g vase maaki sisaldab 87,3 g vaske. Arvuta vase ja lisandite protsendiline sisaldus maagis.<\/td>\n<\/tr>\n<tr>\n<td>2. Jooni tekstis alla k\u00f5ik arvud.<\/td>\n<td><u>150<\/u> g vase maaki sisaldab <u>87,3<\/u> g vaske. Arvuta vase ja lisandite protsendiline sisaldus maagis.<\/td>\n<\/tr>\n<tr>\n<td>3. Omista arvule kindel f\u00fc\u00fcsikaline suurus. Seda on lihtne teha n\u00e4iteks \u00fchiku alusel.<\/td>\n<td>150 g on kogu maagi mass, mis sisaldab 87,3 g on puhast vaske.<\/td>\n<\/tr>\n<tr>\n<td>4. Kirjuta v\u00e4lja andmed ning otsitav suurus.<\/td>\n<td>m(maak) = 150 g<br>m(vask) = 87,3 g<br>P(vask) = ?<br>P(lisand) = ?<\/td>\n<\/tr>\n<tr>\n<td>5. Kirjuta v\u00e4lja valemid ning kui vaja, avalda valemist otsitav suurus.<\/td>\n<td>Lisandi % arvutamisel arvestame, et kogu maak on 100% ja\u00a0 \u00a0P(lisand) = 100 \u2013 P(vask)<\/td>\n<\/tr>\n<tr>\n<td>6. Asenda valemitesse arvud ning arvuta.<\/td>\n<td>\n<p>Puhta vase sisaldus proovis on<br>$P(vask)=\\frac{87,3 g}{150 g}\\times 100=58,2%$.<br>P(lisand) = 100 \u2013 58,2 = 41,8%<\/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-69d5a07e87517-heading\">\n            <button class=\"accordion-button collapsed\" type=\"button\" data-bs-toggle=\"collapse\" data-bs-target=\"#accordion-69d5a07e87517-collapse\" aria-expanded=\"true\" aria-controls=\"accordion-69d5a07e87517-collapse\">\u00dclesanne 1<\/button>\n        <\/h2>\n        <div id=\"accordion-69d5a07e87517-collapse\" class=\"accordion-collapse collapse\" aria-labelledby=\"accordion-69d5a07e87517-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-69d5a07e87525-heading\">\n            <button class=\"accordion-button collapsed\" type=\"button\" data-bs-toggle=\"collapse\" data-bs-target=\"#accordion-69d5a07e87525-collapse\" aria-expanded=\"true\" aria-controls=\"accordion-69d5a07e87525-collapse\">\u00dclesande 1 lahendused<\/button>\n        <\/h2>\n        <div id=\"accordion-69d5a07e87525-collapse\" class=\"accordion-collapse collapse\" aria-labelledby=\"accordion-69d5a07e87525-heading\">\n            <div class=\"accordion-body\">\n<p>1.1. Alumiiniumi sisalduse leidmiseks alumiiniumi ja vase sulamist viidi l\u00e4bi j\u00e4rgmine katse. 7 g sulamile lisati HCl lahust. Kui reaktsioon oli l\u00e4bi viidud siis v\u00e4ike kogus vaske oli alles j\u00e4\u00e4nud ning reaktsiooni tulemusena oli kokku kogutud 1,32 dm<sup>3<\/sup> vesinikku (nt.).<\/p>\n<p style=\"margin-left: 40px;\">2Al + 6HCl \u2192 2AlCl<sub>3<\/sub> + 3H<sub>2<\/sub><\/p>\n<p>Mitu mooli HCl reageeris reaktsiooni k\u00e4igus?<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus: <\/b><b>0.118 mol<\/b><\/p>\n<p>Mitu mooli Al reageeris \u00e4ra?<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus: <\/b><b>0.039 mol<\/b><\/p>\n<p>Mitu grammi Al reageeris \u00e4ra?<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus: 1.06 g<\/b><\/p>\n<p>Mitu grammi Cu oli sulamis?<\/p>\n<p style=\"margin-left: 40px;\"><strong>Vastus: <\/strong><b>5.94<\/b> <strong>g<\/strong><\/p>\n<p>Arvuta lisandi % sulamis.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus<\/b>: 85%<\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus:<\/b><\/p>\n<p style=\"margin-left: 40px;\">L\u00e4htume teadmisest, et H<sub>2<\/sub> tekib 1,32 dm<sup>3<\/sup>. Gaaside puhul kehtib konstant <i>V<\/i><sub>m<\/sub> = 22,4 dm<sup>3<\/sup>\/mol, mille abil saab leida moolide arvu <i>n<\/i>:<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$n={ V \\over V_m}={ 1.32dm^3 \\over 22dm^3\/mol} = 0.059mol$<\/span><\/p>\n<p style=\"margin-left: 40px;\">Reaktsiooniv\u00f5rrandist n\u00e4eme, et vesiniku ja HCl moolsuhe on 3:6=1:2. Seega oli reageeriva HCl moolide arv <i>n<\/i>(HCl)= 0,059\u00d72=0,118 mol.<\/p>\n<p style=\"margin-left: 40px;\">HCl ja alumiiniumi moolsuhe on aga 6:2=3:1, mis t\u00e4hendab, et alumiiniumi moolide arv <i>n<\/i>(Al)=0,118\/3=0,039 mol.<\/p>\n<p style=\"margin-left: 40px;\">N\u00fc\u00fcd saame leida alumiiniumi massi, teades, et M(Al)=27 g\/mol:<\/p>\n<p style=\"margin-left: 80px;\"><i><span lang=\"ET\">m\u00a0=\u00a0n\u00a0\u00d7\u00a0M\u00a0=\u00a0<\/span><\/i><span lang=\"ET\">0,039 mol \u00d7 27 g\/mol = 1,053 g \u2248 1,05 g<\/span><\/p>\n<p style=\"margin-left: 40px;\">Vase mass sulamis on seega 7\u20131,05=5,95 g.<\/p>\n<p style=\"margin-left: 40px;\">Lisandi ehk vase sisalduse protsendi saame n\u00fc\u00fcd samuti leida:<\/p>\n<p style=\"padding-left: 80px;\"><span class=\"math-tex\">$P(lisand)={ 5.95 g \\over 7g}*100=85%$<\/span><\/p>\n<p>\u00a0<\/p>\n<p>1.2. Kaltsiumkloriidi kasutatakse lume sulatamiseks s\u00f5iduteel. 3.51 tonni loodusliku lubjakivi t\u00f6\u00f6tlemisel vesinikkloriidhappega saadi lahus, mis sisaldas 3.00 tonni kaltsiumkloriidi. Mitu protsenti inertseid lisandeid sisaldas lubjakivi? Anna vastus t\u00e4isarvuna.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus: <\/b>23%<\/p>\n<p style=\"margin-left: 40px;\"><b><span lang=\"ET\" style=\"color: black;\">Lahendus:<\/span><\/b><span lang=\"ET\" style=\"color: black;\"> Kirjutame v\u00e4lja ja tasakaalustame reaktsiooniv\u00f5rrandi:<\/span><\/p>\n<p style=\"margin-left: 80px;\"><span style=\"color: black;\">CaCO<sub>3<\/sub> + 2 HCl \u2192 CaCl<sub>2<\/sub> + CO<sub>2<\/sub> <\/span><span lang=\"ET\" style=\"color: black;\">+ <\/span><span style=\"color: black;\">H<sub>2<\/sub>O<\/span><\/p>\n<p style=\"margin-left: 40px;\"><span lang=\"ET\" style=\"color: black;\">Meil on teada, et CaCl<sub>2<\/sub> tekib 3 tonni ehk 3000 kg. Saame leida selle molaarmassi ja arvutada moolide arvu:<\/span><\/p>\n<p style=\"margin-left: 80px;\">M(CaCl<sub>2<\/sub>) = 111 g\/mol = 111 kg\/kmol<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$n={ m \\over M}={ 3000kg \\over111kg\/kmol} = 27.03kmol$<\/span><\/p>\n<p style=\"margin-left: 40px;\">Reaktsiooniv\u00f5rrandist n\u00e4eme, et <span style=\"color: black;\">CaCO<sub>3<\/sub><\/span> ja <span style=\"color: black;\">CaCl<sub>2<\/sub> <\/span><span lang=\"ET\" style=\"color: black;\">moolsuhe on 1:1, seega on ka <\/span><span style=\"color: black;\">CaCO<sub>3<\/sub><\/span> moolide arv 27,03 mol.<span lang=\"ET\" style=\"color: black;\"> Saame leida selle molaarmassi ja arvutada selle massi:<\/span><\/p>\n<p style=\"margin-left: 80px;\">M(CaCO<sub>3<\/sub>) = 100 g\/mol = 100 kg\/kmol<\/p>\n<p style=\"margin-left: 80px;\"><i><span lang=\"ET\">m<\/span><\/i><sub><span lang=\"ET\"><sub>CaCO<\/sub><sub>3<\/sub><\/span><\/sub><i><span lang=\"ET\">=\u00a0n\u00a0\u00d7\u00a0M\u00a0=\u00a0<\/span><\/i><span lang=\"ET\">27,03 kmol \u00d7 100 kg\/kmol = 2703 kg \u2248 2,70 t<\/span><span style=\"text-align: center;\">\u00a0<\/span><\/p>\n<p style=\"margin-left: 40px;\"><span lang=\"ET\" style=\"color: black;\">Kui puhast <\/span>CaCO<sub>3<\/sub> on 3,51 tonnis 2,70 tonni, siis lisandite mass on 3,51\u20132,70=0,81 tonni. N\u00fc\u00fcd saame leida nende sisalduse protsendi lubjakivis:<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$P(lisand)={ 0.81 t \\over 3.5t}*100=23%$<\/span><\/p>\n<p style=\"margin-left: 80px;\">\u00a0<\/p>\n<p>1.3. 5,0 g KClO<sub>3<\/sub>, mille puhtus ei olnud teada, kuumutati ja see lagunes kaaliumkloriidiks ja hapnikuks. Reaktsiooni tulemusena tekkis 1,5 g hapnikku.<\/p>\n<p>2KClO<sub>3<\/sub> \u2192 2KCl + 3O<sub>2<\/sub><\/p>\n<p>Kui suur oli lisandite hulk KClO<sub>3<\/sub>\u2013s? Anna vastus t\u00e4isarvuna.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus: <\/b>23%<\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus:<\/b> <span lang=\"ET\" style=\"color: black;\">Meil on teada, et O<sub>2<\/sub> tekib 1,5 grammi. Saame leida selle molaarmassi ja arvutada moolide arvu:<\/span><\/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\">$n={ m \\over M}={ 1.5g \\over32g\/mol} = 0.046875mol$<\/span><i><\/i><\/p>\n<p style=\"margin-left: 40px;\">Reaktsiooniv\u00f5rrandist n\u00e4eme, et <span style=\"color: black;\">O<\/span><sub><span lang=\"ET\" style=\"color: black;\">2<\/span><\/sub> ja <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> <span lang=\"ET\" style=\"color: black;\">moolsuhe on 3:2, seega on K<\/span><span style=\"color: black;\">Cl<\/span><span lang=\"ET\" style=\"color: black;\">O<sub>3 <\/sub><\/span>moolide arv <span lang=\"ET\">0,046875 mol\u00d7<\/span><span lang=\"ET\">2\/<\/span><span lang=\"ET\">3<\/span><span lang=\"ET\">=0,03125 mol<\/span><span lang=\"ET\" style=\"color: black;\">. Saame n\u00fc\u00fcd leida K<\/span><span style=\"color: black;\">Cl<\/span><span lang=\"ET\" style=\"color: black;\">O<sub>3<\/sub> molaarmassi ja arvutada selle massi:<\/span><\/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;\"><i><span lang=\"ET\">m\u00a0<\/span><\/i><span lang=\"ET\">K<\/span>Cl<span lang=\"ET\">O<sub>3\u00a0<\/sub><\/span><i><span lang=\"ET\">=\u00a0n\u00a0\u00d7\u00a0M\u00a0=\u00a0<\/span><\/i><span lang=\"ET\">0,03125 <\/span><span lang=\"ET\">mol \u00d7 122,5 g\/mol = 3,83 g<\/span><span style=\"text-align: center;\">\u00a0<\/span><\/p>\n<p style=\"margin-left: 40px;\"><span lang=\"ET\" style=\"color: black;\">Kui puhast K<\/span><span style=\"color: black;\">Cl<\/span><span lang=\"ET\" style=\"color: black;\">O<sub>3<\/sub><\/span> on 5 grammis 3,83 g, siis lisandite mass on 5\u20133,83=1,17 g. N\u00fc\u00fcd saame leida nende sisalduse protsendi:<\/p>\n<p style=\"padding-left: 80px;\"><span class=\"math-tex\">$P(lisand)={ 1.17g \\over 5g}*100=23%$<\/span><\/p>\n<p>\u00a0<\/p>\n<p>1.4. 5 mooli l\u00e4mmastiku reageerimisel vesinikuga tekkis 2,5 mol ammoniaaki (NH<sub>3<\/sub>). Arvutage, kui suur oli lisandite hulk l\u00e4hteaines. 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;\">Kuna reaktsiooniv\u00f5rrandi j\u00e4rgi on l\u00e4mmastiku ja tekkiva ammoniaagi moolsuhe 1:2 ja ammoniaagi moolide arv <i>n<\/i>(NH<sub>3<\/sub>) = 2,5 mol on teada, on\u00a0 <i>n<\/i>(N<sub>2<\/sub>) j\u00e4relikult 2,5\/2=1,25 mol. 5 moolis l\u00e4mmastikus oli lisandeid seega 5\u20131,25=3,75 mol ja nende protsendiline sisaldus:<\/p>\n<p style=\"padding-left: 80px;\"><span class=\"math-tex\">$P(lisand)={ 3.75mol \\over 5mol}*100=75%$<\/span><\/p>\n<p><\/p><\/div>\n        <\/div>\n        <\/div>\n    <\/div>\n<h2>N\u00e4idis\u00fclesanne 2. Lisandite arvestamine saaduse hulga arvutamisel<\/h2>\n<table class=\"table table-hover\" style=\"width: 100%;\" border=\"0\" cellspacing=\"1\" cellpadding=\"1\">\n<tbody>\n<tr>\n<td style=\"width: 30%;\"><span lang=\"et\"><span style=\"line-height: 115%;\">1. Loe \u00fclesande tekst m\u00f5ttega l\u00e4bi.<\/span><\/span><\/td>\n<td>\n<p><span lang=\"et\"><span style=\"line-height: 115%;\">Raua tootmiseks kasutatakse punast rauamaaki raud(III)oksiidi. Leiti, et rauamaak sisaldab 27% lisandeid. Mitu kg rauda on v\u00f5imalik saada 500 kg rauamaagist. Raua tootmise t\u00e4psem protsess ei ole teada, aga on teada, et rauamaagi ja raua moolsuhe on<\/span><\/span><\/p>\n<p><span lang=\"et\"><span style=\"line-height: 115%;\">Fe<sub>2<\/sub>O<sub>3<\/sub><\/span><\/span><span lang=\"et\"><span style=\"line-height: 115%;\"> \u2192 2Fe<\/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%;\">Rauda tootmiseks kasutatakse punast rauamaaki raud(III)oksiidi. Leiti, et rauamaak sisaldab <u>27%<\/u> lisandeid. Mitu kg rauda on v\u00f5imalik saada <u>500<\/u> kg rauamaagist. Raua tootmise t\u00e4psem protsess ei ole teada, aga on teada, et rauamaagi ja raua moolsuhe on:<\/span><\/span><\/p>\n<p><span lang=\"et\"><span style=\"line-height: 115%;\">Fe<sub>2<\/sub>O<sub>3<\/sub><\/span><\/span><span lang=\"et\"><span style=\"line-height: 115%;\"> \u2192 2 Fe.<\/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%;\">27% on lisandite hulk esialgses Fe<sub>2<\/sub>O<sub>3<\/sub>, rauamaagi kogus koos lisandiga on 500 kg.<\/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: 12pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">P<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(lisand) = 27%<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">m<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(Fe<sub>2<\/sub>O<sub>3<\/sub>) = 500 kg<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">m<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(Fe) = ?<\/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: 12pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">Kui lisandite hulk on 27% kogu maagist, siis j\u00e4relikult on puhast ainet\u00a073%. Puhta rauamaagi mass:<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\">$m(puhas aine)=\\frac{P(puhas aine)}{100}\\times m(kogu aine koos lisandiga)$<\/p>\n<p style=\"margin: 12pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">Raua massi leidmiseks peame eelnevalt arvutama, mitu mooli puhast rauamaaki on \u00fclesande j\u00e4rgi antud, ning seej\u00e4rel v\u00f5tma arvesse Fe<sub>2<\/sub>O<sub>3<\/sub><\/span><\/span><span lang=\"et\"><span style=\"line-height: 115%;\"> ja Fe moolsuhet: 1 moolist Fe<sub>2<\/sub>O<sub>3<\/sub>-st tekib 2 mooli Fe.<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">Fe<sub>2<\/sub>O<sub>3<\/sub> moolide arvu leidmine:<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\">$n=\\frac{m}{M}$<\/p>\n<p style=\"margin: 12pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">Kui moolide arv on teada, siis saab leida raua massi<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">m\u00a0<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">= <i>n <\/i><span style=\"background: white;\">$\\times $ <\/span><i>M.<\/i><\/span><\/span><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">6. Asenda valemitesse arvud ning arvuta.<\/span><\/span><\/td>\n<td>\n<p style=\"margin: 12pt 0cm;\">$m(Fe_{2}O_{3})=\\frac{73}{100}\\times 500 kg=365 kg$<\/p>\n<p style=\"margin: 12pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">Fe<sub>2<\/sub>O<sub>3<\/sub> moolide leidmiseks tuleb eelnevalt arvutada aine molaarmass:<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">M(Fe<sub>2<\/sub>O<sub>3<\/sub>) = (2 <span style=\"background: white;\">$\\times $ 56 + 3 $\\times $ 16) g\/mol= 160 g\/mol<\/span><\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\">$n(Fe_{2}O_{3})=\\frac{m}{M}=\\frac{365 000 g}{160 g\/mol}=2281 mol$<\/p>\n<p style=\"margin: 12pt 0cm;\"><span lang=\"et\"><span style=\"background: white;\"><span style=\"line-height: 115%;\">Raua moolide leidmiseks arvestame\u00a0<\/span><\/span><span style=\"line-height: 115%;\">Fe<sub>2<\/sub>O<sub>3<\/sub><\/span><\/span><span lang=\"et\"><span style=\"line-height: 115%;\"> ja Fe moolsuhet:<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">1 mooli Fe<sub>2<\/sub>O<sub>3<\/sub>\u00a0 vastab 2 mooli Fe:<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\">$n(Fe)=2\\times 2281 mol=4562,5 mol$<\/p>\n<p style=\"margin: 12pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">Raua massi arvutamiseks kasutame raua molaarmassi:<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">M(Fe) = 56 g\/mol<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">m<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(Fe) = <i>n <\/i><span style=\"background: white;\">$\\times $ <\/span><i>M <\/i>= 4562,5 mol <span style=\"background: white;\">$\\times $ 56 g\/mol = 255500 g = 255,5 kg<\/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-69d5a07e8752a-heading\">\n            <button class=\"accordion-button collapsed\" type=\"button\" data-bs-toggle=\"collapse\" data-bs-target=\"#accordion-69d5a07e8752a-collapse\" aria-expanded=\"true\" aria-controls=\"accordion-69d5a07e8752a-collapse\">\u00dclesanne 2<\/button>\n        <\/h2>\n        <div id=\"accordion-69d5a07e8752a-collapse\" class=\"accordion-collapse collapse\" aria-labelledby=\"accordion-69d5a07e8752a-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-69d5a07e87532-heading\">\n            <button class=\"accordion-button collapsed\" type=\"button\" data-bs-toggle=\"collapse\" data-bs-target=\"#accordion-69d5a07e87532-collapse\" aria-expanded=\"true\" aria-controls=\"accordion-69d5a07e87532-collapse\">\u00dclesande 2 lahendused<\/button>\n        <\/h2>\n        <div id=\"accordion-69d5a07e87532-collapse\" class=\"accordion-collapse collapse\" aria-labelledby=\"accordion-69d5a07e87532-heading\">\n            <div class=\"accordion-body\">\n<p>2.1 S\u00f6\u00f6gisooda (NaHCO<sub>3<\/sub>) sisaldas 9.0% termiliselt stabiilset naatriumkarbonaati (Na<sub>2<\/sub>CO<sub>3<\/sub>). Mitu liitrit (nt.) s\u00fcsinikdioksiidi eraldus 20 g s\u00f6\u00f6gisooda lagunemisel. Anna vastus t\u00e4psusega \u00fcks koht p\u00e4rast koma.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus:<\/b> 2,4 l<\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus:<\/b>Kirjutame v\u00e4lja ja tasakaalustame reaktsiooniv\u00f5rrandi:<\/p>\n<p style=\"margin-left: 80px;\">2 NaHCO<sub>3<\/sub> \u2192 Na<sub>2<\/sub>CO<sub>3<\/sub> + CO<sub>2<\/sub> + H<sub>2<\/sub>O<\/p>\n<p style=\"margin-left: 40px;\">Meil on teada, et 20 g s\u00f6\u00f6gisoodat sisaldas 9,0% lisandit, seega oli seal puhast NaHCO<sub>3<\/sub> 91% ehk 20\u00d70,91=18,2 grammi. Saame leida selle molaarmassi ja arvutada moolide arvu:<\/p>\n<p style=\"margin-left: 80px;\">M(NaHCO<sub>3<\/sub>) = 84 g\/mol<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$n={ m \\over M}={ 18.2g \\over84g\/mol} = 0.217mol$<\/span><\/p>\n<p style=\"margin-left: 40px;\">Reaktsiooniv\u00f5rrandist n\u00e4eme, et NaHCO<sub>3<\/sub> ja CO<sub>2<\/sub> moolsuhe on 2:1. Seega tekib s\u00fcsihappegaasi 0,217\/2=0,108 mol. Gaaside puhul kehtib konstant <i>V<\/i><sub>m<\/sub> = 22,4 dm<sup>3<\/sup>\/mol, mille abil saab leida 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\">0,108 mol \u00d7 22,4 <\/span><span lang=\"ET\">dm<\/span><sup><span lang=\"ET\"><sup>3<\/sup><\/span><\/sup><span lang=\"ET\">\/mol = 2,42 \u2248 2,4 <\/span><span lang=\"ET\">dm<\/span><span lang=\"ET\"><sup>3<\/sup>\u00a0<\/span><span lang=\"ET\">=2,4 l<\/span><\/p>\n<p style=\"margin-left: 80px;\">\u00a0<\/p>\n<p style=\"list-style-type: none;\">2.2 17,5 mooli l\u00e4mmastiku reageerimisel katal\u00fcsaatori juuresolekul piisava hulga vesinikuga tekkis ammoniaak. Arvutuste tulemusena leiti, et lisandite ja kao protsent oli kokku 80 %. Arvuta tekkinud ammoniaagi mass ja ruumala (nt). Anna vastus t\u00e4isarvuna.<\/p>\n<p style=\"list-style-type: none; margin-left: 40px;\"><b>Vastus:<\/b> 119 g; 157 dm<sup>3<\/sup><\/p>\n<p style=\"list-style-type: none; 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=\"list-style-type: none; margin-left: 40px;\">Reaktsiooniv\u00f5rrandist n\u00e4eme, et l\u00e4mmastiku ja ammoniaagi moolsuhe on 1:2. Seega, kui reageeriks 17,5 mooli puhast l\u00e4mmastikku, tekiks 2\u00d717,5=35 mol. Kuna aga lisandite ja kao protsent on 80%, tekib l\u00e4mmastikku 20% eeldatavast ehk 0,2\u00d735=7 mol. Seda teades saame arvutada l\u00e4mmastiku massi ja ruumala:<\/p>\n<p style=\"list-style-type: none; margin-left: 80px;\">M(NH<sub>3<\/sub>)=17 g\/mol<\/p>\n<p style=\"list-style-type: none; margin-left: 80px;\"><i><span lang=\"ET\">m<\/span><\/i><span lang=\"ET\"><sub>NH<\/sub><sub><sub>3<\/sub>\u00a0<\/sub><\/span><i><span lang=\"ET\">=\u00a0n\u00a0\u00d7\u00a0M\u00a0=\u00a0<\/span><\/i><span lang=\"ET\">7 <\/span><span lang=\"ET\">mol \u00d7 17 g\/mol = 119 g<\/span><span style=\"text-align: center;\">\u00a0<\/span><i><\/i><\/p>\n<p style=\"list-style-type: none; 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\">7 mol \u00d7 22,4 <\/span><span lang=\"ET\">dm<\/span><sup><span lang=\"ET\"><sup>3<\/sup><\/span><\/sup><span lang=\"ET\">\/mol = 156,8 \u2248 157 <\/span><span lang=\"ET\">dm<\/span><sup><span lang=\"ET\"><sup>3<\/sup><\/span><\/sup><\/p>\n<p style=\"list-style-type: none; margin-left: 80px;\">\u00a0<\/p>\n<p style=\"list-style-type: none;\">2.3 V\u00e4\u00e4velhappe tootmisel kasutati l\u00e4hteainena p\u00fcriiti (FeS<sub>2<\/sub>). P\u00fcriit sisaldas 16% lisandeid. Mitu kilogrammi v\u00e4\u00e4veldioksiidi saadi 200 kg p\u00fcriidi s\u00e4rdamisel? Anna vastus t\u00e4isarvuna.<\/p>\n<p style=\"list-style-type: none;\">4FeS<sub>2<\/sub> (t) +11 O<sub>2 <\/sub>(g) \u2192 2Fe<sub>2<\/sub>O<sub>3<\/sub>(t) + 8SO<sub>2<\/sub>(g)<\/p>\n<p style=\"list-style-type: none; margin-left: 40px;\"><b>Vastus<\/b>: P\u00fcriidi s\u00e4rdamisel saadi 179 kg v\u00e4\u00e4veldioksiidi<\/p>\n<p style=\"list-style-type: none; margin-left: 40px;\"><b>Lahendus:<\/b> Meil on teada, et p\u00fcriidi mass on 200 kg ja 16% sellest moodustavad lisandid. Puhta FeS<sub>2<\/sub> protsent on seega 84% ja selle mass <i>m<\/i>(FeS<sub>2<\/sub>)=0,84\u00d7200=168 kg. Saame leida FeS<sub>2<\/sub> molaarmassi ja moolide arvu:<\/p>\n<p style=\"list-style-type: none; margin-left: 80px;\">M(FeS<sub>2<\/sub>)=120 g\/mol=120 kg\/kmol<\/p>\n<p style=\"list-style-type: none; margin-left: 80px;\"><span class=\"math-tex\">$n={ m \\over M}={ 168kg \\over120kg\/kmol} = 1.4kmol$<\/span><\/p>\n<p style=\"margin-left: 36.0pt;\">Reaktsiooniv\u00f5rrandist n\u00e4eme, et FeS<sub>2<\/sub> ja SO<sub>2<\/sub> moolsuhe on 4:8=1:2. Seega tekib v\u00e4\u00e4veldioksiidi 2\u00d71,4=2,8 kmol. N\u00fc\u00fcd saab juba molaarmassi kaudu leida v\u00e4\u00e4veldioksiidi massi:<\/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;\"><i><span lang=\"ET\">m<\/span><\/i><span lang=\"ET\"><sub>SO<\/sub><sub><sub>2<\/sub>\u00a0<\/sub><\/span><i><span lang=\"ET\">=\u00a0n\u00a0\u00d7\u00a0M\u00a0=\u00a0<\/span><\/i><span lang=\"ET\">2,8 k<\/span><span lang=\"ET\">mol \u00d7 64 kg\/kmol = 179,2 \u2248 179 kg<\/span><\/p><\/div>\n        <\/div>\n        <\/div>\n    <\/div>\n<h2>N\u00e4idis\u00fclesanne 3. L\u00e4hteainete vajaliku koguse arvutamine<\/h2>\n<table class=\"table table-hover\" style=\"width: 100%;\" border=\"0\" cellspacing=\"1\" cellpadding=\"1\">\n<tbody>\n<tr>\n<td style=\"width: 30%;\"><span lang=\"et\"><span style=\"line-height: 115%;\">1. Loe \u00fclesande tekst m\u00f5ttega l\u00e4bi.<\/span><\/span><\/td>\n<td>\n<p><span lang=\"et\"><span style=\"line-height: 115%;\">10 g kriiti (CaCO<sub>3<\/sub>) reageerib vesinikkloriidhappega vastavalt\u00a0 reaktsioonile:<\/span><\/span><\/p>\n<p><span lang=\"et\"><span style=\"line-height: 115%;\">CaCO<sub>3<\/sub><\/span><\/span><span lang=\"et\"><span style=\"line-height: 115%;\"> + 2HCl \u2192 CaCl<\/span><\/span><sub><span lang=\"et\"><span style=\"line-height: 115%;\">2<\/span><\/span><\/sub><span lang=\"et\"><span style=\"line-height: 115%;\"> + CO<sub>2<\/sub> + H<sub>2<\/sub>O<\/span><\/span><\/p>\n<p><span lang=\"et\"><span style=\"line-height: 115%;\">Mitu ml 0,2 M HCl tuleb v\u00f5tta kogu kriidi \u00e4ra reageerimiseks, kui kriit sisaldab 8% lisandeid?<\/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><u><span lang=\"et\"><span style=\"line-height: 115%;\">10<\/span><\/span><\/u><span lang=\"et\"><span style=\"line-height: 115%;\"> g kriiti (CaCO<sub>3<\/sub>) reageerib vesinikkloriidhappega vastavalt\u00a0 reaktsioonile:<\/span><\/span><\/p>\n<p><span lang=\"et\"><span style=\"line-height: 115%;\">CaCO<sub>3<\/sub><\/span><\/span><span lang=\"et\"><span style=\"line-height: 115%;\"> + 2HCl \u2192 CaCl<\/span><\/span><sub><span lang=\"et\"><span style=\"line-height: 115%;\">2<\/span><\/span><\/sub><span lang=\"et\"><span style=\"line-height: 115%;\"> + CO<sub>2<\/sub> + H<sub>2<\/sub>O<\/span><\/span><\/p>\n<p><span lang=\"et\"><span style=\"line-height: 115%;\">Mitu ml <u>0,2<\/u> M HCl tuleb v\u00f5tta kogu kriidi \u00e4ra reageerimiseks, kui kriit sisaldab <u>8<\/u>% lisandeid?<\/span><\/span><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p><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><\/p>\n<\/td>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">10 g on kogu kriidi mass, 8% on kriidis sisaldavate lisandite hulk, 0,2 M on vesinikkloriidhappe kontsentratsioon.<\/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: 12pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">m<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(kriit) = 10 g<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">P<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(lisand) = 8%<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">c<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(HCl) = 0,2 M<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">V<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(HCl) = ?<\/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: 12pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">Esialgu tuleb arvutada, kui palju on puhast CaCO<sub>3<\/sub> (t), selle saad leida <i>P<\/i>(CaCO<sub>3<\/sub>) = 100 \u2013\u00a0 P(lisand).<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">P<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(CaCO<sub>3<\/sub>) kaudu saab arvutada reaktsioonis osaleva CaCO<sub>3<\/sub> massi.<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\">$m(CaCO_{3})=\\frac{P(puhas aine)}{100}\\times m(kogu aine koos lisandiga)$<\/p>\n<p style=\"margin: 12pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">Seej\u00e4rel arvutame, mitu mooli CaCO<sub>3<\/sub>\u00a0on \u00fclesande j\u00e4rgi antud.<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\">$n=\\frac{m}{M}$<\/p>\n<p style=\"margin: 12pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">Selleks tuleb enne arvutada CaCO<sub>3<\/sub> molaarmass.<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">Reaktsiooniv\u00f5rrandi j\u00e4rgi reageerib 1 mol CaCO<sub>3<\/sub> 2 moli HCl-ga.<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">HCl ainehulgast saab leida reaktsioonis vajaliku lahuse ruumala:<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\">$V=\\frac{n}{c}$<\/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 style=\"margin: 12pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">P<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(CaCO<sub>3<\/sub>) = (100 \u2013 8)% = 92%<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\">$m(CaCO_{3})=\\frac{92}{100}\\times 10 g=9,2 g$<\/p>\n<p style=\"margin: 12pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">M<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(CaCO<sub>3<\/sub>) = (40 + 12 + 3 <span style=\"background: white;\">\u00d7 16)\u00a0g\/mol = 100 g\/mol<\/span><\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\">$n(CaCO_{3})=\\frac{m}{M}=\\frac{9,2 g}{100 g\/mol}=0,092 mol$<\/p>\n<p style=\"margin: 12pt 0cm;\"><span lang=\"et\"><span style=\"background: white;\"><span style=\"line-height: 115%;\">n(HCl) = 2 \u00d7 0,092 mol = 0,184 mol<\/span><\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\">$V(HCl)=\\frac{n}{c}=\\frac{0,184 mol}{0,2 mol\/dm^{3}}=0,92 dm^{3}=920 cm^{3}(ml)$<\/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-69d5a07e8753b-heading\">\n            <button class=\"accordion-button collapsed\" type=\"button\" data-bs-toggle=\"collapse\" data-bs-target=\"#accordion-69d5a07e8753b-collapse\" aria-expanded=\"true\" aria-controls=\"accordion-69d5a07e8753b-collapse\">\u00dclesanne 3<\/button>\n        <\/h2>\n        <div id=\"accordion-69d5a07e8753b-collapse\" class=\"accordion-collapse collapse\" aria-labelledby=\"accordion-69d5a07e8753b-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-69d5a07e87544-heading\">\n            <button class=\"accordion-button collapsed\" type=\"button\" data-bs-toggle=\"collapse\" data-bs-target=\"#accordion-69d5a07e87544-collapse\" aria-expanded=\"true\" aria-controls=\"accordion-69d5a07e87544-collapse\">\u00dclesande 3 lahendused<\/button>\n        <\/h2>\n        <div id=\"accordion-69d5a07e87544-collapse\" class=\"accordion-collapse collapse\" aria-labelledby=\"accordion-69d5a07e87544-heading\">\n            <div class=\"accordion-body\">\n<p style=\"list-style-type: none;\">3.1 Raua tootmisel magnetiidist kasutati redutseerijana s\u00fcsinikoksiidi. Mitu kuupmeetrit gaasi kulus 5 tonni maagi redutseerimiseks (nt), mis sisaldas 82% raud(II,III)oksiidi (Fe<sub>3<\/sub>O<sub>4<\/sub>)? Anna vastus t\u00e4isarvuna.<\/p>\n<p style=\"list-style-type: none; margin-left: 40px;\"><b>Vastus: <\/b>1587 dm<sup>3<\/sup><\/p>\n<p style=\"list-style-type: none; margin-left: 40px;\"><b><span lang=\"ET\" style=\"color: black;\">Lahendus:<\/span><\/b> Kirjutame ja tasakaalustame reaktsiooniv\u00f5rrandi:<\/p>\n<p style=\"list-style-type: none; margin-left: 80px;\">Fe<sub>3<\/sub>O<sub>4<\/sub> + 4CO \u2192 3Fe + 4CO<sub>2<\/sub><\/p>\n<p style=\"list-style-type: none; margin-left: 40px;\">Meil on teada, et magnetiidi mass on 5 tonni ehk 5000 kg, puhta Fe<sub>3<\/sub>O<sub>4<\/sub> protsent on selles 82%. Selle mass on seega <i>m<\/i>(Fe<sub>3<\/sub>O<sub>4<\/sub>)=0,82\u00d75000=4100 kg. Saame leida Fe<sub>3<\/sub>O<sub>4<\/sub> molaarmassi ja moolide arvu:<\/p>\n<p style=\"list-style-type: none; margin-left: 80px;\">M(Fe<sub>3<\/sub>O<sub>4<\/sub>)=231,5 g\/mol=231,5 kg\/kmol<\/p>\n<p style=\"list-style-type: none; margin-left: 80px;\"><span class=\"math-tex\">$n={ m\\over M}={ 4100kg \\over231.5kg\/kmol} = 17.71kmol$<\/span><\/p>\n<p style=\"list-style-type: none; margin-left: 40px;\">Reaktsiooniv\u00f5rrandist n\u00e4eme, et Fe<sub>3<\/sub>O<sub>4<\/sub> ja CO moolsuhe on 1:4. CO moolide arv on seega 4\u00d717,71=70,84 kmol. Gaasi molaarruumala (22,4 dm<sup>3<\/sup>\/mol=22,4 m<sup>3<\/sup>\/kmol) kasutades saame n\u00fc\u00fcd CO ruumala leida:<i><\/i><\/p>\n<p style=\"list-style-type: none; 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\">70,84 kmol \u00d7 22,4 <\/span><span lang=\"ET\">m<\/span><sup><span lang=\"ET\"><sup>3<\/sup><\/span><\/sup><span lang=\"ET\">\/kmol = 1586,86 \u2248 1587 <\/span><span lang=\"ET\">m<\/span><sup><span lang=\"ET\"><sup>3<\/sup><\/span><\/sup><\/p>\n<p style=\"list-style-type: none; margin-left: 80px;\">\u00a0<\/p>\n<p style=\"list-style-type: none;\">3.2 T\u00f6\u00f6stuslikku KClO<sub>3<\/sub> sisaldas 23% lisandeid. Soola kuumutati ja see lagunes kaaliumkloriidiks ja hapnikuks.<\/p>\n<p style=\"list-style-type: none;\">2KClO<sub>3<\/sub> \u2192 2KCl + 3O<sub>2<\/sub><\/p>\n<p style=\"list-style-type: none;\">Reaktsiooni tulemusena tekkis 1,5 g hapnikku. Arvuta mitu grammi t\u00f6\u00f6stuslikku KClO<sub>3<\/sub> v\u00f5eti 1,5 g hapniku saamiseks? Anna vastus t\u00e4isarvuna.<\/p>\n<p style=\"list-style-type: none; margin-left: 40px;\"><b>Vastus:<\/b> 5 g<\/p>\n<p style=\"list-style-type: none; margin-left: 40px;\"><b>Lahendus:<\/b> Meil on teada, et O<sub>2<\/sub> tekib 1,5 grammi. Saame leida selle molaarmassi ja arvutada moolide arvu:<\/p>\n<p style=\"list-style-type: none; margin-left: 80px;\">M(O<sub>2<\/sub>) = 32 g\/mol<\/p>\n<p style=\"list-style-type: none; margin-left: 80px;\"><span class=\"math-tex\">$n={ m\\over M}={ 1.5g \\over32g\/mol} = 0.046875mol$<\/span><i><\/i><\/p>\n<p style=\"list-style-type: none; margin-left: 40px;\">Reaktsiooniv\u00f5rrandist n\u00e4eme, et O<sub>2<\/sub> ja KClO<sub>3<\/sub> moolsuhe on 3:2, seega on KClO<sub>3 <\/sub>moolide arv <span lang=\"ET\">0,046875 mol\u00d7<\/span><span lang=\"ET\">2\/<\/span><span lang=\"ET\">3<\/span><span lang=\"ET\">=0,03125 mol<\/span>. Saame n\u00fc\u00fcd leida KClO<sub>3<\/sub> molaarmassi ja arvutada selle massi:<\/p>\n<p style=\"list-style-type: none; margin-left: 80px;\">M(KClO<sub>3<\/sub>) = 122,5 g\/mol<i><\/i><\/p>\n<p style=\"list-style-type: none; margin-left: 80px;\"><i><span lang=\"ET\">m<\/span><\/i><sub><span lang=\"ET\"><sub>K<\/sub><\/span><sub>Cl<\/sub><span lang=\"ET\"><sub>O3<\/sub><\/span><\/sub><span lang=\"ET\"><sub>\u00a0<\/sub><\/span><i><span lang=\"ET\">=\u00a0n\u00a0\u00d7\u00a0M\u00a0=\u00a0<\/span><\/i><span lang=\"ET\">0,03125 mol \u00d7 122,5 g\/mol =\u00a03,83 g<\/span>\u00a0<\/p>\n<p style=\"list-style-type: none; margin-left: 40px;\">Lisandeid on 23%, seega puhast ainet 77%. Kui sellele vastab 3,83 g, siis t\u00f6\u00f6stusliku aine kogumass on <span lang=\"ET\">3,83 \u00d7\u00a0<\/span><span lang=\"ET\">100% \/\u00a0<\/span><span lang=\"ET\">77%\u00a0<\/span><span lang=\"ET\">= 4,974\u00a0\u2248 5 g<\/span>.<\/p>\n<p style=\"list-style-type: none; margin-left: 40px;\">\u00a0<\/p>\n<p style=\"list-style-type: none;\">3.3 S\u00f6\u00f6gisoola \u00fcks peamisi allikaid on merevesi. Merevee soolasus on 3,5%. Arvuta mitu kg merevett peaks v\u00f5tma, selleks, et saada 5 kg soola. Anna vastus t\u00e4isarvuna.<b> <\/b><\/p>\n<p style=\"list-style-type: none; margin-left: 40px;\"><b>Vastus:<\/b> 143 kg merevett<\/p>\n<p style=\"list-style-type: none; margin-left: 40px;\"><b>Lahendus: <\/b>Kui 3,5% mereveest on 5 kg, siis 100% ehk merevee kogumass vastab <span lang=\"ET\">5 kg \u00d7\u00a0<\/span><span lang=\"ET\">100% \/\u00a0<\/span><span lang=\"ET\">35%\u00a0<\/span><span lang=\"ET\">= 142,85 \u2248 143 kg<\/span>.<\/p>\n<p style=\"list-style-type: none; margin-left: 40px;\">\u00a0<\/p>\n<p style=\"list-style-type: none;\">3.4 Mitu kilogrammi etanooli (C<sub>2<\/sub>H<sub>5<\/sub>OH) saab toota 25g gl\u00fckoosist (C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub>), mis sisaldab 28% lisandeid?<\/p>\n<p style=\"list-style-type: none;\">C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub> \u2192 2 C<sub>2<\/sub>H<sub>5<\/sub>OH + CO<sub>2<\/sub><\/p>\n<p style=\"list-style-type: none;\">Anna vastus t\u00e4psusega \u00fcks koht peale koma.<\/p>\n<p style=\"list-style-type: none; margin-left: 40px;\"><b>Vastus:<\/b> 9,2 g<\/p>\n<p style=\"list-style-type: none; margin-left: 40px;\"><b><span lang=\"ET\" style=\"color: black;\">Lahendus:<\/span><\/b> Meil on teada, et koos lisanditega gl\u00fckoosi mass on 25 g, lisandeid on selles 28%. Puhta gl\u00fckoosi protsent on seega 72% ja selle mass on <i>m<\/i>(gl\u00fckoos) = 0,72 \u00d7 25 = 18 g. Saame leida gl\u00fckoosi molaarmassi ja selle kaudu moolide arvu:<\/p>\n<p style=\"list-style-type: none; margin-left: 80px;\">M(gl\u00fckoos)=180 g\/mol<\/p>\n<p style=\"list-style-type: none; margin-left: 80px;\"><span lang=\"ET\"><span class=\"math-tex\">$n={ m\\over M}={ 18g \\over180g\/mol} = 0.1mol$<\/span><\/span><\/p>\n<p style=\"list-style-type: none; margin-left: 40px;\">Reaktsiooniv\u00f5rrandist n\u00e4eme, et gl\u00fckoosi ja etanooli moolsuhe on 1:2, seega on etanooli moolide arv 2\u00d70,1=0,2 mol. Saame n\u00fc\u00fcd leida etanooli molaarmassi ja arvutada selle massi:<\/p>\n<p style=\"list-style-type: none; margin-left: 80px;\">M(C<sub>2<\/sub>H<sub>5<\/sub>OH) = 46 g\/mol<\/p>\n<p style=\"list-style-type: none; margin-left: 80px;\"><i><span lang=\"ET\">m<\/span><\/i><sub>C<\/sub><sub>2<\/sub><sub>H<\/sub><sub>5<\/sub><sub>OH<\/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;\">0,2 mol \u00d7 46 g\/mol = 9,2 g<\/span><\/p><\/div>\n        <\/div>\n        <\/div>\n    <\/div>\n<p>\u00a0<\/p>","protected":false},"excerpt":{"rendered":"<p>Eri ainete tootmises kasutatakse tihti l\u00e4hteaineid, mis ei ole t\u00e4iesti puhtad\u00a0ehk sisaldavad mingeid lisandeid. Seet\u00f5ttu tekib protsessi k\u00e4igus ka v\u00e4hem saadusi, kui tekiks puhaste ainete korral. J\u00e4rgnevast videost saad vaadata, kuidas arvestada lisandite m\u00f5ju erinevates arvutus\u00fclesannetes. Kolmandate osapoolte sisu n\u00e4gemiseks &#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-11","page","type-page","status-publish","hentry"],"acf":[],"_links":{"self":[{"href":"https:\/\/sisu.ut.ee\/huvitavkeemia\/wp-json\/wp\/v2\/pages\/11","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=11"}],"version-history":[{"count":25,"href":"https:\/\/sisu.ut.ee\/huvitavkeemia\/wp-json\/wp\/v2\/pages\/11\/revisions"}],"predecessor-version":[{"id":982,"href":"https:\/\/sisu.ut.ee\/huvitavkeemia\/wp-json\/wp\/v2\/pages\/11\/revisions\/982"}],"wp:attachment":[{"href":"https:\/\/sisu.ut.ee\/huvitavkeemia\/wp-json\/wp\/v2\/media?parent=11"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}