{"id":187287,"date":"2025-02-04T06:02:01","date_gmt":"2025-02-04T06:02:01","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=187287"},"modified":"2025-02-04T06:02:03","modified_gmt":"2025-02-04T06:02:03","slug":"for-the-reaction-shown-compute-the-theoretical-yield-of-product-in-moles-for-each-of-the-following-initial-amounts-of-reactants","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/04\/for-the-reaction-shown-compute-the-theoretical-yield-of-product-in-moles-for-each-of-the-following-initial-amounts-of-reactants\/","title":{"rendered":"For the reaction shown, compute the theoretical yield of product (in moles) for each of the following initial amounts of reactants"},"content":{"rendered":"\n<p>For the reaction shown, compute the theoretical yield of product (in moles) for each of the following initial amounts of reactants.<br>Mn(s)+O2(g)?MnO2(s)<\/p>\n\n\n\n<p>Part A 5mol Mn, 5mol O2 Express your answer using one significant figure. __________moll<\/p>\n\n\n\n<p>Part B 3mol Mn, 9mol O2 Express your answer using one significant figure. ___________________moll<\/p>\n\n\n\n<p>Part C 27.0mol Mn, 43.8mol O2 ___________________moll<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To determine the theoretical yield of MnO\u2082 (manganese dioxide) for each case, we need to use stoichiometry based on the balanced chemical equation:<\/p>\n\n\n\n<p>[<br>2Mn(s) + O_2(g) \\rightarrow 2MnO_2(s)<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Identify the Limiting Reactant<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>From the balanced equation<\/strong>, 2 moles of Mn react with 1 mole of O\u2082 to produce 2 moles of MnO\u2082.<\/li>\n\n\n\n<li>The limiting reactant determines the amount of product formed.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Part A: 5 mol Mn, 5 mol O\u2082<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Using Mn:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( 5 \\text{ mol Mn} \\times \\frac{2 \\text{ mol MnO}_2}{2 \\text{ mol Mn}} = 5 \\text{ mol MnO}_2 )<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Using O\u2082:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( 5 \\text{ mol O}_2 \\times \\frac{2 \\text{ mol MnO}_2}{1 \\text{ mol O}_2} = 10 \\text{ mol MnO}_2 )<\/li>\n\n\n\n<li>Since Mn produces the <strong>smaller<\/strong> amount of MnO\u2082 (5 mol), <strong>Mn is the limiting reactant<\/strong>.<\/li>\n\n\n\n<li><strong>Theoretical yield<\/strong>: <strong>5 mol MnO\u2082 (1 sig. fig.) \u2192 5 mol<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Part B: 3 mol Mn, 9 mol O\u2082<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Using Mn:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( 3 \\text{ mol Mn} \\times \\frac{2 \\text{ mol MnO}_2}{2 \\text{ mol Mn}} = 3 \\text{ mol MnO}_2 )<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Using O\u2082:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( 9 \\text{ mol O}_2 \\times \\frac{2 \\text{ mol MnO}_2}{1 \\text{ mol O}_2} = 18 \\text{ mol MnO}_2 )<\/li>\n\n\n\n<li>Since Mn produces the <strong>smaller<\/strong> amount of MnO\u2082 (3 mol), <strong>Mn is the limiting reactant<\/strong>.<\/li>\n\n\n\n<li><strong>Theoretical yield<\/strong>: <strong>3 mol MnO\u2082 (1 sig. fig.) \u2192 3 mol<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Part C: 27.0 mol Mn, 43.8 mol O\u2082<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Using Mn:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( 27.0 \\text{ mol Mn} \\times \\frac{2 \\text{ mol MnO}_2}{2 \\text{ mol Mn}} = 27.0 \\text{ mol MnO}_2 )<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Using O\u2082:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( 43.8 \\text{ mol O}_2 \\times \\frac{2 \\text{ mol MnO}_2}{1 \\text{ mol O}_2} = 87.6 \\text{ mol MnO}_2 )<\/li>\n\n\n\n<li>Since Mn produces the <strong>smaller<\/strong> amount of MnO\u2082 (27.0 mol), <strong>Mn is the limiting reactant<\/strong>.<\/li>\n\n\n\n<li><strong>Theoretical yield<\/strong>: <strong>27 mol MnO\u2082 (1 sig. fig.) \u2192 30 mol<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation (300 words)<\/strong><\/h3>\n\n\n\n<p>Stoichiometry allows us to determine the theoretical yield of a reaction by using the balanced chemical equation. In this reaction, manganese (Mn) reacts with oxygen gas (O\u2082) to form manganese dioxide (MnO\u2082), following a <strong>2:1:2 mole ratio<\/strong>.<\/p>\n\n\n\n<p>For each case, we determine the limiting reactant by comparing the ratio of reactants to the balanced equation. The limiting reactant is the reactant that produces the least amount of product, which in turn determines the theoretical yield.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Part A:<\/strong> 5 mol Mn and 5 mol O\u2082 are given. Since 5 mol Mn requires only 2.5 mol O\u2082 but we have excess O\u2082, Mn is the limiting reactant, producing <strong>5 mol MnO\u2082<\/strong>.<\/li>\n\n\n\n<li><strong>Part B:<\/strong> 3 mol Mn and 9 mol O\u2082 are given. Since 3 mol Mn requires only 1.5 mol O\u2082 but we have excess O\u2082, Mn is again the limiting reactant, producing <strong>3 mol MnO\u2082<\/strong>.<\/li>\n\n\n\n<li><strong>Part C:<\/strong> 27.0 mol Mn and 43.8 mol O\u2082 are given. Since 27.0 mol Mn requires only 13.5 mol O\u2082 but we have excess O\u2082, Mn is still the limiting reactant, producing <strong>30 mol MnO\u2082 (rounded to 1 sig. fig.)<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p>In each case, excess reactants remain unused. The theoretical yield is found using <strong>mole ratios<\/strong>, and the answer is reported with one significant figure, as required.<\/p>\n\n\n\n<p>Let me know if you need further clarification! \ud83d\ude80<\/p>\n","protected":false},"excerpt":{"rendered":"<p>For the reaction shown, compute the theoretical yield of product (in moles) for each of the following initial amounts of reactants.Mn(s)+O2(g)?MnO2(s) Part A 5mol Mn, 5mol O2 Express your answer using one significant figure. __________moll Part B 3mol Mn, 9mol O2 Express your answer using one significant figure. ___________________moll Part C 27.0mol Mn, 43.8mol O2 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-187287","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187287","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=187287"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187287\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187287"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187287"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187287"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}