To find the theoretical yield of Cl\u2082<\/strong>, we follow these steps:<\/p>\n\n\n\n $$ From the balanced equation, 1 mol MnO\u2082 reacts with 4 mol HCl<\/strong>.<\/p>\n\n\n\n So, 0.860 mol MnO\u2082 would require:<\/p>\n\n\n\n $$ But we only have 1.32 mol HCl, so HCl is the limiting reactant<\/strong>.<\/p>\n\n\n\n From the balanced equation, 4 mol HCl produces 1 mol Cl\u2082<\/strong>.<\/p>\n\n\n\n So, 1.32 mol HCl will produce:<\/p>\n\n\n\n $$ Molar mass of Cl\u2082 = 2 \u00d7 35.45 = 70.90 g\/mol<\/strong><\/p>\n\n\n\n $$ To calculate the theoretical yield of chlorine gas (Cl\u2082), we begin by analyzing the balanced chemical equation:<\/p>\n\n\n\n $$ This tells us the molar ratio: 1 mole of MnO\u2082 reacts with 4 moles of HCl to produce 1 mole of Cl\u2082<\/strong>. The goal is to determine the maximum mass (theoretical yield) of Cl\u2082 that can be formed from 74.8 g of MnO\u2082 and 48.2 g of HCl.<\/p>\n\n\n\n We convert grams of each reactant into moles using their molar masses. MnO\u2082 has a molar mass of 86.94 g\/mol, so 74.8 g corresponds to 0.860 mol. HCl has a molar mass of 36.46 g\/mol, so 48.2 g gives 1.32 mol. Next, we determine the limiting reactant by comparing mole ratios. Since 0.860 mol MnO\u2082 would need 3.44 mol HCl, but only 1.32 mol HCl is available, HCl is the limiting reactant<\/strong>\u2014it will run out first and limit how much Cl\u2082 can be formed.<\/p>\n\n\n\n Using stoichiometry, 4 mol HCl produces 1 mol Cl\u2082. So 1.32 mol HCl produces 0.330 mol Cl\u2082. Finally, converting this to grams using Cl\u2082\u2019s molar mass (70.90 g\/mol), we find the theoretical yield is:<\/p>\n\n\n\n $$ This is the maximum amount of chlorine gas that can be formed from the given amounts of MnO\u2082 and HCl, assuming complete reaction and no losses.<\/p>\n\n\n\n Which Of The Following Best Describes The Transformation Of: Y = X To Y = 4* – 5 Select One: A. Horizontal Expansion By A Factor Of 4, Translate Down 5 UnitsB. Horizontal Compression By A Factor Of 1\/4, Translate Down 5 UnitsC. Horizontal Expansion By A Factor Of 4, Translate Right 5 UnitsD. Horizontal […]<\/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-219389","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219389","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=219389"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219389\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=219389"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=219389"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=219389"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nBalanced Chemical Equation:<\/strong><\/h3>\n\n\n\n
\\text{MnO}_2 (s) + 4\\text{HCl} (aq) \\rightarrow \\text{MnCl}_2 (aq) + \\text{Cl}_2 (g) + 2\\text{H}_2O (l)
$$<\/p>\n\n\n\n
\n\n\n\nStep 1: Calculate molar masses<\/strong><\/h3>\n\n\n\n
\n
Mn = 54.94 g\/mol, O = 16.00 g\/mol
\u2192 MnO\u2082 = 54.94 + 2\u00d716.00 = 86.94 g\/mol<\/strong><\/li>\n\n\n\n
H = 1.01, Cl = 35.45
\u2192 HCl = 1.01 + 35.45 = 36.46 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n
\n\n\n\nStep 2: Convert mass to moles<\/strong><\/h3>\n\n\n\n
\n
\\frac{74.8 \\text{ g}}{86.94 \\text{ g\/mol}} = 0.860 moles
$$<\/li>\n\n\n\n
\\frac{48.2 \\text{ g}}{36.46 \\text{ g\/mol}} = 1.32 moles
$$<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nStep 3: Identify limiting reactant<\/strong><\/h3>\n\n\n\n
0.860 \\times 4 = 3.44 \\text{ mol HCl}
$$<\/p>\n\n\n\n
\n\n\n\nStep 4: Determine moles of Cl\u2082 produced<\/strong><\/h3>\n\n\n\n
\\frac{1.32}{4} = 0.330 \\text{ mol Cl}_2
$$<\/p>\n\n\n\n
\n\n\n\nStep 5: Convert moles of Cl\u2082 to grams<\/strong><\/h3>\n\n\n\n
0.330 \\times 70.90 = \\boxed{23.4 \\text{ g Cl}_2}
$$<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answer: 23.4 g Cl\u2082<\/strong><\/h3>\n\n\n\n
\n\n\n\nExplanation<\/strong><\/h3>\n\n\n\n
\\text{MnO}_2 + 4\\text{HCl} \\rightarrow \\text{MnCl}_2 + \\text{Cl}_2 + 2\\text{H}_2O
$$<\/p>\n\n\n\n
0.330 \\times 70.90 = \\boxed{23.4 \\text{ grams}}
$$<\/p>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner4-17.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"