Consider the reaction MnO2 + 4HCl ? MnCl2 + Cl2 + H2O If 2.86 mol of MnO2 and 96.2 g of HCl react, which reagent will be used up first?<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n Hydrochloric acid (HCl) is the limiting reagent.<\/p>\n\n\n\n To determine which reagent is the limiting reagent, follow these steps:<\/p>\n\n\n\n [ This equation shows that 1 mole of MnO\u2082 reacts with 4 moles of HCl<\/strong>.<\/p>\n\n\n\n From the balanced equation:<\/p>\n\n\n\n However, we only have 2.64 moles of HCl<\/strong>, which is much less than the required 11.44 moles. This means that HCl will be used up first, making it the limiting reagent<\/strong>.<\/p>\n\n\n\n Since HCl is consumed first, the reaction will stop once all of it is reacted, even though there is still some MnO\u2082 left. This means that HCl is the limiting reagent, and MnO\u2082 is the excess reagent<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":" Consider the reaction MnO2 + 4HCl ? MnCl2 + Cl2 + H2O If 2.86 mol of MnO2 and 96.2 g of HCl react, which reagent will be used up first? The Correct Answer and Explanation is : Answer: Hydrochloric acid (HCl) is the limiting reagent. Explanation: To determine which reagent is the limiting reagent, follow […]<\/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-187476","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187476","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=187476"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187476\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187476"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187476"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187476"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Answer:<\/h3>\n\n\n\n
Explanation:<\/h3>\n\n\n\n
Step 1: Write the balanced chemical equation<\/strong><\/h4>\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\nStep 2: Calculate the moles of each reactant<\/strong><\/h4>\n\n\n\n
\n
Given that we have 2.86 moles<\/strong> of MnO\u2082, this value remains unchanged.<\/li>\n\n\n\n
Given mass = 96.2 g<\/strong>, and the molar mass of HCl = 1.01 (H) + 35.45 (Cl) = 36.46 g\/mol<\/strong>, [
\\text{Moles of HCl} = \\frac{96.2 \\text{ g}}{36.46 \\text{ g\/mol}} = 2.64 \\text{ moles}
]<\/li>\n<\/ul>\n\n\n\nStep 3: Determine the limiting reagent<\/strong><\/h4>\n\n\n\n
\n
2.86 \\times 4 = 11.44 \\text{ moles of HCl}
]<\/li>\n<\/ul>\n\n\n\nStep 4: Conclusion<\/strong><\/h4>\n\n\n\n