To solve this problem, we need to analyze the chemical reaction between potassium superoxide (KO\u2082) and carbon dioxide (CO\u2082). The balanced chemical equation is:<\/p>\n\n\n\n
$$
4KO_2 + 2CO_2 \\rightarrow 2K_2CO_3 + 3O_2
$$<\/p>\n\n\n\n
\n\n\n\n
Step 1: Calculate the moles of each reactant<\/h3>\n\n\n\n
Moles of KO\u2082<\/strong>:<\/p>\n\n\n\n $$ Moles of CO\u2082<\/strong>:<\/p>\n\n\n\n $$ From the equation:<\/p>\n\n\n\n $$ This simplifies to:<\/p>\n\n\n\n $$ So, for every 2 moles of KO\u2082, we need 1 mole of CO\u2082.<\/p>\n\n\n\n Using the mole ratio, check how much CO\u2082 is needed for the given KO\u2082:<\/p>\n\n\n\n $$ We have 1.0227 mol CO\u2082 available<\/strong>, which is much more than 0.1759 mol<\/strong>, so KO\u2082 is the limiting reagent<\/strong>, and CO\u2082 is in excess.<\/p>\n\n\n\n From the balanced reaction:<\/p>\n\n\n\n $$ $$ Convert to grams:<\/p>\n\n\n\n $$ 37.3 grams of CO\u2082 remain unreacted<\/strong><\/p>\n\n\n\n In this problem, we are asked to determine how much of the excess reactant remains unreacted after the completion of a chemical reaction between potassium superoxide (KO\u2082) and carbon dioxide (CO\u2082). The balanced equation for the reaction is:<\/p>\n\n\n\n $$ The first step is to convert the masses of the two reactants into moles using their molar masses. Potassium superoxide has a molar mass of 71.10 g\/mol, and carbon dioxide has a molar mass of 44.01 g\/mol. Dividing the given masses by their respective molar masses gives us approximately 0.3517 moles of KO\u2082 and 1.0227 moles of CO\u2082.<\/p>\n\n\n\n Next, we use the mole ratio from the balanced equation to determine which reactant is limiting. The equation shows that 4 moles of KO\u2082 react with 2 moles of CO\u2082, or more simply, 2 moles of KO\u2082 for every 1 mole of CO\u2082. Based on this ratio, 0.3517 moles of KO\u2082 would need only about 0.1759 moles of CO\u2082 to react completely.<\/p>\n\n\n\n Since we have 1.0227 moles of CO\u2082\u2014much more than needed\u2014it is in excess, and KO\u2082 is the limiting reagent. This means all the KO\u2082 will be consumed, and not all the CO\u2082 will react.<\/p>\n\n\n\n We calculate the amount of CO\u2082 that actually reacts (0.1759 moles), subtract it from the original amount, and convert the leftover CO\u2082 back into grams. This yields approximately 37.3 grams of CO\u2082 remaining unreacted.<\/p>\n\n\n\n Understanding limiting and excess reagents is crucial in stoichiometry, as it allows us to predict the amounts of products and leftovers in a reaction.<\/p>\n\n\n\n In the reaction below, 25.0g of potassium superoxide, KO2, (molar mass is 71.10g\/mole) is used with 45.0g of CO2 (molar mass is 44.01g\/mole). how many grams of the excess reagent are left unreacted? The Correct Answer and Explanation is: To solve this problem, we need to analyze the chemical reaction between potassium superoxide (KO\u2082) and […]<\/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-218507","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218507","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=218507"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218507\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=218507"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=218507"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=218507"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\\frac{25.0 \\text{ g}}{71.10 \\text{ g\/mol}} = 0.3517 \\text{ mol KO\u2082}
$$<\/p>\n\n\n\n
\\frac{45.0 \\text{ g}}{44.01 \\text{ g\/mol}} = 1.0227 \\text{ mol CO\u2082}
$$<\/p>\n\n\n\n
\n\n\n\nStep 2: Use the mole ratio from the balanced equation<\/h3>\n\n\n\n
4 \\text{ mol KO\u2082} : 2 \\text{ mol CO\u2082}
$$<\/p>\n\n\n\n
2 \\text{ mol KO\u2082} : 1 \\text{ mol CO\u2082}
$$<\/p>\n\n\n\n
\n\n\n\nStep 3: Determine the limiting reagent<\/h3>\n\n\n\n
\\frac{0.3517 \\text{ mol KO\u2082}}{2} = 0.1759 \\text{ mol CO\u2082 needed}
$$<\/p>\n\n\n\n
\n\n\n\nStep 4: Calculate how much CO\u2082 is used<\/h3>\n\n\n\n
4 \\text{ mol KO\u2082} : 2 \\text{ mol CO\u2082}
\\Rightarrow \\frac{0.3517 \\text{ mol KO\u2082}}{4} \\times 2 = 0.1759 \\text{ mol CO\u2082 used}
$$<\/p>\n\n\n\n
\n\n\n\nStep 5: Calculate how much CO\u2082 remains<\/h3>\n\n\n\n
1.0227 \\text{ mol CO\u2082} – 0.1759 \\text{ mol CO\u2082} = 0.8468 \\text{ mol CO\u2082 remaining}
$$<\/p>\n\n\n\n
0.8468 \\text{ mol} \\times 44.01 \\text{ g\/mol} = 37.25 \\text{ g CO\u2082 unreacted}
$$<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answer:<\/h3>\n\n\n\n
<\/h3>\n\n\n\n
4KO_2 + 2CO_2 \\rightarrow 2K_2CO_3 + 3O_2
$$<\/p>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-263.png\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"