{"id":189440,"date":"2025-02-09T07:59:46","date_gmt":"2025-02-09T07:59:46","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=189440"},"modified":"2025-02-09T07:59:49","modified_gmt":"2025-02-09T07:59:49","slug":"when-189-4-g-of-ethylene-c2h4-molar-mass-28-05-g-mol-burns-in-oxygen-to-give-carbon-dioxide-and-water","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/09\/when-189-4-g-of-ethylene-c2h4-molar-mass-28-05-g-mol-burns-in-oxygen-to-give-carbon-dioxide-and-water\/","title":{"rendered":"When 189.4 g of ethylene (C2H4) (molar mass= 28.05 g\/mol) burns in oxygen to give carbon dioxide and water"},"content":{"rendered":"\n<p>When 189.4 g of ethylene (C2H4) (molar mass= 28.05 g\/mol) burns in oxygen to give carbon dioxide and water, how many grams of CO2 (molar mass= 44.01 g\/mol) are formed?<br>C2H4 + 3O2 &#8211;&gt; 2 CO2 + 2H2O<\/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>When 189.4 grams of ethylene (C\u2082H\u2084) combust in oxygen, approximately 297.17 grams of carbon dioxide (CO\u2082) are produced.<\/p>\n\n\n\n<p><strong>Balanced Chemical Equation:<\/strong><\/p>\n\n\n\n<p>[ \\text{C}_2\\text{H}_4 + 3\\text{O}_2 \\rightarrow 2\\text{CO}_2 + 2\\text{H}_2\\text{O} ]<\/p>\n\n\n\n<p><strong>Step-by-Step Calculation:<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Determine Moles of Ethylene (C\u2082H\u2084):<\/strong> [ \\text{Moles of C}_2\\text{H}_4 = \\frac{\\text{Mass of C}_2\\text{H}_4}{\\text{Molar Mass of C}_2\\text{H}_4} = \\frac{189.4\\,\\text{g}}{28.05\\,\\text{g\/mol}} \\approx 6.75\\,\\text{mol} ]<\/li>\n\n\n\n<li><strong>Calculate Moles of CO\u2082 Produced:<\/strong> From the balanced equation, 1 mole of C\u2082H\u2084 produces 2 moles of CO\u2082. Therefore: [ \\text{Moles of CO}_2 = 6.75\\,\\text{mol} \\times 2 = 13.5\\,\\text{mol} ]<\/li>\n\n\n\n<li><strong>Convert Moles of CO\u2082 to Grams:<\/strong> [ \\text{Mass of CO}_2 = \\text{Moles of CO}_2 \\times \\text{Molar Mass of CO}_2 = 13.5\\,\\text{mol} \\times 44.01\\,\\text{g\/mol} \\approx 594.14\\,\\text{g} ]<\/li>\n<\/ol>\n\n\n\n<p><strong>Conclusion:<\/strong><\/p>\n\n\n\n<p>Combusting 189.4 grams of ethylene yields approximately 594.14 grams of carbon dioxide.<\/p>\n\n\n\n<p><strong>Explanation:<\/strong><\/p>\n\n\n\n<p>Combustion of ethylene involves a chemical reaction where ethylene (C\u2082H\u2084) reacts with oxygen (O\u2082) to produce carbon dioxide (CO\u2082) and water (H\u2082O). The balanced chemical equation for this reaction is:<\/p>\n\n\n\n<p>[ \\text{C}_2\\text{H}_4 + 3\\text{O}_2 \\rightarrow 2\\text{CO}_2 + 2\\text{H}_2\\text{O} ]<\/p>\n\n\n\n<p>This equation indicates that one mole of ethylene reacts with three moles of oxygen to produce two moles of carbon dioxide and two moles of water.<\/p>\n\n\n\n<p><strong>Stoichiometric Calculations:<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Moles of Ethylene:<\/strong> The molar mass of ethylene (C\u2082H\u2084) is 28.05 g\/mol. Therefore, the number of moles in 189.4 grams of ethylene is: [ \\text{Moles of C}_2\\text{H}_4 = \\frac{189.4\\,\\text{g}}{28.05\\,\\text{g\/mol}} \\approx 6.75\\,\\text{mol} ]<\/li>\n\n\n\n<li><strong>Moles of CO\u2082 Produced:<\/strong> According to the balanced equation, 1 mole of C\u2082H\u2084 produces 2 moles of CO\u2082. Therefore, 6.75 moles of C\u2082H\u2084 will produce: [ \\text{Moles of CO}_2 = 6.75\\,\\text{mol} \\times 2 = 13.5\\,\\text{mol} ]<\/li>\n\n\n\n<li><strong>Mass of CO\u2082 Produced:<\/strong> The molar mass of CO\u2082 is 44.01 g\/mol. Therefore, the mass of CO\u2082 produced is: [ \\text{Mass of CO}_2 = 13.5\\,\\text{mol} \\times 44.01\\,\\text{g\/mol} \\approx 594.14\\,\\text{g} ]<\/li>\n<\/ol>\n\n\n\n<p><strong>Conclusion:<\/strong><\/p>\n\n\n\n<p>Combusting 189.4 grams of ethylene produces approximately 594.14 grams of carbon dioxide.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>When 189.4 g of ethylene (C2H4) (molar mass= 28.05 g\/mol) burns in oxygen to give carbon dioxide and water, how many grams of CO2 (molar mass= 44.01 g\/mol) are formed?C2H4 + 3O2 &#8211;&gt; 2 CO2 + 2H2O The Correct Answer and Explanation is : When 189.4 grams of ethylene (C\u2082H\u2084) combust in oxygen, approximately 297.17 [&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-189440","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189440","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=189440"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189440\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=189440"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=189440"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=189440"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}