{"id":191006,"date":"2025-02-14T05:01:36","date_gmt":"2025-02-14T05:01:36","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=191006"},"modified":"2025-02-14T05:01:39","modified_gmt":"2025-02-14T05:01:39","slug":"the-equation-from-the-combustion-of-hexane-is","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/14\/the-equation-from-the-combustion-of-hexane-is\/","title":{"rendered":"The equation from the combustion of hexane is"},"content":{"rendered":"\n<p>1.The equation from the combustion of hexane is:<br>C6H14(g)+19\/2 O2(g) -(arrow sign) 6CO2(g)+7H2O(g)<br>Calculate the number of moles of<br>a) CO2 produced in the combustion of 1.50 moles of hexane<br>b) H2O produced in the combustion of 0.25 moles of hexane<br>c) O2 consumed in the combustion of 0.40 moles of hexane<\/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<h3 class=\"wp-block-heading\">Solution:<\/h3>\n\n\n\n<p>The balanced chemical equation for the combustion of hexane is:<\/p>\n\n\n\n<p>[<br>C_6H_{14}(g) + \\frac{19}{2} O_2(g) \\rightarrow 6 CO_2(g) + 7 H_2O(g)<br>]<\/p>\n\n\n\n<p>This equation tells us the molar ratio between the reactants and products:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>1 mole of <strong>C\u2086H\u2081\u2084<\/strong> produces <strong>6 moles of CO\u2082<\/strong>.<\/li>\n\n\n\n<li>1 mole of <strong>C\u2086H\u2081\u2084<\/strong> produces <strong>7 moles of H\u2082O<\/strong>.<\/li>\n\n\n\n<li>1 mole of <strong>C\u2086H\u2081\u2084<\/strong> consumes <strong>( \\frac{19}{2} = 9.5 ) moles of O\u2082<\/strong>.<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>(a) Moles of CO\u2082 produced from 1.50 moles of hexane<\/strong><\/h4>\n\n\n\n<p>Using the molar ratio:<br>[<br>\\text{Moles of CO\u2082} = (1.50 \\text{ moles C\u2086H\u2081\u2084}) \\times \\frac{6 \\text{ moles CO\u2082}}{1 \\text{ mole C\u2086H\u2081\u2084}}<br>]<\/p>\n\n\n\n<p>[<br>= 9.00 \\text{ moles CO\u2082}<br>]<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>(b) Moles of H\u2082O produced from 0.25 moles of hexane<\/strong><\/h4>\n\n\n\n<p>Using the molar ratio:<br>[<br>\\text{Moles of H\u2082O} = (0.25 \\text{ moles C\u2086H\u2081\u2084}) \\times \\frac{7 \\text{ moles H\u2082O}}{1 \\text{ mole C\u2086H\u2081\u2084}}<br>]<\/p>\n\n\n\n<p>[<br>= 1.75 \\text{ moles H\u2082O}<br>]<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>(c) Moles of O\u2082 consumed in the combustion of 0.40 moles of hexane<\/strong><\/h4>\n\n\n\n<p>Using the molar ratio:<br>[<br>\\text{Moles of O\u2082} = (0.40 \\text{ moles C\u2086H\u2081\u2084}) \\times \\frac{9.5 \\text{ moles O\u2082}}{1 \\text{ mole C\u2086H\u2081\u2084}}<br>]<\/p>\n\n\n\n<p>[<br>= 3.80 \\text{ moles O\u2082}<br>]<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation:<\/strong><\/h3>\n\n\n\n<p>The calculations above rely on <strong>stoichiometry<\/strong>, which is the method of using balanced chemical equations to determine the quantitative relationships between reactants and products.<\/p>\n\n\n\n<p>In every chemical reaction, substances combine in fixed ratios determined by their molecular structure and bonding. The combustion of hexane follows a <strong>specific molar proportion<\/strong>:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>1 mole of hexane (C\u2086H\u2081\u2084) combusts with 9.5 moles of oxygen<\/strong>.<\/li>\n\n\n\n<li><strong>Each mole of hexane produces 6 moles of CO\u2082<\/strong> and <strong>7 moles of water<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p>These values come directly from the coefficients in the balanced equation.<\/p>\n\n\n\n<p>For <strong>part (a)<\/strong>, the given moles of hexane are multiplied by the molar ratio (6 CO\u2082 per 1 C\u2086H\u2081\u2084), resulting in 9.00 moles of CO\u2082.<\/p>\n\n\n\n<p>For <strong>part (b)<\/strong>, a similar approach is used with water, applying the 7:1 ratio to 0.25 moles of hexane to get 1.75 moles of H\u2082O.<\/p>\n\n\n\n<p>For <strong>part (c)<\/strong>, the oxygen consumption is calculated based on the 9.5:1 molar ratio. Multiplying this by 0.40 moles of hexane results in 3.80 moles of O\u2082.<\/p>\n\n\n\n<p>Understanding these concepts is crucial in chemistry, as they are widely used in <strong>energy calculations, fuel efficiency analysis, and reaction predictions<\/strong> in industries such as automotive, aerospace, and environmental sciences.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>1.The equation from the combustion of hexane is:C6H14(g)+19\/2 O2(g) -(arrow sign) 6CO2(g)+7H2O(g)Calculate the number of moles ofa) CO2 produced in the combustion of 1.50 moles of hexaneb) H2O produced in the combustion of 0.25 moles of hexanec) O2 consumed in the combustion of 0.40 moles of hexane The Correct Answer and Explanation is : Solution: [&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-191006","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191006","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=191006"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191006\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191006"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191006"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191006"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}