{"id":231794,"date":"2025-06-11T11:44:31","date_gmt":"2025-06-11T11:44:31","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=231794"},"modified":"2025-06-11T11:44:33","modified_gmt":"2025-06-11T11:44:33","slug":"the-combustion-of-liquid-octane-c8h18-in-the-presence-of-gaseous-oxygen-yields-carbon-dioxide-and-liquid-water","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/11\/the-combustion-of-liquid-octane-c8h18-in-the-presence-of-gaseous-oxygen-yields-carbon-dioxide-and-liquid-water\/","title":{"rendered":"The combustion of liquid octane, C8H18, in the presence of gaseous oxygen yields carbon dioxide and liquid water"},"content":{"rendered":"\n
The combustion of liquid octane, C8H18, in the presence of gaseous oxygen yields carbon dioxide and liquid water. How many moles of water are produced by the reaction of 1.80 moles of octane? ————–mol H2O<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Let\u2019s go step by step carefully.<\/p>\n\n\n\n
1. Write the balanced chemical equation<\/strong><\/h3>\n\n\n\n
The combustion of octane (C\u2088H\u2081\u2088) can be written as: C8H18(l)+O2(g)\u2192CO2(g)+H2O(l)\\text{C}_8\\text{H}_{18}(l) + \\text{O}_2(g) \\rightarrow \\text{CO}_2(g) + \\text{H}_2\\text{O}(l)C8\u200bH18\u200b(l)+O2\u200b(g)\u2192CO2\u200b(g)+H2\u200bO(l)<\/p>\n\n\n\n
Now we balance it.<\/p>\n\n\n\n
\n
Each octane molecule has 8 carbon atoms \u2192 we need 8 CO\u2082 molecules.<\/li>\n\n\n\n
Each octane molecule has 18 hydrogen atoms \u2192 we need 9 H\u2082O molecules (since each water molecule has 2 hydrogens: 18 \u00f7 2 = 9).<\/li>\n\n\n\n
Total oxygen atoms needed = 16 + 9 = 25.<\/li>\n\n\n\n
Therefore, we need 25\/2 = 12.5 O\u2082 molecules.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n
To eliminate the fraction, multiply the entire equation by 2: 2C8H18(l)+25O2(g)\u219216CO2(g)+18H2O(l)2\\text{C}_8\\text{H}_{18}(l) + 25\\text{O}_2(g) \\rightarrow 16\\text{CO}_2(g) + 18\\text{H}_2\\text{O}(l)2C8\u200bH18\u200b(l)+25O2\u200b(g)\u219216CO2\u200b(g)+18H2\u200bO(l)<\/p>\n\n\n\n
2. Use the mole ratio<\/strong><\/h3>\n\n\n\n
From the balanced equation: 2 moles of C\u2088H\u2081\u2088 produce 18 moles of H\u2082O.<\/p>\n\n\n\n
Thus, the mole ratio is: 18 mol H2O2 mol C8H18=9 mol H2O per mol C8H18\\frac{18\\ \\text{mol H}_2\\text{O}}{2\\ \\text{mol C}_8\\text{H}_{18}} = 9\\ \\text{mol H}_2\\text{O per mol C}_8\\text{H}_{18}2 mol C8\u200bH18\u200b18 mol H2\u200bO\u200b=9 mol H2\u200bO per mol C8\u200bH18\u200b<\/p>\n\n\n\n
3. Calculate moles of water produced<\/strong><\/h3>\n\n\n\n
Given: 1.80 moles of C\u2088H\u2081\u2088.<\/p>\n\n\n\n
4. Explanation (300 words like a textbook)<\/strong><\/h3>\n\n\n\n
The combustion of hydrocarbons is a chemical reaction where the hydrocarbon reacts with oxygen to produce carbon dioxide and water. In this problem, the hydrocarbon is octane (C\u2088H\u2081\u2088), which is commonly found in gasoline.<\/p>\n\n\n\n
To solve stoichiometry problems involving combustion, the first step is to write a balanced chemical equation. For octane, the unbalanced reaction is: C8H18+O2\u2192CO2+H2O\\text{C}_8\\text{H}_{18} + \\text{O}_2 \\rightarrow \\text{CO}_2 + \\text{H}_2\\text{O}C8\u200bH18\u200b+O2\u200b\u2192CO2\u200b+H2\u200bO<\/p>\n\n\n\n
Balancing carbon first, there are 8 carbon atoms in one molecule of octane, requiring 8 CO\u2082 molecules. Next, balancing hydrogen, 18 hydrogen atoms require 9 H\u2082O molecules. Finally, balancing oxygen, we find that 8 CO\u2082 molecules contribute 16 oxygen atoms and 9 H\u2082O molecules contribute 9 oxygen atoms, totaling 25 oxygen atoms. Since each O\u2082 molecule has 2 oxygen atoms, we require 12.5 O\u2082 molecules. To avoid fractions, we multiply the entire equation by 2: 2C8H18+25O2\u219216CO2+18H2O2\\text{C}_8\\text{H}_{18} + 25\\text{O}_2 \\rightarrow 16\\text{CO}_2 + 18\\text{H}_2\\text{O}2C8\u200bH18\u200b+25O2\u200b\u219216CO2\u200b+18H2\u200bO<\/p>\n\n\n\n
The balanced equation shows that 2 moles of octane produce 18 moles of water, establishing a mole ratio of 9:1 (9 moles of H\u2082O per 1 mole of C\u2088H\u2081\u2088). Given 1.80 moles of octane, we multiply by this ratio: 1.80\u00d79=16.2 moles of water1.80 \\times 9 = 16.2 \\text{ moles of water}1.80\u00d79=16.2 moles of water<\/p>\n\n\n\n
Thus, the complete combustion of 1.80 moles of octane produces 16.2 moles of water<\/strong>.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
The combustion of liquid octane, C8H18, in the presence of gaseous oxygen yields carbon dioxide and liquid water. How many moles of water are produced by the reaction of 1.80 moles of octane? ————–mol H2O The Correct Answer and Explanation is: Let\u2019s go step by step carefully. 1. Write the balanced chemical equation The combustion […]<\/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-231794","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231794","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=231794"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231794\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=231794"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=231794"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=231794"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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