{"id":244420,"date":"2025-07-05T06:14:17","date_gmt":"2025-07-05T06:14:17","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=244420"},"modified":"2025-07-05T06:14:19","modified_gmt":"2025-07-05T06:14:19","slug":"hexane-c6h14-burns-in-air-o2-to-give-co2-and-h2o","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/05\/hexane-c6h14-burns-in-air-o2-to-give-co2-and-h2o\/","title":{"rendered":"Hexane (C6H14) burns in air (O2) to give CO2 and H2O"},"content":{"rendered":"\n
Hexane (C6H14) burns in air (O2) to give CO2 and H2O. a) Write the balanced equation for this reaction. b) If 215.0 g of hexane is mixed with 215.0 g of oxygen, is hexane or oxygen the limiting reactant? Show work. c) What mass of excess reactant remains at the end of the reaction? d) How many grams of carbon dioxide are produced? I already solved the equation for the problem: 2C6H14 + 19O2 -> 12CO2 + 14H2O. I need help with parts B-D.<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Let’s work through parts B-D based on the balanced chemical equation you\u2019ve already written: 2C6H14+19O2\u219212CO2+14H2O2C_6H_{14} + 19O_2 \\rightarrow 12CO_2 + 14H_2O2C6\u200bH14\u200b+19O2\u200b\u219212CO2\u200b+14H2\u200bO<\/p>\n\n\n\n
Part B: Identifying the Limiting Reactant<\/strong><\/h3>\n\n\n\n
To find out which reactant is limiting, we\u2019ll need to compare the mole ratios of hexane and oxygen based on the amount of each reactant you have. Here are the steps:<\/p>\n\n\n\n
\n
Calculate the moles of hexane (C6H14):<\/strong>\n
\n
Molar mass of hexane (C6H14): 6\u00d712.01+14\u00d71.008=86.18\u2009g\/mol6 \\times 12.01 + 14 \\times 1.008 = 86.18 \\, \\text{g\/mol}6\u00d712.01+14\u00d71.008=86.18g\/mol<\/li>\n\n\n\n
Compare the ratios from the balanced equation:<\/strong>\n
\n
From the equation, the mole ratio of C6H14 to O2 is 2:19. That means for every 2 moles of hexane, you need 19 moles of oxygen.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
Determine the amount of oxygen needed for 2.49 moles of hexane:<\/strong>
Oxygen needed: 192\u00d72.49=23.71\u2009mol\u00a0O2\\frac{19}{2} \\times 2.49 = 23.71 \\, \\text{mol O2}219\u200b\u00d72.49=23.71mol\u00a0O2<\/li><\/ul>Since we only have 6.72 moles of oxygen available, oxygen is the limiting reactant.<\/strong><\/li>\n<\/ol>\n\n\n\n
Part C: Mass of Excess Reactant Remaining<\/strong><\/h3>\n\n\n\n
Now that we know oxygen is the limiting reactant, we can determine how much hexane is left over:<\/p>\n\n\n\n
\n
Find the moles of hexane that will react with the available oxygen:<\/strong>\n
\n
From the equation, 19 moles of oxygen react with 2 moles of hexane. So, for 6.72 moles of oxygen: 219\u00d76.72=0.71\u2009mol\u00a0C6H14\\frac{2}{19} \\times 6.72 = 0.71 \\, \\text{mol C6H14}192\u200b\u00d76.72=0.71mol\u00a0C6H14<\/li>\n<\/ul>\n<\/li>\n\n\n\n
Calculate the mass of hexane that reacts:<\/strong>\n
\n
Mass of hexane reacted: 0.71\u2009mol\u00d786.18\u2009g\/mol=61.2\u2009g0.71 \\, \\text{mol} \\times 86.18 \\, \\text{g\/mol} = 61.2 \\, \\text{g}0.71mol\u00d786.18g\/mol=61.2g<\/li>\n<\/ul>\n<\/li>\n\n\n\n
Determine the mass of excess hexane:<\/strong>\n
\n
Initial mass of hexane = 215.0 g<\/li>\n\n\n\n
Mass of hexane reacted = 61.2 g<\/li>\n\n\n\n
Mass of excess hexane = 215.0\u2009g\u221261.2\u2009g=153.8\u2009g215.0 \\, \\text{g} – 61.2 \\, \\text{g} = 153.8 \\, \\text{g}215.0g\u221261.2g=153.8g<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n
Thus, 153.8 g of hexane remains<\/strong> after the reaction.<\/p>\n\n\n\n
Part D: Mass of Carbon Dioxide Produced<\/strong><\/h3>\n\n\n\n
To find the mass of carbon dioxide produced, we can use the amount of limiting reactant (oxygen) and the stoichiometry of the reaction:<\/p>\n\n\n\n
\n
Calculate the moles of CO2 produced:<\/strong>\n
\n
From the equation, 19 moles of O2 produce 12 moles of CO2. So, for 6.72 moles of oxygen: 1219\u00d76.72=4.25\u2009mol\u00a0CO2\\frac{12}{19} \\times 6.72 = 4.25 \\, \\text{mol CO2}1912\u200b\u00d76.72=4.25mol\u00a0CO2<\/li>\n<\/ul>\n<\/li>\n\n\n\n
Calculate the mass of CO2 produced:<\/strong>\n
\n
Molar mass of CO2 = 44.01 g\/mol<\/li>\n\n\n\n
Mass of CO2 produced: 4.25\u2009mol\u00d744.01\u2009g\/mol=187.0\u2009g4.25 \\, \\text{mol} \\times 44.01 \\, \\text{g\/mol} = 187.0 \\, \\text{g}4.25mol\u00d744.01g\/mol=187.0g<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n
Therefore, 187.0 g of CO2<\/strong> is produced in the reaction.<\/p>\n\n\n\n
Mass of excess reactant remaining<\/strong>: 153.8 g of hexane<\/li>\n\n\n\n
Mass of CO2 produced<\/strong>: 187.0 g<\/li>\n<\/ul>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
Hexane (C6H14) burns in air (O2) to give CO2 and H2O. a) Write the balanced equation for this reaction. b) If 215.0 g of hexane is mixed with 215.0 g of oxygen, is hexane or oxygen the limiting reactant? Show work. c) What mass of excess reactant remains at the end of the reaction? d) […]<\/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-244420","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244420","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=244420"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244420\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=244420"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=244420"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=244420"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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