Calculate the theoretical air fuel ratio for the combustion of octane and ethanol. If 150% theoretical air and octane are burned to complete combustion, specify the equivalence ratio for the reaction.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the theoretical air-fuel ratio (AFR) for the combustion of octane (C\u2088H\u2081\u2088) and ethanol (C\u2082H\u2085OH), we first need to write the balanced combustion reactions for both fuels.<\/p>\n\n\n\n The balanced combustion reaction for octane is:<\/p>\n\n\n\n [ From this, we can calculate the theoretical air-fuel ratio. The molecular weight of octane is 114 g\/mol, and the molecular weight of oxygen (O\u2082) is 32 g\/mol.<\/p>\n\n\n\n For 2 moles of octane (2 \u00d7 114 = 228 g), 25 moles of oxygen are required (25 \u00d7 32 = 800 g). Since air contains approximately 21% oxygen by mass, the mass of air required is calculated as:<\/p>\n\n\n\n [ Therefore, the theoretical air-fuel ratio for octane is:<\/p>\n\n\n\n [ The balanced combustion reaction for ethanol is:<\/p>\n\n\n\n [ Using the same method as for octane, the molecular weight of ethanol is 46 g\/mol, and oxygen is 32 g\/mol. For 1 mole of ethanol (46 g), 3 moles of oxygen are required (3 \u00d7 32 = 96 g). The mass of air required is:<\/p>\n\n\n\n [ Therefore, the theoretical air-fuel ratio for ethanol is:<\/p>\n\n\n\n [ The equivalence ratio (( \\phi )) is the ratio of the actual fuel-to-air ratio to the stoichiometric fuel-to-air ratio. It can be defined as:<\/p>\n\n\n\n [ If 150% theoretical air is used (meaning 1.5 times the theoretical amount of air), the equivalence ratio can be calculated as:<\/p>\n\n\n\n For octane:<\/p>\n\n\n\n [ Thus, the equivalence ratio for octane is approximately 0.67<\/strong>, which indicates a lean mixture (more air than necessary for complete combustion).<\/p>\n\n\n\n In summary:<\/p>\n\n\n\n This demonstrates the difference in air-fuel ratios for the combustion of different fuels and how the equivalence ratio is affected by the air supply.<\/p>\n","protected":false},"excerpt":{"rendered":" Calculate the theoretical air fuel ratio for the combustion of octane and ethanol. If 150% theoretical air and octane are burned to complete combustion, specify the equivalence ratio for the reaction. The Correct Answer and Explanation is : To calculate the theoretical air-fuel ratio (AFR) for the combustion of octane (C\u2088H\u2081\u2088) and ethanol (C\u2082H\u2085OH), we […]<\/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-189065","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189065","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=189065"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189065\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=189065"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=189065"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=189065"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Combustion of Octane (C\u2088H\u2081\u2088):<\/h3>\n\n\n\n
2C_8H_{18} + 25O_2 \\rightarrow 16CO_2 + 18H_2O
]<\/p>\n\n\n\n
\\text{Mass of air} = \\frac{800}{0.21} \\approx 3809.52 \\, \\text{g}
]<\/p>\n\n\n\n
\\text{AFR (octane)} = \\frac{3809.52}{228} \\approx 16.7 \\, \\text{(mass of air per unit mass of fuel)}
]<\/p>\n\n\n\nCombustion of Ethanol (C\u2082H\u2085OH):<\/h3>\n\n\n\n
C_2H_5OH + 3O_2 \\rightarrow 2CO_2 + 3H_2O
]<\/p>\n\n\n\n
\\text{Mass of air} = \\frac{96}{0.21} \\approx 457.14 \\, \\text{g}
]<\/p>\n\n\n\n
\\text{AFR (ethanol)} = \\frac{457.14}{46} \\approx 9.95 \\, \\text{(mass of air per unit mass of fuel)}
]<\/p>\n\n\n\nEquivalence Ratio:<\/h3>\n\n\n\n
\\phi = \\frac{\\text{Actual fuel-to-air ratio}}{\\text{Theoretical fuel-to-air ratio}}
]<\/p>\n\n\n\n
\\phi_{\\text{octane}} = \\frac{1}{1.5} \\approx 0.67
]<\/p>\n\n\n\n\n