Consider a bipropellant rocket using a fuel-oxidizer combination for which the stoichiometric reaction is written 2H2 + O2 2H2O. The actual reaction occurring in the chamber is a fuel-rich one, written 5H2 + O2 ? 2H2O + 3H2. Determine the corresponding equivalence ratio f for this case and show that the use of either moles or masses for this calculation yields the same value of f.<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To determine the equivalence ratio (denoted ( f )) for this fuel-rich reaction, we need to understand the difference between the stoichiometric and actual reaction and use it to find the ratio of fuel-to-oxidizer.<\/p>\n\n\n\n The stoichiometric reaction for hydrogen and oxygen is: The actual reaction provided in the problem is fuel-rich: The equivalence ratio ( f ) compares the actual fuel-to-oxidizer ratio to the stoichiometric fuel-to-oxidizer ratio.<\/p>\n\n\n\n The equivalence ratio ( f ) is defined as: To calculate the equivalence ratio using masses, we use the molar masses of hydrogen and oxygen. The molar mass of hydrogen (H\u2082) is 2 g\/mol, and the molar mass of oxygen (O\u2082) is 32 g\/mol.<\/p>\n\n\n\n The equivalence ratio ( f ) by mass is: In both cases, whether we use moles or masses, the equivalence ratio ( f ) is 2.5. This confirms that the use of moles or masses in the calculation leads to the same result.<\/p>\n","protected":false},"excerpt":{"rendered":" Consider a bipropellant rocket using a fuel-oxidizer combination for which the stoichiometric reaction is written 2H2 + O2 2H2O. The actual reaction occurring in the chamber is a fuel-rich one, written 5H2 + O2 ? 2H2O + 3H2. Determine the corresponding equivalence ratio f for this case and show that the use of either moles […]<\/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-197577","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197577","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=197577"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197577\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=197577"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=197577"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=197577"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Understand the Stoichiometric Reaction<\/h3>\n\n\n\n
[
2H_2 + O_2 \\rightarrow 2H_2O
]
In this case, 2 moles of hydrogen (H\u2082) react with 1 mole of oxygen (O\u2082) to form 2 moles of water (H\u2082O).<\/p>\n\n\n\nStep 2: The Actual Fuel-Rich Reaction<\/h3>\n\n\n\n
[
5H_2 + O_2 \\rightarrow 2H_2O + 3H_2
]
Here, 5 moles of hydrogen react with 1 mole of oxygen, but unlike the stoichiometric reaction, additional hydrogen (3 moles) is produced.<\/p>\n\n\n\nStep 3: Equivalence Ratio<\/h3>\n\n\n\n
For moles:<\/h4>\n\n\n\n
\n
[
f = \\frac{\\text{Actual fuel-to-oxidizer ratio}}{\\text{Stoichiometric fuel-to-oxidizer ratio}}
]
So,
[
f = \\frac{5}{2} = 2.5
]<\/p>\n\n\n\nFor masses:<\/h4>\n\n\n\n
\n
[
\\text{Fuel mass} = 2 \\text{ mol H}_2 \\times 2 \\text{ g\/mol} = 4 \\text{ g of H}_2
]
[
\\text{Oxidizer mass} = 1 \\text{ mol O}_2 \\times 32 \\text{ g\/mol} = 32 \\text{ g of O}_2
]
Thus, the stoichiometric fuel-to-oxidizer ratio by mass is ( \\frac{4}{32} = \\frac{1}{8} ).<\/li>\n\n\n\n
[
\\text{Fuel mass} = 5 \\text{ mol H}_2 \\times 2 \\text{ g\/mol} = 10 \\text{ g of H}_2
]
[
\\text{Oxidizer mass} = 1 \\text{ mol O}_2 \\times 32 \\text{ g\/mol} = 32 \\text{ g of O}_2
]
Thus, the actual fuel-to-oxidizer ratio by mass is ( \\frac{10}{32} = \\frac{5}{16} ).<\/li>\n<\/ul>\n\n\n\n
[
f = \\frac{\\frac{5}{16}}{\\frac{1}{8}} = 2.5
]<\/p>\n\n\n\nConclusion:<\/h3>\n\n\n\n