{"id":183596,"date":"2025-01-16T17:25:26","date_gmt":"2025-01-16T17:25:26","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=183596"},"modified":"2025-01-16T17:25:29","modified_gmt":"2025-01-16T17:25:29","slug":"given-the-unbalanced-equation-c3h8-o2-co2-h2o","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/16\/given-the-unbalanced-equation-c3h8-o2-co2-h2o\/","title":{"rendered":"Given the unbalanced equation: C3H8 + O2 CO2 + H2O"},"content":{"rendered":"\n
    \n
  1. Given the unbalanced equation: C3H8 + O2 CO2 + H2O a. How many moles of oxygen gas are required to make 4.00 moles of carbon dioxide? b. How many moles of C3H8 must be used to produce 2.00 grams of water?<\/li>\n<\/ol>\n\n\n\n

    The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

    To address the questions based on the unbalanced chemical equation:<\/p>\n\n\n\n

    [ \\text{C}_3\\text{H}_8 + \\text{O}_2 \\rightarrow \\text{CO}_2 + \\text{H}_2\\text{O} ]<\/p>\n\n\n\n

    a. How many moles of oxygen gas are required to make 4.00 moles of carbon dioxide?<\/strong><\/p>\n\n\n\n

    First, balance the chemical equation:<\/p>\n\n\n\n

    [ \\text{C}_3\\text{H}_8 + 5\\text{O}_2 \\rightarrow 3\\text{CO}_2 + 4\\text{H}_2\\text{O} ]<\/p>\n\n\n\n

    From the balanced equation, the stoichiometric ratio between oxygen gas (O\u2082) and carbon dioxide (CO\u2082) is 5:3.<\/p>\n\n\n\n

    This means that for every 3 moles of CO\u2082 produced, 5 moles of O\u2082 are consumed.<\/p>\n\n\n\n

    To find the moles of O\u2082 required for 4.00 moles of CO\u2082:<\/p>\n\n\n\n

    [
    \\text{Moles of O}_2 = \\left( \\frac{5 \\, \\text{moles O}_2}{3 \\, \\text{moles CO}_2} \\right) \\times 4.00 \\, \\text{moles CO}_2 = 6.67 \\, \\text{moles O}_2
    ]<\/p>\n\n\n\n

    Therefore, 6.67 moles of oxygen gas are required to produce 4.00 moles of carbon dioxide.<\/p>\n\n\n\n

    b. How many moles of C\u2083H\u2088 must be used to produce 2.00 grams of water?<\/strong><\/p>\n\n\n\n

    The molar mass of water (H\u2082O) is approximately 18.02 g\/mol.<\/p>\n\n\n\n

    To find the moles of H\u2082O produced from 2.00 grams:<\/p>\n\n\n\n

    [
    \\text{Moles of H}_2\\text{O} = \\frac{2.00 \\, \\text{g}}{18.02 \\, \\text{g\/mol}} \\approx 0.111 \\, \\text{moles H}_2\\text{O}
    ]<\/p>\n\n\n\n

    From the balanced equation, the stoichiometric ratio between propane (C\u2083H\u2088) and water (H\u2082O) is 1:4.<\/p>\n\n\n\n

    This means that for every 1 mole of C\u2083H\u2088, 4 moles of H\u2082O are produced.<\/p>\n\n\n\n

    To find the moles of C\u2083H\u2088 required to produce 0.111 moles of H\u2082O:<\/p>\n\n\n\n

    [
    \\text{Moles of C}_3\\text{H}_8 = \\left( \\frac{1 \\, \\text{mole C}_3\\text{H}_8}{4 \\, \\text{moles H}_2\\text{O}} \\right) \\times 0.111 \\, \\text{moles H}_2\\text{O = 0.0278 moles C}_3\\text{H}_8}
    ]<\/p>\n\n\n\n

    Therefore, approximately 0.0278 moles of C\u2083H\u2088 are required to produce 2.00 grams of water.<\/p>\n\n\n\n

    Explanation:<\/strong><\/p>\n\n\n\n

    Stoichiometry involves using the relationships between reactants and products in a balanced chemical equation to calculate the quantities of substances involved in a reaction.<\/p>\n\n\n\n

    In part (a), the balanced equation indicates that 5 moles of O\u2082 are required to produce 3 moles of CO\u2082.<\/p>\n\n\n\n

    By setting up a proportion, we can determine that 6.67 moles of O\u2082 are needed to produce 4.00 moles of CO\u2082.<\/p>\n\n\n\n

    In part (b), the balanced equation shows that 1 mole of C\u2083H\u2088 produces 4 moles of H\u2082O.<\/p>\n\n\n\n

    By calculating the moles of H\u2082O from its mass and using the stoichiometric ratio, we find that 0.0278 moles of C\u2083H\u2088 are required to produce 2.00 grams of water.<\/p>\n\n\n\n

    These calculations demonstrate how stoichiometric relationships allow chemists to predict the amounts of reactants needed to produce desired amounts of products in chemical reactions.<\/p>\n","protected":false},"excerpt":{"rendered":"

    The Correct Answer and Explanation is : To address the questions based on the unbalanced chemical equation: [ \\text{C}_3\\text{H}_8 + \\text{O}_2 \\rightarrow \\text{CO}_2 + \\text{H}_2\\text{O} ] a. How many moles of oxygen gas are required to make 4.00 moles of carbon dioxide? First, balance the chemical equation: [ \\text{C}_3\\text{H}_8 + 5\\text{O}_2 \\rightarrow 3\\text{CO}_2 + 4\\text{H}_2\\text{O} […]<\/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-183596","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183596","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=183596"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183596\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183596"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183596"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183596"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}