What volume of butane (C4H10) can be produced at STP, from the reaction of 13.45 g of carbon with 17.65 L of hydrogen gas at STP?
Which reactant is in excess and how much of this reactant is left over?
4 C + 5 H2 C4H10<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n To determine the volume of butane ((C_4H_{10})) produced from the reaction between carbon and hydrogen gas, we start with the balanced chemical equation:<\/p>\n\n\n\n [ 4C + 5H_2 \\rightarrow C_4H_{10} ]<\/p>\n\n\n\n Moles of Carbon:<\/strong> [ Moles of Hydrogen:<\/strong> [ From the balanced equation:<\/p>\n\n\n\n Calculate the amount of hydrogen needed to react with the available carbon:<\/p>\n\n\n\n [ We have only 0.79 moles of (H_2) available, while we need 1.40 moles to react with all the carbon. Therefore, (H_2) is the limiting reactant.<\/p>\n\n\n\n From the stoichiometry of the reaction, 5 moles of (H_2) produce 1 mole of (C_4H_{10}). Thus, the moles of butane produced from 0.79 moles of (H_2) are:<\/p>\n\n\n\n [ At STP, 1 mole of gas occupies 22.4 L, so:<\/p>\n\n\n\n [ We started with 1.12 moles of (C). To find out how much carbon is left over after reacting with the available hydrogen:<\/p>\n\n\n\n [ Excess carbon:<\/p>\n\n\n\n [ In grams, this is:<\/p>\n\n\n\n [ The reaction produces approximately 3.54 L<\/strong> of butane, with hydrogen<\/strong> as the limiting reactant. Approximately 5.86 g<\/strong> of carbon is left unreacted.<\/p>\n","protected":false},"excerpt":{"rendered":" What volume of butane (C4H10) can be produced at STP, from the reaction of 13.45 g of carbon with 17.65 L of hydrogen gas at STP?Which reactant is in excess and how much of this reactant is left over?4 C + 5 H2 C4H10 The Correct Answer and Explanation is: To determine the volume of […]<\/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-155973","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/155973","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=155973"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/155973\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=155973"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=155973"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=155973"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Calculate moles of reactants<\/h3>\n\n\n\n
The molar mass of carbon ((C)) is approximately 12.01 g\/mol. To find the moles of carbon in 13.45 g:<\/p>\n\n\n\n
\\text{Moles of } C = \\frac{\\text{mass}}{\\text{molar mass}} = \\frac{13.45 \\, \\text{g}}{12.01 \\, \\text{g\/mol}} \\approx 1.12 \\, \\text{mol}
]<\/p>\n\n\n\n
The molar mass of hydrogen gas ((H_2)) is approximately 2.02 g\/mol. To find the moles of hydrogen in 17.65 L at STP (where 1 mole of gas occupies 22.4 L):<\/p>\n\n\n\n
\\text{Moles of } H_2 = \\frac{17.65 \\, \\text{L}}{22.4 \\, \\text{L\/mol}} \\approx 0.79 \\, \\text{mol}
]<\/p>\n\n\n\nStep 2: Determine the stoichiometry of the reaction<\/h3>\n\n\n\n
\n
\\text{Required } H_2 = 1.12 \\, \\text{mol } C \\times \\frac{5 \\, \\text{mol } H_2}{4 \\, \\text{mol } C} = 1.40 \\, \\text{mol } H_2
]<\/p>\n\n\n\nStep 3: Identify the limiting reactant<\/h3>\n\n\n\n
Step 4: Calculate the volume of butane produced<\/h3>\n\n\n\n
\\text{Moles of } C_4H_{10} = 0.79 \\, \\text{mol } H_2 \\times \\frac{1 \\, \\text{mol } C_4H_{10}}{5 \\, \\text{mol } H_2} = 0.158 \\, \\text{mol } C_4H_{10}
]<\/p>\n\n\n\n
\\text{Volume of } C_4H_{10} = 0.158 \\, \\text{mol} \\times 22.4 \\, \\text{L\/mol} \\approx 3.54 \\, \\text{L}
]<\/p>\n\n\n\nStep 5: Calculate excess reactant<\/h3>\n\n\n\n
\\text{Moles of } C \\text{ reacted} = 0.79 \\, \\text{mol } H_2 \\times \\frac{4 \\, \\text{mol } C}{5 \\, \\text{mol } H_2} = 0.632 \\, \\text{mol } C
]<\/p>\n\n\n\n
\\text{Excess } C = 1.12 \\, \\text{mol} – 0.632 \\, \\text{mol} \\approx 0.488 \\, \\text{mol}
]<\/p>\n\n\n\n
0.488 \\, \\text{mol} \\times 12.01 \\, \\text{g\/mol} \\approx 5.86 \\, \\text{g}
]<\/p>\n\n\n\nConclusion<\/h3>\n\n\n\n