a. Balanced Chemical Equation for the Combustion of Ethanol<\/h3>\n\n\n\n
The combustion of ethanol, C\u2082H\u2085OH, occurs when ethanol reacts with oxygen (O\u2082) to produce carbon dioxide (CO\u2082) and water (H\u2082O). The balanced chemical equation is:<\/p>\n\n\n\n
[
C_2H_5OH (l) + 3O_2 (g) \\rightarrow 2CO_2 (g) + 3H_2O (l)
]<\/p>\n\n\n\n
b. Theoretical Yield of CO\u2082 in Grams<\/h3>\n\n\n\nStep 1: Convert the given amount of ethanol (C\u2082H\u2085OH) to moles.<\/h4>\n\n\n\n
We are given that 17.6 mL of ethanol were used, and the density of ethanol is 0.789 g\/mL. First, calculate the mass of ethanol:<\/p>\n\n\n\n
[
\\text{Mass of ethanol} = 17.6 \\, \\text{mL} \\times 0.789 \\, \\text{g\/mL} = 13.9 \\, \\text{g}
]<\/p>\n\n\n\n
Next, convert the mass of ethanol to moles using the molar mass of ethanol (46.1 g\/mol):<\/p>\n\n\n\n
[
\\text{Moles of ethanol} = \\frac{13.9 \\, \\text{g}}{46.1 \\, \\text{g\/mol}} = 0.301 \\, \\text{mol}
]<\/p>\n\n\n\n
Step 2: Determine the moles of CO\u2082 produced.<\/h4>\n\n\n\n
From the balanced chemical equation, we see that 1 mole of ethanol produces 2 moles of CO\u2082. Therefore, the moles of CO\u2082 produced are:<\/p>\n\n\n\n
[
\\text{Moles of CO\u2082} = 2 \\times 0.301 \\, \\text{mol} = 0.602 \\, \\text{mol}
]<\/p>\n\n\n\n
Step 3: Convert moles of CO\u2082 to grams.<\/h4>\n\n\n\n
To find the mass of CO\u2082, use the molar mass of CO\u2082 (44.0 g\/mol):<\/p>\n\n\n\n
[
\\text{Mass of CO\u2082} = 0.602 \\, \\text{mol} \\times 44.0 \\, \\text{g\/mol} = 26.5 \\, \\text{g}
]<\/p>\n\n\n\n
Thus, the theoretical yield of CO\u2082 is 26.5 grams<\/strong>.<\/p>\n\n\n\n From the balanced equation, 1 mole of ethanol reacts with 3 moles of O\u2082. Therefore, the moles of O\u2082 required for the reaction are:<\/p>\n\n\n\n [ The molar mass of O\u2082 is 32.0 g\/mol, so:<\/p>\n\n\n\n [ We are given that 10.5 L of O\u2082 were used. The density of O\u2082 is 1.43 g\/L, so the mass of O\u2082 available is:<\/p>\n\n\n\n [ Since only 15.0 g of O\u2082 were available, and 28.9 g were required, there will be no O\u2082 remaining<\/strong>. Instead, O\u2082 is the limiting reagent<\/strong> in this reaction.<\/p>\n\n\n\n Since the O\u2082 is the limiting reagent, all of the available O\u2082 will be consumed. The reaction requires 0.903 moles of O\u2082, which will react with 0.301 moles of ethanol. We already calculated that the moles of ethanol required are 0.301 mol, which is exactly the amount of ethanol available. Thus, all of the ethanol will be consumed<\/strong>, and there will be no ethanol remaining<\/strong>.<\/p>\n\n\n\n The combustion of ethanol, C2H5OH (molar mass = 46.1 g\/mol), produces CO2 (molar mass = 44.0 g\/mol) and H2O (molar mass = 18.0 g\/mol). The density of ethanol is 0.789 g\/mL, and the density of oxygen gas is 1.43 g\/L. In one experiment, 17.6 mL of ethanol were combined with 10.5 L O2 and allowed […]<\/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-206151","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/206151","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=206151"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/206151\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=206151"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=206151"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=206151"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}c. Grams of O\u2082 that Will Remain<\/h3>\n\n\n\n
Step 1: Determine the moles of O\u2082 that reacted.<\/h4>\n\n\n\n
\\text{Moles of O\u2082 required} = 3 \\times 0.301 \\, \\text{mol} = 0.903 \\, \\text{mol}
]<\/p>\n\n\n\nStep 2: Convert the moles of O\u2082 to grams.<\/h4>\n\n\n\n
\\text{Mass of O\u2082 required} = 0.903 \\, \\text{mol} \\times 32.0 \\, \\text{g\/mol} = 28.9 \\, \\text{g}
]<\/p>\n\n\n\nStep 3: Convert the volume of O\u2082 used to grams.<\/h4>\n\n\n\n
\\text{Mass of O\u2082 available} = 10.5 \\, \\text{L} \\times 1.43 \\, \\text{g\/L} = 15.0 \\, \\text{g}
]<\/p>\n\n\n\nStep 4: Determine the remaining O\u2082.<\/h4>\n\n\n\n
d. Milliliters of Ethanol that Will Remain<\/h3>\n\n\n\n
\n\n\n\nImage Representation of the Chemical Reaction Process<\/h3>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/03\/image-1527.png\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"