To calculate the mass of oxygen gas (O2) required to completely react with 0.025 g of propane (C3H8), we need to follow a few steps based on stoichiometry.<\/p>\n\n\n\n
Step 1: Write the balanced chemical equation<\/h3>\n\n\n\n
The complete combustion of propane (C3H8) with oxygen (O2) is represented by the following balanced equation:2\u2009C3H8(g)+7\u2009O2(g)\u21926\u2009CO2(g)+8\u2009H2O(g)2 \\, C_3H_8 (g) + 7 \\, O_2 (g) \\rightarrow 6 \\, CO_2 (g) + 8 \\, H_2O (g)2C3\u200bH8\u200b(g)+7O2\u200b(g)\u21926CO2\u200b(g)+8H2\u200bO(g)<\/p>\n\n\n\n
This equation tells us that 2 moles of propane react with 7 moles of oxygen gas.<\/p>\n\n\n\n
Step 2: Convert mass of C3H8 to moles<\/h3>\n\n\n\n
To find the moles of propane, we use its molar mass (44.1 g\/mol):moles of C3H8=mass of C3H8molar mass of C3H8=0.025\u2009g44.1\u2009g\/mol=0.000567\u2009mol C3H8\\text{moles of C}_3H_8 = \\frac{\\text{mass of C}_3H_8}{\\text{molar mass of C}_3H_8} = \\frac{0.025 \\, \\text{g}}{44.1 \\, \\text{g\/mol}} = 0.000567 \\, \\text{mol C}_3H_8moles of C3\u200bH8\u200b=molar mass of C3\u200bH8\u200bmass of C3\u200bH8\u200b\u200b=44.1g\/mol0.025g\u200b=0.000567mol C3\u200bH8\u200b<\/p>\n\n\n\n
Step 3: Use the stoichiometric ratio to find moles of O2 required<\/h3>\n\n\n\n
From the balanced equation, we know that 2 moles of C3H8 react with 7 moles of O2. Using this ratio, we can calculate the moles of O2 required:moles of O2=7\u2009mol O22\u2009mol C3H8\u00d70.000567\u2009mol C3H8=0.00198\u2009mol O2\\text{moles of O}_2 = \\frac{7 \\, \\text{mol O}_2}{2 \\, \\text{mol C}_3H_8} \\times 0.000567 \\, \\text{mol C}_3H_8 = 0.00198 \\, \\text{mol O}_2moles of O2\u200b=2mol C3\u200bH8\u200b7mol O2\u200b\u200b\u00d70.000567mol C3\u200bH8\u200b=0.00198mol O2\u200b<\/p>\n\n\n\n
Step 4: Convert moles of O2 to grams<\/h3>\n\n\n\n
Now, to find the mass of O2 required, we use the molar mass of oxygen gas (32.00 g\/mol):mass of O2=moles of O2\u00d7molar mass of O2=0.00198\u2009mol\u00d732.00\u2009g\/mol=0.0634\u2009g\\text{mass of O}_2 = \\text{moles of O}_2 \\times \\text{molar mass of O}_2 = 0.00198 \\, \\text{mol} \\times 32.00 \\, \\text{g\/mol} = 0.0634 \\, \\text{g}mass of O2\u200b=moles of O2\u200b\u00d7molar mass of O2\u200b=0.00198mol\u00d732.00g\/mol=0.0634g<\/p>\n\n\n\n
Final Answer:<\/h3>\n\n\n\n
The mass of O2 required to completely react with 0.025 g of C3H8 is 0.0634 g<\/strong>.<\/p>\n\n\n\n This solution uses stoichiometric principles to relate the mass of one substance (propane) to the mass of another (oxygen). The key steps involve converting mass to moles, using the stoichiometric coefficients from the balanced chemical equation, and then converting moles of oxygen back to mass. Understanding these conversions allows us to solve for the mass of oxygen required for the reaction to proceed completely.<\/p>\n\n\n\n The molar mass of oxygen gas (O2) is 32.00 g\/mol. The molar mass of C3H8 is 44.1 g\/mol. What mass of O2, in grams, is required to completely react with 0.025 g C3H8 The Correct Answer and Explanation is: To calculate the mass of oxygen gas (O2) required to completely react with 0.025 g 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-249635","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/249635","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=249635"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/249635\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=249635"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=249635"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=249635"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Explanation:<\/h3>\n\n\n\n
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