{"id":37605,"date":"2025-06-26T03:09:29","date_gmt":"2025-06-26T03:09:29","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=37605"},"modified":"2025-06-26T03:09:30","modified_gmt":"2025-06-26T03:09:30","slug":"consider-the-following-balanced-actual-combustion-reaction-for-propane","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/consider-the-following-balanced-actual-combustion-reaction-for-propane\/","title":{"rendered":"Consider the following balanced actual combustion reaction for propane."},"content":{"rendered":"\n<pre id=\"preorder-ask-header-text\" class=\"wp-block-preformatted\">Consider the following balanced actual combustion reaction for propane.<\/pre>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"518\" height=\"329\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-336.png\" alt=\"\" class=\"wp-image-37606\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-336.png 518w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-336-300x191.png 300w\" sizes=\"auto, (max-width: 518px) 100vw, 518px\" \/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>The correct answer is <strong>A. 50%<\/strong>.<\/p>\n\n\n\n<p>To determine the percent theoretical air, we compare the amount of air actually used in the reaction to the amount required for complete combustion. Here is how we break it down:<\/p>\n\n\n\n<p>The balanced chemical equation shows:<\/p>\n\n\n\n<p>C3H8+16.67&nbsp;(air)\u21924H2O+3CO2+13.17N2\\text{C}_3\\text{H}_8 + 16.67\\ \\text{(air)} \\rightarrow 4\\text{H}_2\\text{O} + 3\\text{CO}_2 + 13.17\\text{N}_2<\/p>\n\n\n\n<p>Air is assumed to be 21 percent oxygen and 79 percent nitrogen by volume. So in 16.67 moles of air, the oxygen provided is:<\/p>\n\n\n\n<p>16.67\u00d70.21=3.5007&nbsp;moles&nbsp;of&nbsp;O216.67 \\times 0.21 = 3.5007\\ \\text{moles of } \\text{O}_2<\/p>\n\n\n\n<p>Now, calculate the theoretical oxygen required. In a complete combustion of propane:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each mole of propane yields 3 moles of CO\u2082 and 4 moles of H\u2082O.<\/li>\n\n\n\n<li>Each CO\u2082 needs 1 mole of O atoms (0.5 O\u2082)<\/li>\n\n\n\n<li>Each H\u2082O needs 1 O atom (0.5 O\u2082)<\/li>\n<\/ul>\n\n\n\n<p>So, total oxygen needed:<\/p>\n\n\n\n<p>3+4=7&nbsp;moles&nbsp;of&nbsp;O23 + 4 = 7\\ \\text{moles of O}_2<\/p>\n\n\n\n<p>Next, find how much air would be required to provide 7 moles of O\u2082:<\/p>\n\n\n\n<p>70.21\u224833.33&nbsp;moles&nbsp;of&nbsp;air\\frac{7}{0.21} \\approx 33.33\\ \\text{moles of air}<\/p>\n\n\n\n<p>Now, the actual air used is 16.67 moles. So, percent theoretical air used is:<\/p>\n\n\n\n<p>(16.6733.33)\u00d7100=50%\\left( \\frac{16.67}{33.33} \\right) \\times 100 = 50\\%<\/p>\n\n\n\n<p>Thus, the percent theoretical air is 50 percent.<\/p>\n\n\n\n<p>This problem highlights the importance of stoichiometry and air-fuel ratio calculations in combustion processes. Engineers use these calculations to optimize fuel efficiency and reduce emissions. Knowing whether combustion is complete or limited by air availability influences equipment design in engines, burners, and environmental control systems.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-930.jpeg\" alt=\"\" class=\"wp-image-37607\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-930.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-930-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-930-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Consider the following balanced actual combustion reaction for propane. The Correct Answer and Explanation is: The correct answer is A. 50%. To determine the percent theoretical air, we compare the amount of air actually used in the reaction to the amount required for complete combustion. Here is how we break it down: The balanced chemical [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-37605","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37605","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=37605"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37605\/revisions"}],"predecessor-version":[{"id":37608,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37605\/revisions\/37608"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=37605"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=37605"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=37605"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}