Gaseous C2H4 reacts with O2according to the following: C2H4(g) + 3O2(g) \u2192 2CO2(g) + 2 H2O(g) What volume of oxygen at STP is needed to react with 1.50 mole of C2H4?<\/p>\n\n\n\n
The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n Correct Answer: 1.50 mol C\u2082H\u2084 \u00d7 (3 mol O\u2082 \/ 1 mol C\u2082H\u2084) \u00d7 (22.4 L \/ 1 mol O\u2082) = 100.8 L of O\u2082 at STP<\/strong><\/p>\n\n\n\n The chemical reaction given is: C2H4(g)+3O2(g)\u21922CO2(g)+2H2O(g)\\text{C}_2\\text{H}_4(g) + 3\\text{O}_2(g) \\rightarrow 2\\text{CO}_2(g) + 2\\text{H}_2\\text{O}(g)<\/p>\n\n\n\n This is a balanced combustion reaction of ethene (C\u2082H\u2084). It shows that 1 mole of ethene reacts with 3 moles of oxygen gas<\/strong> to produce carbon dioxide and water vapor.<\/p>\n\n\n\n In this problem, you’re asked to find the volume of oxygen gas at STP<\/strong> (Standard Temperature and Pressure) needed to react with 1.50 moles of C\u2082H\u2084<\/strong>.<\/p>\n\n\n\n From the balanced equation: 1 mol C2H4:3 mol O21\\ \\text{mol C}_2\\text{H}_4 : 3\\ \\text{mol O}_2<\/p>\n\n\n\n So if you have 1.50 mol of C\u2082H\u2084: 1.50 mol C2H4\u00d73 mol O21 mol C2H4=4.50 mol O21.50\\ \\text{mol C}_2\\text{H}_4 \\times \\frac{3\\ \\text{mol O}_2}{1\\ \\text{mol C}_2\\text{H}_4} = 4.50\\ \\text{mol O}_2<\/p>\n\n\n\n At STP (0\u00b0C and 1 atm), 1 mole of any ideal gas occupies 22.4 liters<\/strong>. 4.50 mol O2\u00d722.4 L1 mol=100.8 L4.50\\ \\text{mol O}_2 \\times \\frac{22.4\\ \\text{L}}{1\\ \\text{mol}} = 100.8\\ \\text{L}<\/p>\n\n\n\n This means that to completely combust 1.50 moles of ethene, you would need 100.8 liters of oxygen gas<\/strong> at standard temperature and pressure.<\/p>\n\n\n\n This calculation illustrates how stoichiometry and molar gas volume principles are applied to real-world chemical reactions. In chemistry, such conversions are crucial in predicting gas volumes in industrial or laboratory settings, ensuring the correct proportions of reactants are used for efficient and safe reactions.<\/p>\n","protected":false},"excerpt":{"rendered":" Gaseous C2H4 reacts with O2according to the following: C2H4(g) + 3O2(g) \u2192 2CO2(g) + 2 H2O(g) What volume of oxygen at STP is needed to react with 1.50 mole of C2H4? The correct answer and explanation is: Correct Answer: 1.50 mol C\u2082H\u2084 \u00d7 (3 mol O\u2082 \/ 1 mol C\u2082H\u2084) \u00d7 (22.4 L \/ 1 […]<\/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-16497","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16497","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=16497"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16497\/revisions"}],"predecessor-version":[{"id":16498,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16497\/revisions\/16498"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=16497"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=16497"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=16497"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nDetailed Explanation (Approx. 300 words):<\/strong><\/h3>\n\n\n\n
\n\n\n\nStep 1: Use the Mole Ratio<\/strong><\/h3>\n\n\n\n
\n\n\n\nStep 2: Convert Moles of O\u2082 to Volume at STP<\/strong><\/h3>\n\n\n\n
\n\n\n\nFinal Answer: 100.8 liters of O\u2082<\/strong><\/h3>\n\n\n\n