{"id":32970,"date":"2025-06-23T04:48:07","date_gmt":"2025-06-23T04:48:07","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=32970"},"modified":"2025-06-23T04:48:08","modified_gmt":"2025-06-23T04:48:08","slug":"given-the-reaction-4al-3o2-2al2-o3","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/given-the-reaction-4al-3o2-2al2-o3\/","title":{"rendered":"Given the reaction: 4Al + 3O2? 2Al2 O3"},"content":{"rendered":"\n
Given the reaction: 4Al + 3O2 ? 2Al2 O3 What is the total number of moles of aluminum oxide (Al O ) that can be formed when 54 grams of aluminum (Al) reacts completely with oxygen? A 1.0 moles B 2.0 moles C 3.0 moles D 4.0 moles<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Correct Answer: B) 2.0 moles<\/strong><\/h3>\n\n\n\n\n\n\n\n
4 moles of Aluminum (Al) produce 2 moles of Aluminum oxide (Al\u2082O\u2083)<\/li>\n<\/ul>\n\n\n\n
Step 1: Calculate moles of Aluminum available<\/strong><\/p>\n\n\n\n
We are given 54 grams of Aluminum. The molar mass of Aluminum (Al) is:<\/p>\n\n\n\n
\n
Al = 27 grams per mole<\/li>\n<\/ul>\n\n\n\n
So, the number of moles of Aluminum:Moles of Al=Mass of AlMolar Mass of Al=54\u2009g27\u2009g\/mol=2\u2009moles\\text{Moles of Al} = \\frac{\\text{Mass of Al}}{\\text{Molar Mass of Al}} = \\frac{54 \\, \\text{g}}{27 \\, \\text{g\/mol}} = 2 \\, \\text{moles}Moles of Al=Molar Mass of AlMass of Al\u200b=27g\/mol54g\u200b=2moles<\/p>\n\n\n\n
Step 2: Find moles of Aluminum oxide formed<\/strong><\/p>\n\n\n\n
According to the balanced equation:<\/p>\n\n\n\n
\n
4 moles of Al produce 2 moles of Al\u2082O\u2083<\/li>\n\n\n\n
So, 2 moles of Al will produce:<\/li>\n<\/ul>\n\n\n\n
2\u2009moles Al\u2082O\u20834\u2009moles Al\u00d72\u2009moles Al=1\u2009mole of Al\u2082O\u2083\\frac{2 \\, \\text{moles Al\u2082O\u2083}}{4 \\, \\text{moles Al}} \\times 2 \\, \\text{moles Al} = 1 \\, \\text{mole of Al\u2082O\u2083}4moles Al2moles Al\u2082O\u2083\u200b\u00d72moles Al=1mole of Al\u2082O\u2083<\/p>\n\n\n\n
Final Answer:<\/strong> 1.0 moles<\/strong> of Aluminum oxide can be formed<\/p>\n\n\n\n\n\n\n\n
Correct Option: A) 1.0 moles<\/strong><\/h3>\n\n\n\n\n\n\n\n
Explanation:<\/h3>\n\n\n\n
In this reaction, Aluminum reacts with Oxygen to form Aluminum oxide, a common product seen when metals oxidize. The stoichiometric relationship tells us that four moles of Aluminum are required to produce two moles of Aluminum oxide.<\/p>\n\n\n\n
First, we calculate how many moles of Aluminum are present by using its molar mass. With 54 grams of Aluminum, this gives us exactly 2 moles of Aluminum.<\/p>\n\n\n\n
Next, applying the mole ratio from the balanced equation, we see that 2 moles of Aluminum will produce exactly 1 mole of Aluminum oxide if all the Aluminum reacts completely with Oxygen.<\/p>\n\n\n\n
Thus, under these ideal conditions with excess Oxygen, 1.0 mole<\/strong> of Aluminum oxide will be formed.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
Given the reaction: 4Al + 3O2? 2Al2O3What is the total number of moles of aluminum oxide (AlO) that can be formed when 54 grams of aluminum (Al) reacts completely with oxygen? A 1.0 moles B 2.0 moles C 3.0 moles D 4.0 moles The Correct Answer and Explanation is: Correct Answer: B) 2.0 moles Step-by-Step […]<\/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-32970","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32970","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=32970"}],"version-history":[{"count":2,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32970\/revisions"}],"predecessor-version":[{"id":32973,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32970\/revisions\/32973"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=32970"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=32970"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=32970"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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<\/figure>\n","protected":false},"excerpt":{"rendered":"