{"id":32982,"date":"2025-06-23T04:58:35","date_gmt":"2025-06-23T04:58:35","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=32982"},"modified":"2025-06-23T04:58:37","modified_gmt":"2025-06-23T04:58:37","slug":"4al-3o2-2al2o3-the-molar-mass-of-o2-is-32-0-g-mol","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/4al-3o2-2al2o3-the-molar-mass-of-o2-is-32-0-g-mol\/","title":{"rendered":"4Al + 3O2 —> 2Al2O3 The molar mass of O2 is 32.0 g\/mol."},"content":{"rendered":"\n

4Al + 3O2 —> 2Al2O3 The molar mass of O2 is 32.0 g\/mol. What mass, in grams, of O2 must react to form 3.80 mol of Al2O3? A.) 60.8 B.) 81.1 C.) 122 D.) 182<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Correct Answer: D) 182 grams<\/strong><\/h3>\n\n\n\n

Step-by-Step Solution:<\/h3>\n\n\n\n

The balanced chemical equation is:<\/p>\n\n\n\n

4Al + 3O\u2082 \u2192 2Al\u2082O\u2083<\/strong><\/p>\n\n\n\n

Step 1: Find the mole ratio between O\u2082 and Al\u2082O\u2083<\/strong><\/p>\n\n\n\n

From the equation:<\/p>\n\n\n\n

    \n
  • 3 moles of O\u2082 produce 2 moles of Al\u2082O\u2083<\/li>\n<\/ul>\n\n\n\n

    So, to find how many moles of O\u2082 are needed to produce 3.80 moles of Al\u2082O\u2083, set up the ratio:3 moles O\u20822 moles Al\u2082O\u2083=x moles O\u20823.80 moles Al\u2082O\u2083\\frac{3 \\text{ moles O\u2082}}{2 \\text{ moles Al\u2082O\u2083}} = \\frac{x \\text{ moles O\u2082}}{3.80 \\text{ moles Al\u2082O\u2083}}2 moles Al\u2082O\u20833 moles O\u2082\u200b=3.80 moles Al\u2082O\u2083x moles O\u2082\u200b<\/p>\n\n\n\n

    Step 2: Solve for x<\/strong>x=3\u00d73.802=5.70 moles of O\u2082x = \\frac{3 \\times 3.80}{2} = 5.70 \\text{ moles of O\u2082}x=23\u00d73.80\u200b=5.70 moles of O\u2082<\/p>\n\n\n\n

    Step 3: Find the mass of O\u2082 required<\/strong><\/p>\n\n\n\n

    The molar mass of O\u2082 is given as 32.0 g\/mol. So:Mass of O\u2082=5.70 moles\u00d732.0 g\/mol=182.4 grams\\text{Mass of O\u2082} = 5.70 \\text{ moles} \\times 32.0 \\text{ g\/mol} = 182.4 \\text{ grams}Mass of O\u2082=5.70 moles\u00d732.0 g\/mol=182.4 grams<\/p>\n\n\n\n

    Rounded to three significant figures, the mass is 182 grams<\/strong>, which matches option D<\/strong>.<\/p>\n\n\n\n

    \n\n\n\n

    Explanation:<\/h3>\n\n\n\n

    This problem involves stoichiometric calculations based on a balanced chemical reaction. The reaction provided shows how aluminum reacts with oxygen to form aluminum oxide. The key to solving this is understanding the mole ratios from the balanced equation.<\/p>\n\n\n\n

    In this equation, 4 moles of aluminum react with 3 moles of oxygen gas to produce 2 moles of aluminum oxide. The coefficients in the equation indicate the proportions in which substances react and are produced. Using these ratios, one can determine how much of one reactant is needed to produce a given amount of product.<\/p>\n\n\n\n

    We are asked to find the mass of oxygen gas needed to produce 3.80 moles of aluminum oxide. From the equation, every 2 moles of aluminum oxide require 3 moles of oxygen gas. Using simple cross multiplication, we calculated that 3.80 moles of aluminum oxide need 5.70 moles of oxygen gas.<\/p>\n\n\n\n

    Next, using the molar mass of oxygen gas (32.0 g\/mol), we convert moles of oxygen to grams. The calculation shows that 5.70 moles of oxygen corresponds to 182.4 grams, which rounds to 182 grams. This matches option D.<\/p>\n\n\n\n

    Stoichiometry like this is essential for predicting the amounts of substances needed or produced in chemical reactions, allowing chemists to plan and control reactions efficiently.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    4Al + 3O2 —> 2Al2O3 The molar mass of O2 is 32.0 g\/mol. What mass, in grams, of O2 must react to form 3.80 mol of Al2O3? A.) 60.8 B.) 81.1 C.) 122 D.) 182 The Correct Answer and Explanation is: Correct Answer: D) 182 grams Step-by-Step Solution: The balanced chemical equation is: 4Al + […]<\/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-32982","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32982","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=32982"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32982\/revisions"}],"predecessor-version":[{"id":32984,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32982\/revisions\/32984"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=32982"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=32982"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=32982"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}