Aluminum reacts with oxygen to produce aluminum

General, Organic, and Biological Chemistry: Structures of Life | 4th Edition | ISBN: 9780321750891 | Authors: Karen Timberlake

Problem 61QP Chapter 6

General, Organic, and Biological Chemistry: Structures of Life | 4th Edition

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General, Organic, and Biological Chemistry: Structures of Life | 4th Edition | ISBN: 9780321750891 | Authors: Karen Timberlake

General, Organic, and Biological Chemistry: Structures of Life | 4th Edition

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Problem 61QP

Aluminum reacts with oxygen to produce aluminum oxide.4Al(s) + 3O2(g) ? 2Al2O3(s)Calculate the mass of Al2O3 that can be produced if the reaction of 50.0 g of aluminum and sufficient oxygen has a 75.0% yield.

Step-by-Step Solution:

Solution 61 QPHere, we have to calculate the actual yield of Aluminium oxide.Step 1 of 4The given chemical reaction is as follows.4Al(s) + 3O2(g) 2Al2O3(s)Given:Mass of Aluminium - 50.0 gOxygen yield - 75%.______________________________________________________________________________Step 2 of 4Let’s write the equality and conversion factors for molar mass and mole of Aluminium and Al2O3Molar mass of Aluminium = 27.0g/mole1 mole of Al = 27.0 g of AlSo, the conversion factors are and Molar mass of Al2O3 = 102.0g/mole.1 mole of Al2O3 = 102.0 g/mole1 mole of Al2O3 = 102 g of Al2O3.So, the conversion factors are and Let’s write the equality and conversion factors for Al2O3 and Al.From the equation,4 moles of Al = 2 moles of Al2O3So, the conversion factors are and ____________________________________________________________________________

Step 3 of 4

Chapter 6, Problem 61QP is Solved
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Textbook: General, Organic, and Biological Chemistry: Structures of Life
Edition: 4
Author: Karen Timberlake
ISBN: 9780321750891

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