Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 3 - Chemical Reactions - Questions for Review and Thought - Topical Questions - Page 149e: 69b


$5.08g (Al_2Cl_6)$

Work Step by Step

As we determined in the last exercise, $Cl_2$ is the limiting reactant, so we have to do the calculations based on its mass. 1. Calculate the number of moles of $Cl_2$: 35.45* 2 = 70.9g/mol $4.05g \times \frac{1 mol}{ 70.9g} = 0.0571mol (Cl_2)$ - The balanced reaction is: $2Al + 3Cl_2 -- \gt Al_2Cl_6$ The ratio of $Cl_2$ to $Al_2Cl_6$ is 3 to 1: $0.0571 mol (Cl_2) \times \frac{ 1 mol(Al_2Cl_6)}{ 3 mol (Cl_2)} = 0.0190mol (Al_2Cl_6)$ 2. Calculate the mass of $Al_2Cl_6$: 26.98* 2 + 35.45* 6 = 266.66g/mol $0.0190 mol \times \frac{ 266.66 g}{ 1 mol} = 5.08g (Al_2Cl_6)$
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