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: 71b


$188g (H_2)$ can be produced.

Work Step by Step

As we determined on the last exercise, $CH_4$ is the limiting reactant, so the calculations are based on its mass: 1. Calculate the number of moles of $CH_4$: 12.01* 1 + 1.007* 4 = 16.05g/mol $500g \times \frac{1 mol}{ 16.05g} = 31.17mol (CH_4)$ - The balanced reaction is: $CH_4 + H_2O -- \gt CO + 3H_2$ The ratio of $CH_4$ to $H_2$ is 1 to 3: $31.17 mol (CH_4) \times \frac{ 3 mol(H_2)}{ 1 mol (CH_4)} = 93.5mol (H_2)$ 2. Calculate the mass of $H_2$: 1.007* 2 = 2.014g/mol $93.5 mol \times \frac{ 2.014 g}{ 1 mol} = 188g (H_2)$
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