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: 63b

Answer

7.98 g of $Fe_2O_3$

Work Step by Step

1. Calculate the number of moles of $Fe$: 55.85* 1 = 55.85g/mol $5.58g \times \frac{1 mol}{ 55.85g} = 0.100mol (Fe)$ - As we determined in the last exercise, the balanced reaction is: $4Fe + 3O_2 -- \gt 2Fe_2O_3$ The ratio of $Fe$ to $Fe_2O_3$ is 4 to 2: $0.100 mol (Fe) \times \frac{ 2 mol(Fe_2O_3)}{ 4 mol (Fe)} = 0.050mol (Fe_2O_3)$ 2. Calculate the mass of $Fe_2O_3$: 55.85* 2 + 16* 3 = 159.7g/mol $0.050 mol \times \frac{ 159.7 g}{ 1 mol} = 7.98g (Fe_2O_3)$
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