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: 63c

Answer

2.40 g of $O_2$

Work Step by Step

1. Calculate the number of moles of $Fe$: 55.85* 1 = 55.85g/mol $5.58g \times \frac{1 mol}{ 55.85g} = 0.100mol (Fe)$ - The balanced reaction is: $4Fe + 3O_2 -- \gt 2Fe_2O_3$ The ratio of $Fe$ to $O_2$ is 4 to 3: $0.100 mol (Fe) \times \frac{ 3 mol(O_2)}{ 4 mol (Fe)} = 0.0750mol (O_2)$ 2. Calculate the mass of $O_2$: 16.00* 2 = 32.00g/mol $0.0750 mol \times \frac{ 32.00 g}{ 1 mol} = 2.40g (O_2)$
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