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: 67a


We can obtain 699g of $Fe$ using 1 kg of $Fe_2O_3$

Work Step by Step

1. Calculate the number of moles of $ Fe_2O_3$: 55.85* 2 + 16* 3 = 159.7g/mol $1000g \times \frac{1 mol}{ 159.7g} = 6.26mol (Fe_2O_3)$ The ratio of $Fe_2O_3$ to $Fe$ is 1 to 2 (According to the reaction of the question): $6.26 mol (Fe_2O_3) \times \frac{ 2 mol(Fe)}{ 1 mol (Fe_2O_3)} = 12.52mol (Fe)$ 2. Calculate the mass of $Fe$: 55.85* 1 = 55.85g/mol $12.52 mol \times \frac{ 55.85 g}{ 1 mol} = 699g (Fe)$
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