Chemistry: An Introduction to General, Organic, and Biological Chemistry (12th Edition)

Published by Prentice Hall
ISBN 10: 0321908449
ISBN 13: 978-0-32190-844-5

Chapter 7 - Section 7.7 - Mass Calculations for Reactions - Questions and Problems - Page 239: 7.56c

Answer

$4.20$ g of $Fe$

Work Step by Step

1. Calculate the molar mass of $Fe_2O_3$: $Fe: 55.85g * 2= 111.7g $ $O: 16.00g * 3= 48.00g $ 111.7g + 48.00g = 159.7g $ \frac{1 mole (Fe_2O_3)}{ 159.7g (Fe_2O_3)}$ and $ \frac{ 159.7g (Fe_2O_3)}{1 mole (Fe_2O_3)}$ 2. The balanced reaction is: $Fe_2O_3 + 3C --\gt 2Fe + 3CO$ According to the coefficients, the ratio of $Fe_2O_3$ to $Fe$ is 1 to 2: $ \frac{ 2 moles(Fe)}{ 1 mole (Fe_2O_3)}$ and $ \frac{ 1 mole (Fe_2O_3)}{ 2 moles(Fe)}$ 3. Calculate the molar mass for $Fe$: $Fe: 55.85g$ $ \frac{1 mole (Fe)}{ 55.85g (Fe)}$ and $ \frac{ 55.85g (Fe)}{1 mole (Fe)}$ 4. Use the conversion factors to find the mass of $Fe$ $6.00g(Fe_2O_3) \times \frac{1 mole(Fe_2O_3)}{ 159.7g( Fe_2O_3)} \times \frac{ 2 moles(Fe)}{ 1 moles (Fe_2O_3)} \times \frac{ 55.85 g (Fe)}{ 1 mole (Fe)} = 4.20g (Fe)$
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