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 240: 7.60d

Answer

149 g of $O_2$

Work Step by Step

1. Calculate the molar mass of $H_2O$: $H: 1.008g * 2= 2.016g $ $O: 16.00g$ 2.016g + 16.00g = 18.02g $ \frac{1 mole (H_2O)}{ 18.02g (H_2O)}$ and $ \frac{ 18.02g (H_2O)}{1 mole (H_2O)}$ 2. The balanced reaction is: $2H_2S(g) + 3O_2(g) --\gt 2SO_2(g) + 2H_2O(g)$ According to the coefficients, the ratio of $H_2O$ to $O_2$ is 2 to 3: $ \frac{ 3 moles(O_2)}{ 2 moles (H_2O)}$ and $ \frac{ 2 moles (H_2O)}{ 3 moles(O_2)}$ 3. Calculate the molar mass for $O_2$: $O: 16.00g * 2= 32.00g $ $ \frac{1 mole (O_2)}{ 32.00g (O_2)}$ and $ \frac{ 32.00g (O_2)}{1 mole (O_2)}$ 4. Use the conversion factors to find the mass of $O_2$ $55.8g(H_2O) \times \frac{1 mole(H_2O)}{ 18.02g( H_2O)} \times \frac{ 3 moles(O_2)}{ 2 moles (H_2O)} \times \frac{ 32.00 g (O_2)}{ 1 mole (O_2)} = 149g (O_2)$
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