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: 7.54b

Answer

$0.604$ g of $H_2$

Work Step by Step

1. Calculate the molar mass of $N_2$: $N: 14.01g * 2= 28.02g $ $ \frac{1 mole (N_2)}{ 28.02g (N_2)}$ and $ \frac{ 28.02g (N_2)}{1 mole (N_2)}$ 2. The balanced reaction is: $N_2 + 3H_2 --\gt 2NH_3$ According to the coefficients, the ratio of $N_2$ to $H_2$ is 1 to 3: $ \frac{ 3 moles(H_2)}{ 1 mole (N_2)}$ and $ \frac{ 1 mole (N_2)}{ 3 moles(H_2)}$ 3. Calculate the molar mass for $H_2$: $H: 1.008g * 2= 2.016g $ $ \frac{1 mole (H_2)}{ 2.016g (H_2)}$ and $ \frac{ 2.016g (H_2)}{1 mole (H_2)}$ 4. Use the conversion factors to find the mass of $H_2$ $2.80g(N_2) \times \frac{1 mole(N_2)}{ 28.02g( N_2)} \times \frac{ 3 moles(H_2)}{ 1 mole (N_2)} \times \frac{ 2.016 g (H_2)}{ 1 mole (H_2)} = 0.604g (H_2)$
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