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 7.8 - Energy in Chemical Reactions - Challenge Questions - Page 250: 7.103a

Answer

$242$ g of glucose $(C_6H_{12}O_6)$ are required to form 124 g of ethanol.

Work Step by Step

1. Calculate the molar mass of $C_2H_6O$: $C: 12.01g * 2= 24.02g $ $H: 1.008g * 6= 6.048g $ $O: 16.00g$ 24.02g + 6.048g + 16.00g = 46.07g $ \frac{1 mole (C_2H_6O)}{ 46.07g (C_2H_6O)}$ and $ \frac{ 46.07g (C_2H_6O)}{1 mole (C_2H_6O)}$ 2. The balanced reaction is: $C_6H_{12}O_6 --\gt 2C_2H_6O + 2CO_2$ According to the coefficients, the ratio of $C_2H_6O$ to $C_6H_{12}O_6$ is 2 to 1: $ \frac{ 1 mole(C_6H_{12}O_6)}{ 2 moles (C_2H_6O)}$ and $ \frac{ 2 moles (C_2H_6O)}{ 1 mole(C_6H_{12}O_6)}$ 3. Calculate the molar mass for $C_6H_{12}O_6$: $C: 12.01g * 6= 72.06g $ $H: 1.008g * 12= 12.10g $ $O: 16.00g * 6= 96.00g $ 72.06g + 12.10g + 96.00g = 180.16g $ \frac{1 mole (C_6H_{12}O_6)}{ 180.156g (C_6H_{12}O_6)}$ and $ \frac{ 180.156g (C_6H_{12}O_6)}{1 mole (C_6H_{12}O_6)}$ 4. Use the conversion factors to find the mass of $C_6H_{12}O_6$ $124g(C_2H_6O) \times \frac{1 mole(C_2H_6O)}{ 46.07g( C_2H_6O)} \times \frac{ 1 mole(C_6H_{12}O_6)}{ 2 moles (C_2H_6O)} \times \frac{ 180.16 g (C_6H_{12}O_6)}{ 1 mole (C_6H_{12}O_6)} = 242g (C_6H_{12}O_6)$
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