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: 7.105e

Answer

17.0 g of $Al_2O_3$ can be produced.

Work Step by Step

1. Calculate the molar mass of $O_2$: $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)}$ 2. The balanced reaction is: $4Al(s) + 3O_2(g) --\gt 2Al_2O_3$ According to the coefficients, the ratio of $O_2$ to $Al_2O_3$ is 3 to 2: $ \frac{ 2 moles(Al_2O_3)}{ 3 moles (O_2)}$ and $ \frac{ 3 moles (O_2)}{ 2 moles(Al_2O_3)}$ 3. Calculate the molar mass for $Al_2O_3$: $Al: 26.98g * 2= 53.96g $ $O: 16.00g * 3= 48.00g $ 53.96g + 48.00g = 101.96g $ \frac{1 mole (Al_2O_3)}{ 101.96g (Al_2O_3)}$ and $ \frac{ 101.96g (Al_2O_3)}{1 mole (Al_2O_3)}$ 4. Use the conversion factors to find the mass of $Al_2O_3$ $8.00g(O_2) \times \frac{1 mole(O_2)}{ 32.00g( O_2)} \times \frac{ 2 moles(Al_2O_3)}{ 3 moles (O_2)} \times \frac{ 101.96 g (Al_2O_3)}{ 1 mole (Al_2O_3)} = 17.0g (Al_2O_3)$
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