Chemistry (7th Edition)

Published by Pearson
ISBN 10: 0321943171
ISBN 13: 978-0-32194-317-0

Chapter 3 - Mass Relationships in Chemical Reactions - Worked Example - Page 86: 9

Answer

3.33 g of acetic anhydride.

Work Step by Step

1. Calculate the number of moles of $C_7H_6O_3$: Molar mass: 12.01* 7 + 1.008* 6 + 16.00* 3 = 138.12g/mol $4.50g \times \frac{1 mol}{ 138.12g} = 0.0326mol (C_7H_6O_3)$ Since the given equation is already balanced: The ratio of $C_7H_6O_3$ to $C_4H_6O_3$ is 1 to 1: $0.0326 mol (C_7H_6O_3) \times \frac{ 1 mol(C_4H_6O_3)}{ 1 mol (C_7H_6O_3)} = 0.0326mol (C_4H_6O_3)$ 2. Calculate the mass of $C_4H_6O_3$: Molar mass: 12.01* 4 + 1.008* 6 + 16.00* 3 = 102.09g/mol $0.0326 mol \times \frac{ 102.09 g}{ 1 mol} = 3.33g (C_4H_6O_3)$
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