Chemistry (7th Edition)

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

Chapter 3 - Mass Relationships in Chemical Reactions - Section Problems - Page 107: 55

Answer

(a) 0.034 g (b) 0.31 g (c) 0.43 g

Work Step by Step

(a) Molar mass (Sodium) = 22.99 g/mol $0.0015 mol \times \frac{22.99g}{1mol} = 0.034$ g (b) Molar mass (Lead) = 207.2 g/mol $0.0015 mol \times \frac{207.2g}{1mol} = 0.31$ g (c) Molar Mass ($C_{16}H_{13}ClN_2O$): 12.01* 16 + 1.008* 13 + 35.45* 1 + 14.01* 2 + 16.00* 1 = 284.7 g/mol $0.0015 mol \times \frac{284.7g}{1mol} = 0.43$ g
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