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: 58

Answer

166.8 kg

Work Step by Step

1. Calculate the number of moles of $Ti$: 47.87* 1 = 47.87g/mol $100.0 kg \times \frac{1 mol}{ 47.87g} = 2.089 \times 10^{3}mol (Ti)$ Since each $TiO_2$ has 1 $Ti$ molecule: The ratio of $Ti$ to $TiO_2$ is 1 to 1: $2.089 \times 10^{3} mol (Ti) \times \frac{ 1 mol(TiO_2)}{ 1 mol (Ti)} = 2.089mol \times 10^{3} (TiO_2)$ 2. Calculate the mass of $TiO_2$: 47.87* 1 + 16.00* 2 = 79.87g/mol $2.089 \times 10^3mol \times \frac{ 79.87 g}{ 1 mol} = 166.8 \times 10^3 g (TiO_2) = 166.8 kg$
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