Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 4 - Sections 4.1-4.9 - Exercises - Problems by Topic - Page 187: 46a

Answer

The theorectical yield of the product for that initial amount of reactants is equal to 8.1 g $TiF_4$.

Work Step by Step

1. Calculate the number of moles of $Ti$: 47.87* 1 = 47.87g/mol $5.0g \times \frac{1 mol}{ 47.87g} = 0.10mol (Ti)$ - The balanced reaction is: $Ti(s) + 2F_2(g) -- \gt TiF_4$ - Therefore: The ratio of $Ti$ to $TiF_4$ is 1 to 1: $0.10 mol (Ti) \times \frac{ 1 mol(TiF_4)}{ 1 mol (Ti)} = 0.10mol (TiF_4)$ 2. Calculate the number of moles of $F_2$: 19.00* 2 = 38.00g/mol $5.0g \times \frac{1 mol}{ 38.00g} = 0.13mol (F_2)$ - The balanced reaction is: $Ti(s) + 2F_2(g) -- \gt TiF_4$ - Therefore: The ratio of $F_2$ to $TiF_4$ is 2 to 1: $0.13 mol (F_2) \times \frac{ 1 mol(TiF_4)}{ 2 mol (F_2)} = 0.065mol (TiF_4)$ Since the number of moles of $TiF_4$ produced by $F_2$ is lower, it is the limiting reactant, so that reaction will produce: $ 0.065$ moles of $TiF_4$. 3. Calculate the mass of $TiF_4$: 47.87* 1 + 19.00* 4 = 123.87g/mol $0.065 mol \times \frac{ 123.87 g}{ 1 mol} = 8.1g (TiF_4)$
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