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: 45c

Answer

The theorectical yield of product for that initial amount of reactants is equal to 1.16 g $AlCl_3$.

Work Step by Step

1. Calculate the number of moles of $Al$: 26.98* 1 = 26.98g/mol $0.235g \times \frac{1 mol}{ 26.98g} = 8.71 \times 10^{-3}mol (Al)$ - The balanced reaction is: $2Al(s) + 3Cl_2(g) --\gt 2AlCl_3(s)$ - Therefore: The ratio of $Al$ to $AlCl_3$ is 2 to 2: $8.71 \times 10^{-3} mol (Al) \times \frac{ 2 mol(AlCl_3)}{ 2 mol (Al)} = 8.71 \times 10^{-3}mol (AlCl_3)$ 2. Calculate the number of moles of $Cl_2$: 35.45* 2 = 70.90g/mol $1.15g \times \frac{1 mol}{ 70.90g} = 0.0162mol (Cl_2)$ - The balanced reaction is: $2Al(s) + 3Cl_2(g) --\gt 2AlCl_3(s)$ - Therefore: The ratio of $Cl_2$ to $AlCl_3$ is 3 to 2: $0.0162 mol (Cl_2) \times \frac{ 2 mol(AlCl_3)}{ 3 mol (Cl_2)} = 0.0108mol (AlCl_3)$ Since the number of moles of $AlCl_3$ produced by $Al$ is lower, it is the limiting reactant, so that reaction will produce: $ 0.00871$ moles of $AlCl_3$. 3. Calculate the mass of $AlCl_3$: 26.98* 1 + 35.45* 3 = 133.33g/mol $8.71 \times 10^{-3} mol \times \frac{ 133.33 g}{ 1 mol} = 1.16g (AlCl_3)$
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