Answer
The theorectical yield of product for that initial amount of reactants is equal to 31.1 g $AlCl_3$.
Work Step by Step
1. Calculate the number of moles of $Al$:
26.98* 1 = 26.98g/mol
$7.5g \times \frac{1 mol}{ 26.98g} = 0.278mol (Al)$
- 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:
$0.278 mol (Al) \times \frac{ 2 mol(AlCl_3)}{ 2 mol (Al)} = 0.278mol (AlCl_3)$
2. Calculate the number of moles of $Cl_2$:
35.45* 2 = 70.90g/mol
$24.8g \times \frac{1 mol}{ 70.90g} = 0.350mol (Cl_2)$
- 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:
$0.350 mol (Cl_2) \times \frac{ 2 mol(AlCl_3)}{ 3 mol (Cl_2)} = 0.233mol (AlCl_3)$
Since the number of moles of $AlCl_3$ produced by $Cl_2$ is lower, it is the limiting reactant, so that reaction will produce: $ 0.233$ moles of $AlCl_3$.
3. Calculate the mass of $AlCl_3$:
26.98* 1 + 35.45* 3 = 133.33g/mol
$0.233 mol \times \frac{ 133.33 g}{ 1 mol} = 31.1g (AlCl_3)$
