Answer :
To determine the percentage yield given the reaction:
[tex]\[ \text{Al}_2\text{O}_3(\text{s}) + 6\text{HCl}(\text{aq}) \rightarrow 2\text{AlCl}_3(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \][/tex]
Let's break down the problem step-by-step.
Step 1: Calculate moles of [tex]\(\text{Al}_2\text{O}_3\)[/tex]
Given:
- Mass of [tex]\(\text{Al}_2\text{O}_3\)[/tex] = 16 g
- Molar mass of [tex]\(\text{Al}_2\text{O}_3\)[/tex] ≈ 101.96 g/mol
[tex]\[ \text{Moles of Al}_2\text{O}_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{16 \text{ g}}{101.96 \text{ g/mol}} \approx 0.1569 \text{ mol} \][/tex]
Step 2: Calculate moles of HCl
Given:
- Volume of HCl = 250 mL = 0.250 L
- Molarity of HCl = 5.00 M
[tex]\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume} = 5.00 \text{ M} \times 0.250 \text{ L} = 1.25 \text{ mol} \][/tex]
Step 3: Determine the limiting reactant
From the balanced chemical equation, 1 mole of [tex]\(\text{Al}_2\text{O}_3\)[/tex] reacts with 6 moles of HCl to produce 2 moles of [tex]\(\text{AlCl}_3\)[/tex].
Let's verify the limiting reactant:
- Moles of HCl required for the available moles of [tex]\(\text{Al}_2\text{O}_3\)[/tex] (which is 0.1569 mol):
[tex]\[ \text{Required moles of HCl} = 0.1569 \text{ mol } \times 6 = 0.9415 \text{ mol} \][/tex]
Since we have 1.25 moles of HCl (which is greater than 0.9415 moles), [tex]\(\text{Al}_2\text{O}_3\)[/tex] is the limiting reactant.
Step 4: Calculate the moles of [tex]\(\text{AlCl}_3\)[/tex] produced
From the stoichiometry of the balanced equation:
1 mole of [tex]\(\text{Al}_2\text{O}_3\)[/tex] produces 2 moles of [tex]\(\text{AlCl}_3\)[/tex].
So, the moles of [tex]\(\text{AlCl}_3\)[/tex] produced are:
[tex]\[ \text{Moles of AlCl}_3 = 0.1569 \text{ mol of Al}_2\text{O}_3 \times 2 = 0.3138 \text{ mol} \][/tex]
Step 5: Calculate the theoretical yield of [tex]\(\text{AlCl}_3\)[/tex]
Given:
- Molar mass of [tex]\(\text{AlCl}_3\)[/tex] ≈ 133.34 g/mol
[tex]\[ \text{Theoretical yield} = \text{Moles of AlCl}_3 \times \text{Molar mass of AlCl}_3 = 0.3138 \text{ mol} \times 133.34 \text{ g/mol} \approx 41.85 \text{ g} \][/tex]
Step 6: Calculate the percentage yield
Given:
- Actual yield of [tex]\(\text{AlCl}_3\)[/tex] recovered = 15.3 g
[tex]\[ \text{Percentage yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100 = \left( \frac{15.3 \text{ g}}{41.85 \text{ g}} \right) \times 100 \approx 36.56\% \][/tex]
Thus, the percentage yield is approximately [tex]\(36.56\%\)[/tex].
[tex]\[ \text{Al}_2\text{O}_3(\text{s}) + 6\text{HCl}(\text{aq}) \rightarrow 2\text{AlCl}_3(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \][/tex]
Let's break down the problem step-by-step.
Step 1: Calculate moles of [tex]\(\text{Al}_2\text{O}_3\)[/tex]
Given:
- Mass of [tex]\(\text{Al}_2\text{O}_3\)[/tex] = 16 g
- Molar mass of [tex]\(\text{Al}_2\text{O}_3\)[/tex] ≈ 101.96 g/mol
[tex]\[ \text{Moles of Al}_2\text{O}_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{16 \text{ g}}{101.96 \text{ g/mol}} \approx 0.1569 \text{ mol} \][/tex]
Step 2: Calculate moles of HCl
Given:
- Volume of HCl = 250 mL = 0.250 L
- Molarity of HCl = 5.00 M
[tex]\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume} = 5.00 \text{ M} \times 0.250 \text{ L} = 1.25 \text{ mol} \][/tex]
Step 3: Determine the limiting reactant
From the balanced chemical equation, 1 mole of [tex]\(\text{Al}_2\text{O}_3\)[/tex] reacts with 6 moles of HCl to produce 2 moles of [tex]\(\text{AlCl}_3\)[/tex].
Let's verify the limiting reactant:
- Moles of HCl required for the available moles of [tex]\(\text{Al}_2\text{O}_3\)[/tex] (which is 0.1569 mol):
[tex]\[ \text{Required moles of HCl} = 0.1569 \text{ mol } \times 6 = 0.9415 \text{ mol} \][/tex]
Since we have 1.25 moles of HCl (which is greater than 0.9415 moles), [tex]\(\text{Al}_2\text{O}_3\)[/tex] is the limiting reactant.
Step 4: Calculate the moles of [tex]\(\text{AlCl}_3\)[/tex] produced
From the stoichiometry of the balanced equation:
1 mole of [tex]\(\text{Al}_2\text{O}_3\)[/tex] produces 2 moles of [tex]\(\text{AlCl}_3\)[/tex].
So, the moles of [tex]\(\text{AlCl}_3\)[/tex] produced are:
[tex]\[ \text{Moles of AlCl}_3 = 0.1569 \text{ mol of Al}_2\text{O}_3 \times 2 = 0.3138 \text{ mol} \][/tex]
Step 5: Calculate the theoretical yield of [tex]\(\text{AlCl}_3\)[/tex]
Given:
- Molar mass of [tex]\(\text{AlCl}_3\)[/tex] ≈ 133.34 g/mol
[tex]\[ \text{Theoretical yield} = \text{Moles of AlCl}_3 \times \text{Molar mass of AlCl}_3 = 0.3138 \text{ mol} \times 133.34 \text{ g/mol} \approx 41.85 \text{ g} \][/tex]
Step 6: Calculate the percentage yield
Given:
- Actual yield of [tex]\(\text{AlCl}_3\)[/tex] recovered = 15.3 g
[tex]\[ \text{Percentage yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100 = \left( \frac{15.3 \text{ g}}{41.85 \text{ g}} \right) \times 100 \approx 36.56\% \][/tex]
Thus, the percentage yield is approximately [tex]\(36.56\%\)[/tex].
