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[tex]\[ Al_2O_3(s) + 6HCl(aq) \rightarrow 2AlCl_3(aq) + 3H_2O(l) \][/tex]

A mixture containing 16 g of aluminum oxide and 250 mL of 5.00 M hydrochloric acid produces aluminum chloride.

Determine the percentage yield if 15.3 g of aluminum chloride is recovered.

Sagot :

GinnyAnswer
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].
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