Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Get quick and reliable answers to your questions from a dedicated community of professionals on our platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
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].
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.