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Sagot :
To solve this problem, let’s go through each part step-by-step:
### a. Determining the Limiting Reagent
Given data:
- Mass of CuSO₄ (Copper Sulfate): [tex]\(2.548 \, \text{g}\)[/tex]
- Mass of Fe (Iron): [tex]\(3.094 \, \text{g}\)[/tex]
We need their molar masses:
- Molar mass of CuSO₄ = [tex]\(159.609 \, \text{g/mol}\)[/tex]
- Molar mass of Fe = [tex]\(55.845 \, \text{g/mol}\)[/tex]
First, we calculate the moles of each reactant:
1. Moles of CuSO₄:
[tex]\[ \text{Moles of CuSO}_4 = \frac{\text{Mass of CuSO}_4}{\text{Molar mass of CuSO}_4} = \frac{2.548 \, \text{g}}{159.609 \, \text{g/mol}} \approx 0.015964 \, \text{mol} \][/tex]
2. Moles of Fe:
[tex]\[ \text{Moles of Fe} = \frac{\text{Mass of Fe}}{\text{Molar mass of Fe}} = \frac{3.094 \, \text{g}}{55.845 \, \text{g/mol}} \approx 0.055403 \, \text{mol} \][/tex]
The balanced chemical equation [tex]\( \text{CuSO}_4 (\text{aq}) + \text{Fe} (\text{s}) \rightarrow \text{Cu} (\text{s}) + \text{FeSO}_4 (\text{aq}) \)[/tex] shows a 1:1 molar ratio between CuSO₄ and Fe.
To find the limiting reagent, we compare the number of moles of each reactant. The reactant with fewer moles will limit the reaction:
[tex]\[ \text{Moles of CuSO}_4 \approx 0.015964 \, \text{mol} < \text{Moles of Fe} \approx 0.055403 \, \text{mol} \][/tex]
Since the moles of CuSO₄ are less, it is the limiting reagent.
### b. Determining the Theoretical Yield
The limiting reagent determines the maximum amount of product that can be formed. Since CuSO₄ is the limiting reagent with approximately [tex]\(0.015964 \, \text{mol}\)[/tex], it will dictate the theoretical yield of Cu (Copper).
Using the molar mass of Cu (63.546 g/mol):
[tex]\[ \text{Theoretical yield of Cu} = \text{Moles of limiting reagent} \times \text{Molar mass of Cu} \][/tex]
[tex]\[ \text{Theoretical yield of Cu} = 0.015964 \, \text{mol} \times 63.546 \, \text{g/mol} \approx 1.014 \, \text{g} \][/tex]
### c. Calculating the Percent Yield
Given that the actual mass of Cu obtained is [tex]\(0.784 \, \text{g}\)[/tex]:
[tex]\[ \text{Percent Yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100 \][/tex]
[tex]\[ \text{Percent Yield} = \left( \frac{0.784 \, \text{g}}{1.014 \, \text{g}} \right) \times 100 \approx 77.283\% \][/tex]
### Summary:
a. The limiting reagent is CuSO₄.
b. The theoretical yield of Cu is approximately [tex]\(1.014 \, \text{g}\)[/tex].
c. The percent yield is approximately [tex]\(77.283\%\)[/tex].
### a. Determining the Limiting Reagent
Given data:
- Mass of CuSO₄ (Copper Sulfate): [tex]\(2.548 \, \text{g}\)[/tex]
- Mass of Fe (Iron): [tex]\(3.094 \, \text{g}\)[/tex]
We need their molar masses:
- Molar mass of CuSO₄ = [tex]\(159.609 \, \text{g/mol}\)[/tex]
- Molar mass of Fe = [tex]\(55.845 \, \text{g/mol}\)[/tex]
First, we calculate the moles of each reactant:
1. Moles of CuSO₄:
[tex]\[ \text{Moles of CuSO}_4 = \frac{\text{Mass of CuSO}_4}{\text{Molar mass of CuSO}_4} = \frac{2.548 \, \text{g}}{159.609 \, \text{g/mol}} \approx 0.015964 \, \text{mol} \][/tex]
2. Moles of Fe:
[tex]\[ \text{Moles of Fe} = \frac{\text{Mass of Fe}}{\text{Molar mass of Fe}} = \frac{3.094 \, \text{g}}{55.845 \, \text{g/mol}} \approx 0.055403 \, \text{mol} \][/tex]
The balanced chemical equation [tex]\( \text{CuSO}_4 (\text{aq}) + \text{Fe} (\text{s}) \rightarrow \text{Cu} (\text{s}) + \text{FeSO}_4 (\text{aq}) \)[/tex] shows a 1:1 molar ratio between CuSO₄ and Fe.
To find the limiting reagent, we compare the number of moles of each reactant. The reactant with fewer moles will limit the reaction:
[tex]\[ \text{Moles of CuSO}_4 \approx 0.015964 \, \text{mol} < \text{Moles of Fe} \approx 0.055403 \, \text{mol} \][/tex]
Since the moles of CuSO₄ are less, it is the limiting reagent.
### b. Determining the Theoretical Yield
The limiting reagent determines the maximum amount of product that can be formed. Since CuSO₄ is the limiting reagent with approximately [tex]\(0.015964 \, \text{mol}\)[/tex], it will dictate the theoretical yield of Cu (Copper).
Using the molar mass of Cu (63.546 g/mol):
[tex]\[ \text{Theoretical yield of Cu} = \text{Moles of limiting reagent} \times \text{Molar mass of Cu} \][/tex]
[tex]\[ \text{Theoretical yield of Cu} = 0.015964 \, \text{mol} \times 63.546 \, \text{g/mol} \approx 1.014 \, \text{g} \][/tex]
### c. Calculating the Percent Yield
Given that the actual mass of Cu obtained is [tex]\(0.784 \, \text{g}\)[/tex]:
[tex]\[ \text{Percent Yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100 \][/tex]
[tex]\[ \text{Percent Yield} = \left( \frac{0.784 \, \text{g}}{1.014 \, \text{g}} \right) \times 100 \approx 77.283\% \][/tex]
### Summary:
a. The limiting reagent is CuSO₄.
b. The theoretical yield of Cu is approximately [tex]\(1.014 \, \text{g}\)[/tex].
c. The percent yield is approximately [tex]\(77.283\%\)[/tex].
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