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To determine the percent yield of lead(II) chloride (PbCl₂), let's go through a detailed, step-by-step solution.
1. Calculate the moles of lead(II) nitrate (Pb(NO₃)₂) used:
Given:
- Mass of Pb(NO₃)₂ = 870 grams
- Molar mass of Pb(NO₃)₂ = 331.2 g/mol
Moles of Pb(NO₃)₂ = (Mass of Pb(NO₃)₂) / (Molar mass of Pb(NO₃)₂)
[tex]\[ \text{Moles of Pb(NO₃)₂} = \frac{870 \text{ g}}{331.2 \text{ g/mol}} \approx 2.6268115942028984 \text{ mol} \][/tex]
2. Determine the moles of PbCl₂ produced:
From the balanced chemical equation:
[tex]\[ \text{Pb(NO₃)₂} + 2 \text{HCl} \rightarrow \text{PbCl₂} + 2 \text{HNO₃} \][/tex]
The molar ratio of Pb(NO₃)₂ to PbCl₂ is 1:1. Therefore, the moles of PbCl₂ produced is the same as the moles of Pb(NO₃)₂ used.
[tex]\[ \text{Moles of PbCl₂ produced} = 2.6268115942028984 \text{ mol} \][/tex]
3. Calculate the theoretical yield of PbCl₂:
Given:
- Molar mass of PbCl₂ = 278.1 g/mol
Theoretical yield of PbCl₂ = (Moles of PbCl₂) * (Molar mass of PbCl₂)
[tex]\[ \text{Theoretical yield of PbCl₂} = 2.6268115942028984 \text{ mol} \times 278.1 \text{ g/mol} \approx 730.5163043478261 \text{ grams} \][/tex]
4. Calculate the percent yield:
Given:
- Actual yield of PbCl₂ = 650 grams
- Theoretical yield of PbCl₂ ≈ 730.5163043478261 grams
Percent yield = [tex]\(\left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100\)[/tex]
[tex]\[ \text{Percent yield} = \left( \frac{650 \text{ grams}}{730.5163043478261 \text{ grams}} \right) \times 100 \approx 88.97816463936317\% \][/tex]
5. Round the percent yield to two significant figures:
Final rounded percent yield: 88.98%
Thus, the percent yield of lead(II) chloride is 88.98%.
So, the correct answer is:
The percent yield of lead chloride is [tex]\(\boxed{88.98}\)[/tex] %.
1. Calculate the moles of lead(II) nitrate (Pb(NO₃)₂) used:
Given:
- Mass of Pb(NO₃)₂ = 870 grams
- Molar mass of Pb(NO₃)₂ = 331.2 g/mol
Moles of Pb(NO₃)₂ = (Mass of Pb(NO₃)₂) / (Molar mass of Pb(NO₃)₂)
[tex]\[ \text{Moles of Pb(NO₃)₂} = \frac{870 \text{ g}}{331.2 \text{ g/mol}} \approx 2.6268115942028984 \text{ mol} \][/tex]
2. Determine the moles of PbCl₂ produced:
From the balanced chemical equation:
[tex]\[ \text{Pb(NO₃)₂} + 2 \text{HCl} \rightarrow \text{PbCl₂} + 2 \text{HNO₃} \][/tex]
The molar ratio of Pb(NO₃)₂ to PbCl₂ is 1:1. Therefore, the moles of PbCl₂ produced is the same as the moles of Pb(NO₃)₂ used.
[tex]\[ \text{Moles of PbCl₂ produced} = 2.6268115942028984 \text{ mol} \][/tex]
3. Calculate the theoretical yield of PbCl₂:
Given:
- Molar mass of PbCl₂ = 278.1 g/mol
Theoretical yield of PbCl₂ = (Moles of PbCl₂) * (Molar mass of PbCl₂)
[tex]\[ \text{Theoretical yield of PbCl₂} = 2.6268115942028984 \text{ mol} \times 278.1 \text{ g/mol} \approx 730.5163043478261 \text{ grams} \][/tex]
4. Calculate the percent yield:
Given:
- Actual yield of PbCl₂ = 650 grams
- Theoretical yield of PbCl₂ ≈ 730.5163043478261 grams
Percent yield = [tex]\(\left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100\)[/tex]
[tex]\[ \text{Percent yield} = \left( \frac{650 \text{ grams}}{730.5163043478261 \text{ grams}} \right) \times 100 \approx 88.97816463936317\% \][/tex]
5. Round the percent yield to two significant figures:
Final rounded percent yield: 88.98%
Thus, the percent yield of lead(II) chloride is 88.98%.
So, the correct answer is:
The percent yield of lead chloride is [tex]\(\boxed{88.98}\)[/tex] %.
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