Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Discover comprehensive solutions to your questions from a wide network of experts on our user-friendly platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
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] %.
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.