To solve this problem, we need to find the percent yield of lead(II) chloride (PbCl₂) from the given reaction. Here's a step-by-step solution:
### Step 1: Determine the molar masses
- Molar mass of lead(II) nitrate (Pb(NO₃)₂):
- Pb: 207.2 g/mol
- N: 14.0 g/mol × 2 = 28.0 g/mol
- O: 16.0 g/mol × 6 = 96.0 g/mol
- Total = 207.2 + 28.0 + 96.0 = 331.2 g/mol
- Molar mass of lead(II) chloride (PbCl₂):
- Pb: 207.2 g/mol
- Cl: 35.5 g/mol × 2 = 71.0 g/mol
- Total = 207.2 + 71.0 = 278.2 g/mol
### Step 2: Calculate the moles of lead(II) nitrate (Pb(NO₃)₂) used
- Lead(II) nitrate initial mass: 870 grams
- Molar mass of Pb(NO₃)₂: 331.2 g/mol
[tex]\[
\text{Moles of Pb(NO₃)₂} = \frac{870 \text{ g}}{331.2 \text{ g/mol}} \approx 2.63 \text{ mol}
\][/tex]
### Step 3: Using stoichiometry to find moles of PbCl₂ produced
- According to the balanced equation, 1 mole of Pb(NO₃)₂ produces 1 mole of PbCl₂.
- Therefore, moles of PbCl₂ = moles of Pb(NO₃)₂ = 2.63 mol
### Step 4: Calculate the theoretical yield of PbCl₂
- Moles of PbCl₂: 2.63 mol
- Molar mass of PbCl₂: 278.2 g/mol
[tex]\[
\text{Theoretical yield of PbCl₂} = 2.63 \text{ mol} \times 278.2 \text{ g/mol} \approx 731.66 \text{ g}
\][/tex]
### Step 5: Calculate the percent yield
- Actual yield of PbCl₂: 650 grams
- Theoretical yield of PbCl₂: 731.66 grams
[tex]\[
\text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100 = \left( \frac{650 \text{ g}}{731.66 \text{ g}} \right) \times 100 \approx 88.8\%
\][/tex]
To express the answer to two significant figures:
The percent yield of lead chloride is [tex]\( \boxed{89} \% \)[/tex].
