To determine the percent yield of lead(II) chloride, we need to follow these steps:
1. Calculate the number of moles of lead(II) nitrate used in the reaction:
- Molar mass of lead(II) nitrate (Pb(NO₃)₂) = 331.2 g/mol.
- Mass of lead(II) nitrate used = 870 grams.
[tex]\[
\text{Moles of Pb(NO}_3\text{)₂} = \frac{\text{Mass of Pb(NO}_3\text{)₂}}{\text{Molar mass of Pb(NO}_3\text{)₂}} = \frac{870 \text{ g}}{331.2 \text{ g/mol}} = 2.63 \text{ moles}
\][/tex]
2. Determine the theoretical yield of PbCl₂ (lead(II) chloride):
- From the balanced equation, 1 mole of Pb(NO₃)₂ produces 1 mole of PbCl₂.
- Molar mass of PbCl₂ = 278.1 g/mol.
[tex]\[
\text{Theoretical yield of PbCl₂} = \text{Moles of PbCl₂} \times \text{Molar mass of PbCl₂} = 2.63 \text{ moles} \times 278.1 \text{ g/mol} = 730.52 \text{ grams}
\][/tex]
3. Calculate the percent yield:
- Actual yield of PbCl₂ = 650 grams.
- Theoretical yield of PbCl₂ = 730.52 grams.
[tex]\[
\text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100\% = \left( \frac{650 \text{ g}}{730.52 \text{ g}} \right) \times 100\% = 89\%
\][/tex]
The percent yield of lead chloride is [tex]\( 89\% \)[/tex].
