Type the correct answer in the box. Express your answer to two significant figures.

An industrial vat contains 650 grams of solid lead(II) chloride formed from a reaction of 870 grams of lead(II) nitrate with excess hydrochloric acid. This is the equation of the reaction:

[tex]\[ 2 \text{HCl} + \text{Pb} \left( \text{NO}_3\right)_2 \rightarrow 2 \text{HNO}_3 + \text{PbCl}_2 \][/tex]

What is the percent yield of lead(II) chloride?

The percent yield of lead(II) chloride is [tex]$\square$[/tex] \%.



Answer :

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] %.