answer
78.6
explanation
Sure, I can help you with that.
The first step is to calculate the theoretical yield of lead(II) chloride. The molar mass of lead(II) chloride is 278.10 g/mol, so 650 g of lead(II) chloride is equivalent to 650 / 278.10 = 2.33 moles.
The balanced equation shows that 1 mole of lead(II) nitrate produces 1 mole of lead(II) chloride. So, if 870 g of lead(II) nitrate is used, the theoretical yield of lead(II) chloride is 870 / 278.10 = 3 moles.
The percent yield is calculated as follows:
```
percent yield = actual yield / theoretical yield * 100%
```
In this case, the actual yield is 650 g and the theoretical yield is 3 moles, or 824 g. So, the percent yield is:
```
percent yield = 650 / 824 * 100% = 78.6%
```
To two significant figures, the percent yield is **78.6%**.
Is there anything else I can help you with?