Question

TType 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:

2HCl + Pb(NO3)2 → 2HNO3 + PbCl2.

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

The percent yield of lead chloride is
%.

Answer 1

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%**.

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