Question

For the following chemical reaction, how many moles of lead (II) iodide will be produced from 136 g of potassium iodide?

2Kl +Pb(NO3)2 ---> PbI2 + 2KNO3

Answer 1

Answer on Question #40468 - Chemistry - Other

Question

For the following chemical reaction, how many moles of lead (II) iodide will be produced from 136 g of potassium iodide?

$$2 \mathrm{KI} + \mathrm{Pb}(\mathrm{NO}_3)_2 \rightarrow \mathrm{PbI}_2 + 2\mathrm{KNO}_3$$

Answer:

Number of moles equals:

$$n = \frac{m}{M}$$

m – Mass of KI, m = 136 g.

M – Molar mass of KI, equals:

$$M = M(K) + M(I) = 39.1 + 126.9 = 166.0 \frac{g}{mole}$$

Then number of moles in 136 g of KI equals:

$$n = \frac{136}{166} = 0.819 \text{ mol}$$

According to the reaction:

- 2 mol of KI produces 1 mol of $\mathrm{PbI}_2$ (lead (II) iodide)

- 0.819 mol of KI – x moles of $\mathrm{PbI}_2$

So, the number of moles of lead (II) iodide produced from 136 g of potassium iodide equals:

$$x = \frac{0.819 \cdot 1}{2} = 0.41 \text{ mol}$$

Answer: $\mathrm{n}(\mathrm{PbI}_2) = 0.41$ mol.