The chemical equation for the reaction of aqueous lead(II) nitrate with aqueous sodium chloride is given below. $\text{Pb(NO}_3)_2\,(aq) + 2\text{NaCl}\,(aq) \rightarrow \text{PbCl}_2\,(s) + 2\text{NaNO}_3\,(aq)$ When $100\,\text{cm}^3$ of aqueous lead(II) nitrate at a concentration of $0.1\,\text{mol dm}^{-3}$ is treated with excess aqueous sodium chloride, what mass of lead(II) chloride is formed?
- A$1.16\,\text{g}$
- B$2.42\,\text{g}$
- C$2.78\,\text{g}$
- D$3.31\,\text{g}$