Using the balanced chemical equation below, calculate the mass of lead$($II$)$ iodide, $Pb{I}_2$ produced by the reaction of $67\mathrm{.}0g$ of lead$($II$)$ nitrate, $Pb(N{O}_3{)}_2,$ and $53\mathrm{.}5g$ of potassium iodide, $KI.$

$2KI(aq)+Pb(N{O}_3{)}_2(aq)Pb{I}_2(s)+2KN{O}_3(aq)$

The molar mass of $Kl$ is $165\mathrm{.}998\frac{g}{m}ol,$ the molar mass of $Pb(N{O}_3{)}_2$ is $331\mathrm{.}21\frac{g}{m}ol,$ and the molar mass of $Pb{I}_2$ is $461\frac{g}{m}ol.$

$7\mathrm{.}4\mathrm{.}3g$

$93\mathrm{.}3g$

$297g$

$0\mathrm{.}161g$