Which reactant in the reaction to create the $Pb{I}_2(s)$ is the limiting reactant?

dotsPb$(N{O}_3{)}_2(aq)+$NaI$(aq)dotsPb{I}_2(s)+dotsNaN{O}_3(aq)$

Given you have $2\mathrm{.}0mL$ of $0\mathrm{.}10MPb(N{O}_3{)}_2$ and $2\mathrm{.}0mL$ of $0\mathrm{.}15$MNal. $($Hint: balance the reaction first.$)$

$NaN{O}_3$ is the limiting reactant.

$Pb{I}_2$ is the limiting reactant.

Nal is the limiting reactant.

$Pb(N{O}_3{)}_2$ is the limiting reactant.