\(Pb \left( NO _{3}\right)_{2}\) reacts with both \(Br\)
\(\ln rxn T\).
\begin{aligned}
&\operatorname{In} rXn I . \\
& Pb \left( NO _{3}\right)_{2(a q)}+2 NaBr _{(\text {aq })} \rightarrow PbBr _{2(0)}+2 NaNO _{\text {(3a) }} \\
&\eta_{P b\left( NO _{3}\right)_{2}}: \eta_{ NaBr }=1: 2 \\
&\text { i.e. } \eta_{P b\left(N O_{3}\right)_{2}}=\frac{\eta_{ NaBr }}{2}
\end{aligned}
\begin{aligned}
\eta_{ NaBr } &=0.20 mol \\
\therefore \eta_{P b\left(N O_{3}\right)_{2}} &=\frac{0.20 mol }{2} \\
&=0.10 mol
\end{aligned}
for reaction II
\(Pb \left( NO _{3}\right)_{2(\text { (av) }}+ MgBr r_{2(a)} \rightarrow\)
\(PbBr _{2(0)}+ Mg \left( NO _{3}\right)_{2(\omega)}\)
\(\eta_{ Pt \left( NO _{5}\right)_{2}}: \eta_{ MgBR _{2}}=1: 1\)
i.c. \(\eta_{P b\left( NO _{3}\right)_{2}}=\eta_{ MgBr }\)
\(\eta_{ MgBr _{2}}=0.20 mol\)
\(\eta_{P b\left( NO _{3}\right)_{2}}=0.20 mol\)
Total no of mole of \(Pb \left( NO _{3}\right)_{2}\)
that reacted,
\(\eta_{P E\left( NO _{3}\right)_{2}}^{T}\) is given by
\(\eta_{P b\left(N O_{3}\right)_{2}}^{T}=\eta_{P b\left( NO _{3}\right)_{2}}+\eta_{P b\left( NO _{3}\right)_{2}}\)
\(=0.10 mol +0.20 mol\)
\(=0.30 mol\)