Since \(HCl\) is in excess, \(CaCO _{3}\) (in limestone) is the limiting reagent.
The equation of the rxn is:
\begin{gathered}
\underset{\downarrow}{2 HCl _{(2 q)}}+ CaCO _{3( s )} \\aCl _{2( s )}+ H _{2} O _{(t)}+ CO _{2(s)} \\
\eta_{H C l}: \eta_{ CaCO }=2: 1 \\
\eta_{H C l}=2 \eta_{ CaCO _{3}}
\end{gathered}
No of mole \(=\frac{\text { mass }}{\text { molar mass }}\)
\begin{aligned}
&=40+12+(3 \times 16) \\
&=100 g / mol \\
\eta_{\text {Cuco }} &=\frac{2 g }{100 g / mol }=0.02 mol \\
& \therefore \eta_{n c y}=2 \times 0.02 mol \\
&=0.04 mol
\end{aligned}
The reaction of the unreacted acid with \(K _{2} CO _{3}\) is \(2 HCl _{(a q)}+ K _{2} CO _{3} \rightarrow 2 KCl _{\text {(aq) }}+ H _{2} O _{(n}\)
\begin{aligned}
&\eta_{i 1 C 1}: \eta_{K_{2}} CO _{3}=2: 1 \\
&\eta_{u C 4}=2 \eta_{\kappa_{2} CO } \\
&\text { Using the relation } \\
&n=C \times V
\end{aligned}
Where: \(n-\) no of mole
\(c\) - molar concentration
\(v\) - volume in \(dm ^{3}\)
we calculate the no of mole of \(K _{2} CO _{3}\) that reacted thus:
that reacted:
\begin{aligned}
\eta_{H C L}^{T} &=\eta_{H C T}+\eta_{H C L} \\
&=0.04 mol +5.0 \times 10^{-3} mol \\
&=0.045 mol
\end{aligned}
Volume \(\left(V_{n c i}\right)\) of acid \(\left( dm ^{-3}\right)\) \(=\frac{\text { Total no of mole of } HCl }{\text { concentration in moldm }}\) i.e. \(V_{H C T}=\frac{\eta_{H C}^{T}}{C_{M C}}=\frac{0.045}{0.10}=0.45 dm ^{3}\) \(0.45 \times 1000 cm ^{3}=450 cm ^{3}\)