\(n_{CO_2^{}}^{} = \frac{{2,24}}{{22,4}} = 0,1\;{\text{mol}}\)

\({\text{n}}_{CaCl_2^{}}^{} = \frac{{111.30\% }}{{111}} = 0,3\;{\text{mol}}\)

\({\text{BTCacbon: n}}_{CaCO_3^{}}^{} = n_{CO_2^{}}^{} = 0,1\;{\text{mol}}\)

\({\text{n}}_{CaCl_2^{}}^{} = n_{CaCO_3^{}}^{} + n_{Ca(OH)_2^{}}^{} = 0,3\)

\( \to n_{Ca(OH)_2^{}}^{} = 0,3 - 0,1 = 0,2{\text{ mol}}\)

\({\text{\% m}}_{CaCO_3^{}}^{} = \frac{{0,1.100}}{{0,1.100 + 0,2.74}}.100 = 40,32\% \)

\(\% m_{Ca(OH)_2^{}}^{} = 100 - 40,32 = 59,68\% \)

\(b)n_{HCl}^{} = 2n_{CaCl_2^{}}^{} = 2.0,3 = 0,6{\text{ mol}}\)

\(m_{dd_{HCl}^{}}^{} = \frac{{0,6.36,5}}{{20\% }} = 109,5g\)

\(V = \frac{m}{D} = \frac{{109,5}}{{1,1}} = 99,54g\)

\(\begin{array}{l}
a)\\
CaCO_3+2HCl\to CaCl_2+CO_2+H_2O\\
Ca(OH)_2+2HCl\to CaCl_2+2H_2O\\
n_{CO_2}=n_{CaCO_3}=\frac{2,24}{22,4}=0,1(mol)\\
\sum n_{CaCl_2}=\frac{111.30\%}{111}=0,3(mol)\\
\to n_{CaCO_3}+n_{Ca(OH)_2}=n_{CaCl_2}=0,3\\
\to n_{Ca(OH)_2}=0,3-0,1=0,2(mol)\\
\to m_{hh}=0,1.100+0,2.74=24,8(g)\\
\%m_{CaCO_3}=\frac{0,1.100}{24,8}.100\%\approx 40,32\%\\
\%m_{Ca(OH)_2}=100-40,32=59,68\%\\
b)\\
Theo\,PT:\,n_{HCl}=2n_{CaCl_2}=0,6(mol)\\
\to m_{dd\,HCl}=\frac{0,6.36,5}{20\%}=109,5(g)\\
\to V_{dd\,HCl}=\frac{109,5}{1,1}\approx 99,5(cm^3)
\end{array}\)