Đáp án:

\(\begin{array}{l}
b)\\
\% CaC{O_3} = 47,17\% \\
\% CaO = 52,83\% \\
c)\\
{C_{{M_{HCl}}}} = 2M\\
d)\\
{C_{{M_{CaC{l_2}}}}} = 1M
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
CaO + 2HCl \to CaC{l_2} + {H_2}O\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
b)\\
{n_{C{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{1,12}}{{22,4}} = 0,05mol\\
{n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,05mol\\
{m_{CaC{O_3}}} = n \times M = 0,05 \times 100 = 5g\\
{m_{CaO}} = 10,6 - 5 = 5,6g\\
\% CaC{O_3} = \dfrac{5}{{10,6}} \times 100\%  = 47,17\% \\
\% CaO = 100 - 47,17 = 52,83\% \\
c)\\
{n_{CaO}} = \dfrac{m}{M} = \dfrac{{5,6}}{{56}} = 0,1mol\\
{n_{HCl}} = 2{n_{CaO}} + 2{n_{CaC{O_3}}} = 0,3mol\\
{C_{{M_{HCl}}}} = \dfrac{n}{V} = \dfrac{{0,3}}{{0,15}} = 2M\\
d)\\
{n_{CaC{l_2}}} = {n_{CaO}} + {n_{CaC{O_3}}} = 0,15mol\\
{C_{{M_{CaC{l_2}}}}} = \dfrac{n}{V} = \dfrac{{0,15}}{{0,15}} = 1M
\end{array}\)