Đáp án:

\(\begin{array}{l}
a)\\
\% {m_{CaC{O_3}}} = 47,17\% \\
\% {m_{CaO}} = 52,83\% \\
b)\\
{C_M}HCl = 0,6M
\end{array}\)

Giải thích các bước giải:

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

CaCO3+2HCl→CaCl2+H2O+CO2

CaO+2HCl→CaCl2+H2O

nCO2=1,12/22,4=0,05mol

mCaCO3=0,05.100=5g

%mCaCO3=5/10,6.100%=47,17%
%mCaO=100-47,17=52,83%
mCaO=10,6-5=5,6mol

nCaO=5,6/56=0,1mol

nHCl=0,05.2+0,1.2=0,3mol

CM HCl=0,3/0,5=0,6M