CaO + 2HCl --> CaCl2 + H2O

  0,3       0,6          0,3

CaCO3 + 2HCl --> CaCl2 + CO2 + H2O

   0,2          0,4           0,2       0,2

mCaCO3 = 0,2.100 = 20 g

mCaO = 36,8-20 = 16,8 g

%mCaO = 16,8.100/36,8 = 45,65%

%mCaCO3 = 100%-45,65% = 54,35%

b)

nCaO = 16,8/56 = 0,3 mol

nCaCl2 tổng = 0,3+0,2 = 0,5 mol

=> CmCaCl2 = 0,5/5 = 0,1M

Em tham khảo nha :

\(\begin{array}{l}
a)\\
CaO + 2HCl \to CaC{l_2} + {H_2}O\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
{n_{C{O_2}}} = \dfrac{{4,032}}{{22,4}} = 0,18mol\\
{n_{C{O_2}th}} = \dfrac{{0,18 \times 100}}{{90}} = 0,2mol\\
{n_{CaC{O_3}}} = {n_{C{O_2}th}} = 0,2mol\\
{m_{CaC{O_3}}} = 0,2 \times 100 = 20g\\
\% CaC{O_3} = \dfrac{{20}}{{36,8}} \times 100\%  = 54,35\% \\
\% CaO = 100 - 54,35 = 45,65\% \\
b)\\
{n_{CaO}} = \dfrac{{36,8 - 20}}{{56}} = 0,3mol\\
{n_{HCl}} = 2{n_{CaC{O_3}}} + 2{n_{CaO}} = 1mol\\
{C_{{M_{HCl}}}} = \dfrac{1}{5} = 0,2M\\
{n_{CaC{l_2}}} = {n_{CaO}} + {n_{CaC{O_3}}} = 0,5mol\\
{C_{{M_{CaC{l_2}}}}} = \dfrac{{0,5}}{5} = 0,1M
\end{array}\)