Em tham khảo nha :
\(\begin{array}{l}
MgC{O_3} + 2HCl \to MgC{l_2} + C{O_2} + {H_2}O\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
{n_{C{O_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
hh:MgC{O_3}(a\,mol),CaC{O_3}(b\,mol)\\
a + b = 0,1(1)\\
84a + 100b = 9,04(2)\\
\text{Từ (1) và (2)}\Rightarrow a = 0,06mol;b = 0,04mol\\
{m_{MgC{O_3}}} = 0,06 \times 84 = 5,04g\\
\% MgC{O_3} = \dfrac{{5,04}}{{9,04}} \times 100\% = 55,75\% \\
\% CaC{O_3} = 100 - 55,75 = 44,25\% \\
{n_{HCl}} = 2{n_{C{O_2}}} = 0,2mol\\
{m_{HCl}} = 0,2 \times 36,5 = 7,3g\\
{m_{{\rm{dd}}HCl}} = \dfrac{{7,3 \times 100}}{{20}} = 36,5g\\
{m_{{\rm{dd}}spu}} = 9,04 + 36,5 - 44 \times 0,1 = 41,14\% \\
{n_{MgC{l_2}}} = {n_{MgC{O_3}}} = 0,06mol\\
{m_{MgC{l_2}}} = 0,06 \times 95 = 5,7g\\
{n_{CaC{l_2}}} = {n_{CaC{O_3}}} = 0,04mol\\
{m_{CaC{l_2}}} = 0,04 \times 111 = 4,44g\\
C{\% _{MgC{l_2}}} = \dfrac{{5,7}}{{41,14}} \times 100\% = 13,86\% \\
C{\% _{CaC{l_2}}} = \dfrac{{4,44}}{{41,14}} \times 100\% = 10,8\%
\end{array}\)
