$\begin{gathered}
{n_{HCl}} = 0.08\,\,mol \hfill \\
{n_{KOH}} = 0.02\,\,mol \hfill \\
\left\{ \begin{gathered}
MgC{O_3}\,x\,mol \hfill \\
CaC{O_3}\,y\,mol \hfill \\
\end{gathered} \right. \hfill \\
MgC{O_3} + 2HCl \to MgC{l_2} + C{O_2} + {H_2}O \hfill \\
x\,\,\xrightarrow{{}}\,2x\,\xrightarrow{{}}\,\,\,\,x\,\,\,\,\,(mol) \hfill \\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O \hfill \\
y\,\,\xrightarrow{{}}\,2y\,\xrightarrow{{}}\,\,\,\,y\,\,\,\,(mol) \hfill \\
HC{l_{du}}\, + \,KOH\, \to \,KCl\, + \,{H_2}O \hfill \\
\,\,0.02\, \leftarrow 0.02\,\,\,\,(mol) \hfill \\
Ta\,co: \hfill \\
\left\{ \begin{gathered}
84x\, + \,100y\, = \,2.84 \hfill \\
2x\, + \,2y\, + 0.02 = 0.08 \hfill \\
\end{gathered} \right. \hfill \\
\to \left\{ \begin{gathered}
x = 0.01 \hfill \\
y = 0.02 \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
{m_{MgC{O_3}}} = 0.84\,g \hfill \\
{m_{CaC{O_3}}} = 2\,g \hfill \\
\end{gathered} \right. \hfill \\
Dung\,dich\,\,gom\,\left\{ \begin{gathered}
MgC{l_2}\,0.01\,mol \hfill \\
CaC{l_2}\,0.02\,mol \hfill \\
\end{gathered} \right.co\,V = 0.4 + 0.2 = 0.6l \hfill \\
\to \left\{ \begin{gathered}
{C_{{M_{MgC{l_2}}}}} = 0.01/0.6 = 1/60\,mol/l \hfill \\
{C_{{M_{CaC{l_2}}}}} = 0.02/0.6 = 1/30\,mol/l \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered} $
