\(n_{CO_2}=\dfrac{2,0608}{22,4}=0,092mol\)
Gọi \(\left\{{}\begin{matrix}n_{CaCO_3}=x\\n_{MgCO_3}=y\end{matrix}\right.\)
\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
x x ( mol )
\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)
y y ( mol )
Ta có:
\(\left\{{}\begin{matrix}100x+84y=8\\x+y=0,092\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,017\\y=0,075\end{matrix}\right.\)
\(\Rightarrow m_{CaCO_3}=0,017.100=1,7g\)
\(\Rightarrow m_{MgCO_3}=0,075.84=6,3g\)
\(\%m_{CaCO_3}=\dfrac{1,7}{8}.100=21,25\%\)
\(\%m_{MgCO_3}=100\%-21,25\%=78,75\%\)
