PTHH:

\(Mg + 2HCl \to MgC{l_2} + {H_2} \uparrow \)                               (1)

\(CaC{O_3} + 2HCl \to CaC{l_2} + {H_2}O + C{O_2} \uparrow \)     (2)

Khi đó khí thoát ra là \(C{O_2},{H_2}\)

Gọi số mol \(Mg;CaC{O_3}\) lần lượt là x, y mol.

Theo PTHH (1); (2) ta có: \({n_{{H_2}}} = a;{n_{C{O_2}}} = b(mol)\)

HPT: \(\left\{ \begin{array}{l}
24a + 100b = 27,2\\
\overline M  = \dfrac{{2a + 44b}}{{a + b}} = 6,25.2
\end{array} \right. \to \left\{ \begin{array}{l}
a \approx 0,474\\
b \approx 0,158
\end{array} \right.mol\)

a) \( \Rightarrow {m_{Mg}} = 11,376g;{m_{CaC{O_3}}} = 15,824g\)

b) \({n_{HCl{\rm{ pu}}}} = 2a + 2b = 1,264mol \to \sum {{n_{HCl}} = 1,264 + 1,264.10\%  = 1,3904mol} \)

\( \to {m_{HCl}} = 50,75g \to {m_{{\rm{ddHCl}}}}{\rm{ = }}\dfrac{{50,75.100}}{{25}} = 203g\)

c) \({m_{{\rm{ddsau}}}} = {m_{hh}} + {m_{{\rm{ddHCl}}}} - {m_{{H_2}}} - {m_{C{O_2}}}\)

\( = 27,2 + 203 - 0,474.2 - 0,158.44 = 222,3g\)

⇒\(C\% (MgC{l_2}) = \dfrac{{0,474.95}}{{222,3}}.100\%  = 20,26\% \)

\(C\% (CaC{l_2}) = \dfrac{{0,158.111}}{{222,3}}.100\%  = 7,89\% \)