Giải thích các bước giải:
\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{Mg}} = 0,1mol\\
\to {n_{MgC{l_2}}} = {n_{Mg}} = 0,1mol\\
\to {m_{MgC{l_2}}} = 9,5g\\
\to {n_{{H_2}}} = {n_{Mg}} = 0,1mol \to {V_{{H_2}}} = 2,24l\\
{n_{HCl}} = 2{n_{Mg}} = 0,2mol\\
\to {m_{HCl}} = 7,3g \to {m_{HCl{\rm{dd}}}} = \dfrac{{7,3 \times 100}}{{14,6}} = 50g
\end{array}\)