Giải thích các bước giải:
\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
{n_{{H_2}}} = 0,05mol\\
\to {n_{Mg}} = {n_{{H_2}}} = 0,05mol \to {m_{Mg}} = 1,2g\\
\to {m_{MgO}} = 8g \to {n_{MgO}} = 0,2mol\\
a)\\
\% {m_{Mg}} = \dfrac{{1,2}}{{9,2}} \times 100\% = 13,04\% \\
\% {m_{MgO}} = 100\% - 13,04\% = 86,96\% \\
b)\\
{n_{HCl}} = 2{n_{Mg}} + 2{n_{MgO}} = 0,5mol \to {m_{HCl}} = 18,25g\\
\to m = {m_{{\rm{dd}}HCl}} = 125g\\
c)\\
{n_{MgC{l_2}}} = {n_{Mg}} + {n_{MgO}} = 0,25mol \to {m_{MgC{l_2}}} = 23,75g\\
{m_{{\rm{dd}}}} = {m_{hh}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 134,1g\\
\to C{\% _{MgC{l_2}}} = \dfrac{{23,75}}{{134,1}} \times 100\% = 17,7\%
\end{array}\)