Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2R + 2nHCl \to 2RC{l_n} + n{H_2}\\
{m_{{\rm{dd}}A}} = {m_R} + {m_{{\rm{dd}}HCl}} - {m_B}\\
\to {m_B} = {m_R} + {m_{{\rm{dd}}HCl}} - {m_{{\rm{dd}}A}} = 1,2 + 200 - 201,1 = 0,1g\\
\to {n_B} = 0,05mol\\
\to {n_R} = {n_B} = \frac{{0,1}}{n}mol\\
\to {M_R} = \dfrac{{1,2}}{{\dfrac{{0,1}}{n}}} = 12n\\
\to R = Mg\\
b)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{HCl}} = 2{n_B} = 0,1mol\\
{m_{HCl}} = 0,1 \times 36,5 = 3,65g\\
\to C{\% _{HCl}} = \dfrac{{3,65}}{{200}} \times 100\% = 1,825\% \\
{n_{MgC{l_2}}} = {n_B} = 0,05mol\\
\to {m_{MgC{l_2}}} = 0,05 \times 95 = 4,75g\\
\to C{\% _{MgC{l_2}}} = \dfrac{{4,75}}{{201,1}} \times 100\% = 2,36\%
\end{array}\)
