Đáp án:

\(\begin{array}{l}
a)\\
{V_{{H_2}}} = 3,36l\\
{m_{{\rm{dd}}HCl}} = 75g\\
b)\\
C{\% _{MgC{l_2}}} = 18,2\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{Mg}} = \dfrac{m}{M} = \dfrac{{3,6}}{{24}} = 0,15mol\\
{n_{{H_2}}} = {n_{Mg}} = 0,15mol\\
{V_{{H_2}}} = n \times 22,4 = 0,15 \times 22,4 = 3,36l\\
{n_{HCl}} = 2{n_{Mg}} = 0,3mol\\
{m_{HCl}} = 0,3 \times 36,5 = 10,95g\\
{m_{{\rm{dd}}HCl}} = \dfrac{{10,95 \times 100}}{{14,6}} = 75g\\
b)\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,15mol\\
{m_{MgC{l_2}}} = 0,15 \times 95 = 14,25g\\
{m_{{\rm{dd}}spu}} = 3,6 + 75 - 0,15 \times 2 = 78,3g\\
C{\% _{MgC{l_2}}} = \dfrac{{14,25}}{{78,3}} \times 100\%  = 18,2\% 
\end{array}\)