Đáp án:

\(\begin{array}{l}
a)\\
\% {m_{Mg}} = 80\% \\
\% {m_{MgO}} = 20\% \\
b)\\
{C_M}MgC{l_2} = 0,23M\\
{C_M}HCl = 0,54M\\
c)\\
{C_\% }MgC{l_2} = 1,61\% \\
{C_\% }HCl = 1,45\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_{Mg}} = {n_{{H_2}}} = 0,1\,mol\\
\% {m_{Mg}} = \dfrac{{0,1 \times 24}}{3} \times 100\%  = 80\% \\
\% {m_{MgO}} = 100 - 80 = 20\% \\
b)\\
{n_{MgO}} = \dfrac{{3 - 2,4}}{{40}} = 0,015\,mol\\
{n_{MgC{l_2}}} = 0,1 + 0,015 = 0,115\,mol\\
{n_{HCl}} = 0,5 - 0,1 \times 2 - 0,015 \times 2 = 0,27\,mol\\
{C_M}MgC{l_2} = \dfrac{{0,115}}{{0,5}} = 0,23M\\
{C_M}HCl = \dfrac{{0,27}}{{0,5}} = 0,54M\\
c)\\
{m_{{\rm{dd}}HCl}} = 1,35 \times 500 = 675g\\
{m_{{\rm{dd}}spu}} = 3 + 675 - 0,1 \times 2 = 677,8g\\
{C_\% }MgC{l_2} = \dfrac{{0,115 \times 95}}{{677,8}} \times 100\%  = 1,61\% \\
{C_\% }HCl = \dfrac{{0,27 \times 36,5}}{{677,8}} \times 100\%  = 1,45\% 
\end{array}\)