Đáp án:

\(\% {m_{Mg}} = 30,77\%; \% {m_{Al}} = 69,23\% \)

\({C_{M{\text{ HCl}}}}  = 4,8M\)

\({C_{M{\text{ MgC}}{{\text{l}}_2}}} = 0,6M;{C_{M{\text{ AlC}}{{\text{l}}_3}}}= 1,2M\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(Mg + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}\)

\(2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}\)

Gọi số mol \(Mg;Al\) lần lượt là \(x;y\)

\( \to 24x + 27y = 7,8\)

Ta có:

\({m_{dd{\text{ tăng}}}} = {m_{kl}} - {m_{{H_2}}} = 7,8 - {m_{{H_2}}} = 7{\text{ gam}}\)

\( \to {m_{{H_2}}} = 0,8{\text{ gam}} \to {{\text{n}}_{{H_2}}} = \frac{{0,8}}{2} = 0,4{\text{ mol}}\)

\( \to {n_{{H_2}}} = {n_{Mg}} + \frac{3}{2}{n_{Al}} = x + 1,5y = 0,4\)

Giải được: \(x=0,1;y=0,2\)

\( \to {m_{Mg}} = 0,1.24 = 2,4{\text{ gam}}\)

\( \to \% {m_{Mg}} = \frac{{2,4}}{{7,8}} = 30,77\%  \to \% {m_{Al}} = 69,23\% \)

Ta có:

\({V_{dd{\text{ HCl}}}} = \frac{{200}}{{1,2}} = 166,67{\text{ ml = 0}}{\text{,16667 lít}}\)

\( \to {n_{HCl}} = 2{n_{{H_2}}} = 0,8{\text{ mol}}\)

\( \to {C_{M{\text{ HCl}}}} = \frac{{0,8}}{{0,1667}} = 4,8M\)

\({C_{M{\text{ MgC}}{{\text{l}}_2}}} = \frac{{0,1}}{{0,1667}} = 0,6M;{C_{M{\text{ AlC}}{{\text{l}}_3}}} = \frac{{0,2}}{{0,1667}} = 1,2M\)