Đáp án:

\({C_{M{\text{ FeC}}{{\text{l}}_3}}} = 1,6M;{C_{M{\text{MgC}}{{\text{l}}_2}}} = 1,2M;{C_{M{\text{ HCl dư}}}} = 2,784M\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(F{e_2}{O_3} + 6HCl\xrightarrow{{}}2FeC{l_3} + 3{H_2}O\)

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

Ta có: 

\({m_{HCl}} = 260.35,04\%  = 91,1{\text{ gam}} \to {{\text{n}}_{HCl}} = \frac{{91,104}}{{36,5}} = 2,496{\text{ mol}}\)

Tổng khối lượng giảm là do khí hidro thoát ra.

\( \to {n_{{H_2}}} = \frac{{0,6}}{2} = 0,3{\text{ mol = }}{{\text{n}}_{Mg}} \to {m_{Mg}} = 0,3.24 = 7,2{\text{ gam}} \to {{\text{n}}_{F{e_2}{O_3}}} = \frac{{39,2 - 7,2}}{{56.2 + 16.3}} = 0,2{\text{ mol}}\)

\( \to {n_{FeC{l_3}}} = 2{n_{F{e_2}{O_3}}} = 0,4{\text{ mol;}}{{\text{n}}_{MgC{l_2}}} = {n_{Mg}} = 0,3{\text{ mol;}}{{\text{n}}_{HCl{\text{dư}}}} = 2,496 - 0,4.3 - 0,3.2 = 0,696{\text{ mol}}\)

\({V_{dd}} = \frac{{260}}{{1,04}} = 250{\text{ ml = 0}}{\text{,25 lít}}\)

\( \to {C_{M{\text{ FeC}}{{\text{l}}_3}}} = 1,6M;{C_{M{\text{MgC}}{{\text{l}}_2}}} = 1,2M;{C_{M{\text{ HCl dư}}}} = 2,784M\)