Đáp án:

 C% MgCl2=12,16%

Giải thích các bước giải:

Giả sử X chứa 1 mol Fe và x mol Mg.

\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)

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

Ta có: \({n_{HCl}} = 2{n_{Fe}} + 2{n_{Mg}} = 1.2 + 2x = 2 + 2x \to {m_{HCl}} = 36,5.(2 + 2x) = 73 + 73x \to {m_{dd{\text{HCl}}}} = \frac{{73 + 73x}}{{20\% }} = 365 + 365x\)

Ta có: \({n_{{H_2}}} = {n_{Fe}} + {n_{Mg}} = 1 + x{\text{ mol}}\)

BTKL: \({m_{dd\;{\text{Y}}}} = 56 + 24x + 365 + 365x - (1 + x).2 = 419 + 387x\)

Ta có:  \({m_{FeC{l_2}}} = 1.(56 + 35,5.2) = 127 \to {m_{dd{\text{Y}}}} = \frac{{127}}{{15,76\% }} = 418 + 387x \to x = 1\)

\(\to {m_{MgC{l_2}}} = 1.(27 + 35,5.2) = 98{\text{ gam}} \to {\text{C}}{{\text{\% }}_{MgC{l_2}}} = \frac{{98}}{{419 + 387}} = 12,16\% \)