Đáp án:
\( \% {m_{Mg}} = 30\% ; \% {m_{Fe}} = 70\% \)
\(\% {m_{MgC{l_2}}} = 2,33\% ;\% {m_{FeC{l_2}}} = 3,11\% \)
\(C{\% _{HCl}} = 4,015\% \)
Giải thích các bước giải:
Gọi số mol \(Mg;Fe\) lần lượt là \(x;y\)
\( \to 24x + 56y = 4{\text{ gam}}\)
Cho hỗn hợp tác dụng với \(HCl\)
\(Mg + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}\)
\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)
Ta có:
\({n_{{H_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol = }}{{\text{n}}_{Mg}} + {n_{Mg}} = x + y\)
Giải được: \(x=y=0,05\)
\( \to {m_{Mg}} = 0,05.24 = 1,2{\text{ gam}}\)
\( \to \% {m_{Mg}} = \frac{{1,2}}{4} = 30\% \to \% {m_{Fe}} = 70\% \)
BTKL:
\({m_{kl}} + {m_{dd{\text{ HCl}}}} = {m_{muối}} + {m_{{H_2}}}\)
\( \to 4 + 200 = {m_{muối}} + 0,1.2\)
\( \to {m_{muối}} = 203,8{\text{ gam}}\)
Ta có:
\({n_{MgC{l_2}}} = {n_{FeC{l_2}}} = 0,05{\text{ mol}}\)
\( \to {m_{MgC{l_2}}} = 0,05.(24 + 35,5.2) = 4,75{\text{ gam}}\)
\({m_{FeC{l_2}}} = 0,05.(56 + 35,5.2) = 6,35{\text{ gam}}\)
\( \to \% {m_{MgC{l_2}}} = \frac{{4,75}}{{203,8}} = 2,33\% ;\% {m_{FeC{l_2}}} = \frac{{6,35}}{{203,8}} = 3,11\% \)
\({n_{HCl}} = 2{n_{{H_2}}} = 0,1.2 = 0,2{\text{ mol}}\)
\( \to {n_{HCl{\text{ tham gia}}}} = 0,2.(100\% + 10\% ) = 0,22{\text{ mol}}\)
\( \to {m_{HCl}} = 0,22.36,5 = 8,03{\text{ gam}}\)
\( \to C{\% _{HCl}} = \frac{{8,03}}{{200}} = 4,015\% \)
