Đáp án:
a) \({\text{\% }}{{\text{m}}_{Fe}} = 60,87\% ;{\text{ }}\% {m_{Mg}} = 39,13\% \)
b) \({C_{M{\text{ FeC}}{{\text{l}}_2}}} = {C_{M{\text{ HCl dư}}}} = 0,333M;{\text{ }}{{\text{C}}_{M{\text{ MgC}}{{\text{l}}_2}}} = 0,5M\)
Giải thích các bước giải:
Gọi số mol Fe là x; Mg là y.
Phản ứng xảy ra:
\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)
\(Mg + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}\)
\(\to {n_{{H_2}}} = {n_{Fe}} + {n_{Mg}} = x + y = \frac{{5,6}}{{22,4}} = 0,25{\text{ mol}}\)
Giải được: x=0,1; y=0,15.
\(\to {m_{Fe}} = 0,1.56 = 5,6{\text{ gam}} \to {\text{\% }}{{\text{m}}_{Fe}} = \frac{{5,6}}{{9,2}} = 60,87\% \to \% {m_{Mg}} = 39,13\% \)
Ta có: \({n_{HCl{\text{ phản ứng}}}} = 2{n_{{H_2}}} = 0,5{\text{ mol}} \to {{\text{n}}_{HCl{\text{ tham gia}}}} = 0,5.120\% = 0,6{\text{ mol}} \to {{\text{V}}_{dd{\text{ HCl}}}} = {V_A} = \frac{{0,6}}{2} = 0,3{\text{ lít}}\)
\(\to {n_{FeC{l_2}}} = {n_{Fe}} = 0,1{\text{ mol}} \to {{\text{C}}_{M{\text{ FeC}}{{\text{l}}_2}}} = \frac{{0,1}}{{0,3}} = 0,333M\)
\({n_{MgC{l_2}}} = {n_{Mg}} = 0,15{\text{ mol}} \to {{\text{C}}_{M{\text{ MgC}}{{\text{l}}_2}}} = \frac{{0,15}}{{0,3}} = 0,5M\)
\({n_{HCl{\text{ dư}}}} = 0,1{\text{ mol}} \to {{\text{C}}_{M{\text{ HCl dư}}}} = \frac{{0,1}}{{0,3}} = 0,333M\)
