Đáp án:
\( \% {m_{Fe}} = 50,91\% ; \% {m_{Al}} = 49,09\% \)
\( {C_{M{\text{ }}{{\text{H}}_2}S{O_4}}} = 0,8M\)
\( {C_{M{\text{ FeS}}{{\text{O}}_4}}} = {C_{M{\text{ A}}{{\text{l}}_2}{{(S{O_4})}_3}}}= 0,2M\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Fe + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}\)
\(2Al + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}\)
Gọi số mol \(Fe;Al\) lần lượt là \(x;y\)
\( \to 56x + 27y = 11{\text{ gam}}\)
\( \to {n_{{H_2}}} = {n_{Fe}} + \frac{3}{2}{n_{Al}} = x + 1,5y = \frac{{8,96}}{{22,4}} = 0,4{\text{ mol}}\)
Giải được: \(x=0,1;y=0,2\)
\( \to {m_{Fe}} = 0,1.56 = 5,6{\text{ gam}} \to {{\text{m}}_{Al}} = 5,4{\text{ gam}}\)
\( \to \% {m_{Fe}} = \frac{{5,6}}{{11}}.100\% = 50,91\% \to \% {m_{Al}} = 49,09\% \)
\( \to {n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,4{\text{ mol}}\)
\( \to {C_{M{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{{0,4}}{{0,5}} = 0,8M\)
\({n_{FeS{O_4}}} = {n_{Fe}} = 0,1{\text{ mol;}}{{\text{n}}_{A{l_2}{{(S{O_4})}_3}}} = \frac{1}{2}{n_{Al}} = 0,1{\text{ mol}}\)
\({V_{dd}} = 500{\text{ ml = 0}}{\text{,5 lít}}\)
\( \to {C_{M{\text{ FeS}}{{\text{O}}_4}}} = {C_{M{\text{ A}}{{\text{l}}_2}{{(S{O_4})}_3}}} = \frac{{0,1}}{{0,5}} = 0,2M\)
