Em tham khảo nha:
\(\begin{array}{l}
6)\\
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
hh:Fe(a\,mol),Al(b\,mol)\\
\left\{ \begin{array}{l}
56a + 27b = 11\\
a + 1,5b = 0,4
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,2\\
\% {m_{Fe}} = \dfrac{{56 \times 0,1}}{{11}} \times 100\% = 50,91\% \\
\% {m_{Al}} = 100 - 50,91 = 49,09\% \\
b)\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,4\,mol\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,4 \times 98}}{{9,8\% }} = 400g\\
{m_{{\rm{dd}}X}} = 11 + 400 - 0,4 \times 2 = 410,2g\\
{C_\% }FeS{O_4} = \dfrac{{0,1 \times 152}}{{410,2}} \times 100\% = 3,71\% \\
{C_\% }A{l_2}{(S{O_4})_3} = \dfrac{{0,1 \times 342}}{{410,2}} \times 100\% = 8,34\% \\
2)\\
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
hh:Fe(a\,mol),Mg(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,3\\
56a + 24b = 13,6
\end{array} \right.\\
\Rightarrow a = 0,2;b = 0,1\\
\% {m_{Fe}} = \dfrac{{0,2 \times 56}}{{13,6}} \times 100\% = 82,35\% \\
\% {m_{Mg}} = 100 - 82,35 = 17,65\% \\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,6\,mol\\
{m_{{\rm{dd}}HCl}} = \dfrac{{0,6 \times 36,5}}{{14,6\% }} = 150g\\
{m_{{\rm{dd}}X}} = 13,6 + 150 - 0,3 \times 2 = 163g\\
{C_\% }FeC{l_2} = \dfrac{{0,2 \times 127}}{{163}} \times 100\% = 15,58\% \\
{C_\% }MgC{l_2} = \dfrac{{0,1 \times 95}}{{163}} \times 100\% = 5,83\%
\end{array}\)
