Đáp án:
\(\begin{array}{l}
1)\\
\% {m_{Mg}} = 57,14\% \\
\% {m_{Al}} = 42,86\% \\
2)\\
{C_\% }MgS{O_4} = 8,75\% \\
{C_\% }A{l_2}{(S{O_4})_3} = 8,31\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{13,44}}{{22,4}} = 0,6\,mol\\
hh:Mg(a\,mol),Al(b\,mol)\\
\left\{ \begin{array}{l}
24a + 27b = 12,6\\
a + 1,5b = 0,6
\end{array} \right.\\
\Rightarrow a = 0,3;b = 0,2\\
\% {m_{Mg}} = \dfrac{{0,3 \times 24}}{{12,6}} \times 100\% = 57,14\% \\
\% {m_{Al}} = 100 - 57,14 = 42,86\% \\
2)\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,6\,mol\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,6 \times 98}}{{14,7\% }} = 400g\\
{m_{{\rm{dd}}spu}} = 12,6 + 400 - 0,6 \times 2 = 411,4g\\
{n_{MgS{O_4}}} = {n_{Mg}} = 0,3\,mol\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{0,2}}{2} = 0,1\,mol\\
{C_\% }MgS{O_4} = \dfrac{{0,3 \times 120}}{{411,4}} \times 100\% = 8,75\% \\
{C_\% }A{l_2}{(S{O_4})_3} = \dfrac{{0,1 \times 342}}{{411,4}} \times 100\% = 8,31\%
\end{array}\)
