Em tham khảo nha:

\(\begin{array}{l}
4)\\
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{1,792}}{{22,4}} = 0,08\,mol\\
hh:Zn(a\,mol),Al(b\,mol)\\
\left\{ \begin{array}{l}
65a + 27b = 3,79\\
a + 1,5b = 0,08
\end{array} \right.\\
 \Rightarrow a = 0,05;b = 0,02\\
\% {m_{Zn}} = \dfrac{{0,05 \times 65}}{{3,79}} \times 100\%  = 85,75\% \\
\% {m_{Al}} = 100 - 85,75 = 14,25\% \\
5)
\end{array}\)

\(\begin{array}{l}
2Al + 3CuC{l_2} \to 2AlC{l_3} + 3Cu\\
{n_{CuC{l_2}}} = \dfrac{{30 \times 13,5\% }}{{135}} = 0,03\,mol\\
{n_{Al}} = 0,03 \times \frac{2}{3} = 0,02\,mol\\
{m_{Al}} = 0,02 \times 27 = 0,54g\\
{n_{Cu}} = {n_{CuC{l_2}}} = 0,03\,mol\\
{m_{{\rm{dd}}spu}} = 0,54 + 30 - 0,03 \times 64 = 28,62g\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,02\,mol\\
{C_\% }AlC{l_3} = \dfrac{{0,02 \times 133,5}}{{28,62}} \times 100\%  = 9,33\% 
\end{array}\)

Đáp án

2Al+3CuCl2→2AlCl3+3Cu

nCuCl2=30×13,5%135=0,03mol

nAl=0,03×23=0,02mol

mAl=0,02×27=0,54g

nCu=nCuCl2=0,03mol

mddspu=0,54+30−0,03×64=28,62g

nAlCl3=nAl=0,02mol

C%AlCl3=(0,02×133,5/28,62)×100%=9,33%