Em tham khảo nha :

\(\begin{array}{l}
1)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{m_{HCl}} = \dfrac{{100 \times 18,25}}{{100}} = 18,25g\\
{n_{HCl}} = \dfrac{{18,25}}{{36,5}} = 0,5mol\\
hh:Al(a\,mol);Fe(b\,mol)\\
\left\{ \begin{array}{l}
3a + 2b = 0,5\\
27a + 56b = 8,3
\end{array} \right.\\
 \Rightarrow a = 0,1;b = 0,1\\
{n_{{H_2}}} = \dfrac{3}{2}{n_{Al}} + {n_{Fe}} = 0,25mol\\
{m_{{\rm{dd}}spu}} = 8,3 + 100 - 0,25 \times 2 = 107,8g\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,1mol\\
{m_{AlC{l_3}}} = 0,1 \times 133,5 = 13,35g\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,1mol\\
{m_{FeC{l_2}}} = 0,1 \times 127 = 12,7g\\
C{\% _{AlC{l_3}}} = \dfrac{{13,35}}{{107,8}} \times 100\%  = 12,4\% \\
C{\% _{FeC{l_2}}} = \dfrac{{12,7}}{{107,8}} \times 100\%  = 11,8\% \\
2)\\
2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
{n_{KMn{O_4}}} = \dfrac{{15,8}}{{158}} = 0,1mol\\
{n_{{O_2}}} = \dfrac{{{n_{KMn{O_4}}}}}{2} = 0,05mol\\
{V_{{O_2}}} = 0,05 \times 22,4 = 1,12l\\
\text{Thể tích khí Oxi thu được là :}\\
{V_{{O_2}}} = \dfrac{{1,12 \times 90}}{{100}} = 1,008l
\end{array}\)