\(\begin{array}{l}
2)\\
a)\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
n{H_2} = \dfrac{{13,44}}{{22,4}} = 0,6\,mol\\
hh:Mg(a\,mol),Fe(b\,mol)\\
24a + 56b = 25,6\\
a + b = 0,6\\
\Rightarrow a = 0,25;b = 0,35\\
\% mMg = \dfrac{{0,25 \times 24}}{{25,6}} \times 100\% = 23,4375\% \\
\% mFe = 100 - 23,4375 = 76,5625\% \\
b)\\
CM{H_2}S{O_4} = \dfrac{{0,6}}{{0,2}} = 3M\\
3)\\
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
n{H_2} = \dfrac{{11,2}}{{22,4}} = 0,5\,mol\\
hh:Al(a\,mol),Fe(b\,mol)\\
27a + 56b = 22\\
1,5a + b = 0,5\\
\Rightarrow a = 0,11;b = 0,34\\
\% mAl = \dfrac{{0,11 \times 27}}{{22}} \times 100\% = 13,5\% \\
\% mFe = 100 - 13,5 = 86,5\% \\
b)\\
C\% {H_2}S{O_4} = \dfrac{{0,5 \times 98}}{{500}} \times 100\% = 9,8\%
\end{array}\)
