Đáp án:

 b) 8,4l

c) 59,25g

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
n{H_2} = \dfrac{{20,16}}{{22,4}} = 0,9\,mol\\
hh:Al(a\,mol),Fe(b\,mol)\\
27a + 56b = 33,3\\
1,5a + b = 0,9\\
 \Rightarrow a = 0,3;b = 0,45\\
\% mAl = 24,32\% \\
\% mFe = 75,68\% \\
b)\\
nAl = 0,1\,mol\\
nFe = 0,15\,mol\\
2Al + 6{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
nS{O_2} = 0,1 \times 1,5 + 0,15 \times 1,5 = 0,375\,mol\\
VS{O_2} = 0,375 \times 22,4 = 8,4l\\
c)\\
S{O_2} + 2KOH \to {K_2}S{O_3} + {H_2}O\\
n{K_2}S{O_3} = nS{O_2} = 0,375\,mol\\
m{K_2}S{O_3} = 0,375 \times 158 = 59,25g
\end{array}\)