Đáp án:

 b) 8,96l

c) 0,7l

d) 10,08l

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
b)\\
n{H_2}S{O_4} = 0,4 \times 1 = 0,4\,mol\\
n{H_2} = n{H_2}S{O_4} = 0,4\,mol\\
V = 0,4 \times 22,4 = 8,96l\\
c)\\
hh:Fe(a\,mol),Al(b\,mol)\\
a + 1,5b = 0,4\\
56a + 27b = 11\\
 \Rightarrow a = 0,1;b = 0,2\\
FeS{O_4} + BaC{l_2} \to FeC{l_2} + BaS{O_4}\\
A{l_2}{(S{O_4})_3} + 3BaC{l_2} \to 2AlC{l_3} + 3BaS{O_4}\\
nBaC{l_2} = 0,1 + 0,2 \times 3 = 0,7\,mol\\
VBaC{l_2} = \dfrac{{0,7}}{1} = 0,7l\\
d)\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,15\\
2Al + 6{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
0,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,3\\
VS{O_2} = (0,3 + 0,15) \times 22,4 = 10,08l
\end{array}\)