\(\begin{array}{l}
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
FeO + {H_2}S{O_4} \to FeS{O_4} + {H_2}O\\
n{H_2} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol \Rightarrow nFe = n{H_2} = 0,1\,mol\\
\% mFe = \dfrac{{0,1 \times 56}}{{20}} \times 100\% = 28\% \\
\% mFeO = 100 - 28 = 72\% \\
b)\\
nFeO = \dfrac{{20 - 5,6}}{{72}} = 0,2\,mol\\
n{H_2}S{O_4} = 0,2 + 0,1 = 0,3\,mol\\
C\% {H_2}S{O_4} = \dfrac{{0,3 \times 98}}{{250}} \times 100\% = 11,76\% \\
c)\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
2FeO + 4{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + S{O_2} + 4{H_2}O\\
nS{O_2} = 0,025 \times 1,5 + 0,5 \times 0,5 = 0,2875\,mol\\
\Rightarrow VS{O_2} = 0,2875 \times 22,4 = 6,44l
\end{array}\)
