Em tham khảo nha:

\(\begin{array}{l}
a)\\
F{e_3}{O_4} + 4{H_2}S{O_4} \to FeS{O_4} + F{e_2}{(S{O_4})_3} + 4{H_2}O\\
FeO + {H_2}S{O_4} \to FeS{O_4} + {H_2}O\\
hh:F{e_3}{O_4}(a\,mol),FeO(b\,mol)\\
232a + 72b = 26\\
552a + 152b = 58\\
 \Rightarrow a = 0,05;b = 0,2\\
\% mF{e_3}{O_4} = \dfrac{{0,05 \times 232}}{{26}} \times 100\%  = 44,615\% \\
\% mFeO = 100 - 44,615 = 55,385\% \\
b)\\
n{H_2}S{O_4} = (0,05 \times 4 + 0,2) \times 115\%  = 0,46\,mol\\
m{\rm{dd}}{H_2}S{O_4} = \dfrac{{0,46 \times 98}}{{16,1\% }} = 280g\\
m{\rm{dd}}spu = 26 + 280 = 306g\\
n{H_2}S{O_4} = 0,4 \times 15\%  = 0,06\,mol\\
nFeS{O_4} = 0,05 + 0,2 = 0,25\,mol\\
nF{e_2}{(S{O_4})_3} = nF{e_3}{O_4} = 0,05\,mol\\
C\% {H_2}S{O_4} = \dfrac{{0,06 \times 98}}{{306}} \times 100\%  = 1,92\% \\
C\% FeS{O_4} = \dfrac{{0,25 \times 152}}{{306}} \times 100\%  = 12,42\% \\
C\% F{e_2}{(S{O_4})_3} = \dfrac{{0,05 \times 400}}{{306}} \times 100\%  = 6,54\% 
\end{array}\)