$$\begin{gathered} n_{H_{2}S{O}_4}=0,3mol \\ gọi x,y là số mol của Fe và FeO trog hh \\ PTHH: Fe+H_{2}S{O}_4-> FeS{O}_4+H_2 \\ x-->x->x------->x \\ FeO+H_{2}S{O}_4->FeS{O}_4+H_{2}O \\ y-> y---->y---->y \\ Có x+y=0,3 \left(1\right) \\ 56x+72y=18,4 \left(2\right) \\ (1,2)=> x=0,2 , y=0,1 \\ {m}_{Fe}=11,2g \\ {m}_{FeO}=7,2g \\ b, n_{FeS{O}_4}=n_{Fe}+n_{FeO}=0,3 mol \\ {m}_{FeS{O}_4}=45,6 g \\ mdd=m_{hh}+m_{dd H_{2}S{O}_4}-m_{H_2}=18,4+300.1,65-0,2.2=513g \\ C\% FeS{O}_4=45,6:513.100=8,9\% \end{gathered}$$

.

$$\begin{gathered} {\mathrm{nH}}_2{\mathrm{SO}}_4=0,3 \mathrm{mol} \\ \mathrm{a}, \mathrm{Fe}+{\mathrm{H}}_2{\mathrm{SO}}_4->{\mathrm{FeSO}}_4+{\mathrm{H}}_2 \\ \mathrm{x} \mathrm{x} \mathrm{x} \mathrm{x} \left(\mathrm{mol}\right) \\ \mathrm{FeO}+{\mathrm{H}}_2{\mathrm{SO}}_4->{\mathrm{FeSO}}_4+{\mathrm{H}}_2\mathrm{O} \\ \mathrm{y} \mathrm{y} \mathrm{y} \left(\mathrm{mol}\right) \\ \mathrm{Ta} \mathrm{có}: \left\{\begin{array}{l}56\mathrm{x}+72\mathrm{y}=18,4\\ \mathrm{x}+\mathrm{y}=0,3\end{array}<=>\left\{\begin{array}{l}\mathrm{x}=0,2\\ \mathrm{y}=0,1\end{array}\right.\right. \\ =>\mathrm{mFe}=0,2.56=11,2\mathrm{g} \\ \mathrm{mFeO}=0,1.72=7,2\mathrm{g} \\ \mathrm{Vậy} \mathrm{có}: \mathrm{mFe}=11,2\mathrm{g}; \mathrm{mFeO}=7,2\mathrm{g} \\ \mathrm{b}, {\mathrm{nH}}_2=\mathrm{x}=0,2 \mathrm{mol} \\ {\mathrm{mdd}}_{{\mathrm{H}}_2{\mathrm{SO}}_4}=300.1,65=495\mathrm{g} \\ =>\mathrm{mdd} \mathrm{sau}=18,4+495-0,2.2=513\mathrm{g} \\ {\mathrm{nFeSO}}_4=\mathrm{x}+\mathrm{y}=0,3 \mathrm{mol} \\ =>{\mathrm{mFeSO}}_4=45,6\mathrm{g} \\ =>\mathrm{C}\%{\mathrm{FeSO}}_4=45,6 : 513.100\%=8,89\% \end{gathered}$$

Vậy C%=8,89%