Em tham khảo nha :
\(\begin{array}{l}
a)\\
FeO + {H_2}S{O_4} \to FeS{O_4} + {H_2}O\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,1mol\\
{m_{Fe}} = 0,1 \times 56 = 5,6g\\
{m_{FeO}} = 12,8 - 5,6 = 7,2g\\
\% Fe = \dfrac{{5,6}}{{12,8}} \times 100\% = 43,75\% \\
\% FeO = 100 - 43,75 = 56,25\% \\
c)\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
2FeO + 4{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + S{O_2} + 4{H_2}O\\
{n_{S{O_2}}} = \dfrac{3}{2}{n_{Fe}} + \dfrac{{{n_{FeO}}}}{2} = 0,1mol\\
S{O_2} + Ca{(OH)_2} \to CaS{O_3} + {H_2}O\\
{n_{CaS{O_3}}} = {n_{S{O_2}}} = 0,1mol\\
{m_{CaS{O_3}}} = 0,1 \times 120 = 12g\\
{m_{CaS{O_3}}} - {m_{S{O_2}}} = 12 - 0,1 \times 64 = 5,6g\\
\text{Vậy sau phản ứng dung dịch nước vôi trong giảm đi 5,6g}
\end{array}\)
