Em tham khảo nha :
\(\begin{array}{l}
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
BaO + {H_2}S{O_4} \to BaS{O_4} + {H_2}O\\
{n_{{H_2}}} = \dfrac{{1,456}}{{22,4}} = 0,065mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,065mol\\
{m_{Fe}} = 0,065 \times 56 = 3,64g\\
{m_{BaO}} = 11,29 - 3,64 = 7,65g\\
\% Fe = \dfrac{{3,64}}{{11,29}} \times 100\% = 32,2\% \\
\% BaO = 100 - 32,2 = 67,8\% \\
b)\\
{n_{BaO}} = \dfrac{{7,65}}{{153}} = 0,05mol\\
{n_{BaS{O_4}}} = {n_{BaO}} = 0,05mol\\
{m_{BaS{O_4}}} = 0,05 \times 233 = 11,65g\\
{n_{FeS{O_4}}} = {n_{Fe}} = 0,065mol\\
{m_{FeS{O_4}}} = 0,065 \times 152 = 9,88g\\
C{\% _{FeS{O_4}}} = \dfrac{{9,88}}{{11,29 + 100 - 11,65 - 0,065 \times 2}} \times 100\% = 12,84\%
\end{array}\)
