Giải thích các bước giải:

1.

\(\begin{array}{l}
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{Fe}} = 0,06mol\\
 \to {n_{{H_2}}} = {n_{Fe}} = 0,06mol\\
 \to {V_{{H_2}}} = 0,06 \times 22,4 = 1,344l\\
 \to {n_{{H_2}S{O_4}}} = {n_{Fe}} = 0,06mol\\
 \to {m_{{H_2}S{O_4}}} = 0,06 \times 98 = 5,88g\\
 \to C{\% _{{H_2}S{O_4}}} = \dfrac{{5,88}}{{200}} \times 100\%  = 2,94\% \\
{n_{FeS{O_4}}} = {n_{Fe}} = 0,06mol\\
 \to {m_{FeS{O_4}}} = 0,06 \times 152 = 9,12g\\
 \to {m_{{\rm{dd}}FeS{O_4}}} = {m_{Fe}} + {m_{{\rm{dd}}{H_2}S{O_4}}} - {m_{{H_2}}} = 203,24g\\
 \to C{\% _{FeS{O_4}}} = \dfrac{{9,12}}{{203,24}} \times 100\%  = 4,5\% 
\end{array}\)

2.

\(\begin{array}{l}
ZnO + {H_2} \to Zn + {H_2}O\\
{n_{ZnO}} = 0,1mol\\
 \to {n_{{H_2}}} = {n_{ZnO}} = 0,1mol\\
 \to {V_{{H_2}}} = 0,1 \times 22,4 = 2,24l\\
{n_{Zn}} = {n_{ZnO}} = 0,1mol\\
 \to {m_{Zn}} = 0,1 \times 65 = 6,5g\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{m_{HCl}} = \dfrac{{200 \times 7,3\% }}{{100\% }} = 14,6g\\
 \to {n_{HCl}} = 0,4mol\\
 \to {n_{{H_2}}} = \dfrac{1}{2}{n_{HCl}} = 0,2mol\\
 \to {V_{{H_2}}} = 0,2 \times 22,4 = 4,48l
\end{array}\)