Em tham khảo nha :
\(\begin{array}{l}
1)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{Fe}} = \dfrac{{5,6}}{{56}} = 0,1mol\\
{n_{{H_2}}} = {n_{Fe}} = 0,1mol\\
{V_{{H_2}}} = 0,1 \times 22,4 = 2,24l\\
{n_{{H_2}S{O_4}}} = {n_{Fe}} = 0,1mol\\
{m_{{H_2}S{O_4}}} = 0,1 \times 98 = 9,8g\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{9,8 \times 100}}{{40}} = 24,5g\\
{n_{FeS{O_4}}} = {n_{Fe}} = 0,1mol\\
{m_{FeS{O_4}}} = 0,1 \times 152 = 15,2g\\
C{\% _{FeS{O_4}}} = \dfrac{{15,2}}{{24,5 + 5,6 - 0,1 \times 2}} \times 100\% = 50,4\% \\
2)\\
{H_2}S{O_4} + BaC{l_2} \to BaS{O_4} + 2HCl\\
{n_{{H_2}S{O_4}}} = 0,05 \times 2 = 0,1mol\\
{n_{HCl}} = 2{n_{{H_2}S{O_4}}} = 0,2mol\\
{C_{{M_{HCl}}}} = \dfrac{{0,2}}{{0,05}} = 4M\\
3)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{Zn}} = \dfrac{{12,8}}{{65}} = 0,2mol\\
{n_{{H_2}}} = {n_{Zn}} = 0,2mol\\
2{H_2} + {O_2} \to 2{H_2}O\\
{n_{{H_2}O}} = {n_{{H_2}}} = 0,2mol\\
{m_{{H_2}O}} = 0,2 \times 18 = 3,6g
\end{array}\)
