Em tham khảo nha :
\(\begin{array}{l}
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
b)\\
{n_{Fe}} = \dfrac{{5,6}}{{56}} = 0,1mol\\
{m_{{H_2}S{O_4}}} = \dfrac{{100 \times 20}}{{100}} = 20g\\
{n_{{H_2}S{O_4}}} = \dfrac{{20}}{{98}} = 0,204mol\\
\dfrac{{0,1}}{1} < \dfrac{{0,204}}{1} \Rightarrow {H_2}S{O_4}\text{ dư}\\
{n_{{H_2}S{O_4}pu}} = {n_{Fe}} = 0,1mol\\
{m_{{H_2}S{O_4}pu}} = 0,1 \times 98 = 9,8g\\
{m_{{H_2}S{O_4}d}} = 20 - 9,8 = 10,2g\\
{n_{FeS{O_4}}} = {n_{Fe}} = 0,1mol\\
{m_{FeS{O_4}}} = 0,1 \times 152 = 15,2g\\
C{\% _{{H_2}S{O_4}d}} = \dfrac{{10,2}}{{5,6 + 100 - 0,1 \times 2}} \times 100\% = 9,68\% \\
C{\% _{FeS{O_4}}} = \dfrac{{15,2}}{{5,6 + 100 - 0,1 \times 2}} \times 100\% = 14,4\%
\end{array}\)
