Em tham khảo nha :
\(\begin{array}{l}
a)\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_{{H_2}S{O_4}}} = 0,2 \times 2 = 0,4mol\\
{n_{{H_2}}} < {n_{{H_2}S{O_4}}} \Rightarrow {H_2}S{O_4}\text{ dư}\\
hh:Mg(a\,mol);Fe(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,2\\
24a + 56b = 8
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,1\\
{m_{Mg}} = 0,1 \times 24 = 2,4g\\
\% Mg = \dfrac{{2,4}}{8} \times 100\% = 30\% \\
\% Fe = 100 - 30 = 70\% \\
c)\\
{n_{{H_2}S{O_4}d}} = {n_{{H_2}S{O_4}}} - {n_{{H_2}}} = 0,4 - 0,2 = 0,2mol\\
{n_{MgS{O_4}}} = {n_{Mg}} = 0,1mol\\
{n_{FeS{O_4}}} = {n_{Fe}} = 0,1mol\\
BaC{l_2} + {H_2}S{O_4} \to BaS{O_4} + 2HCl\\
BaC{l_2} + MgS{O_4} \to BaS{O_4} + MgC{l_2}\\
BaC{l_2} + FeS{O_4} \to BaS{O_4} + FeC{l_2}\\
{n_{BaS{O_4}}} = {n_{{H_2}S{O_{4d}}}} + {n_{FeS{O_4}}} + {n_{MgS{O_4}}} = 0,4mol\\
{m_{BaS{O_4}}} = 0,4 \times 233 = 93,2g
\end{array}\)
