Em tham khảo nha :
\(\begin{array}{l}
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}(1)\\
Fe + CuS{O_4} \to FeS{O_4} + Cu(2)\\
{n_{{H_2}}} = \dfrac{{0,56}}{{22,4}} = 0,025mol\\
{n_{Fe(1)}} = {n_{{H_2}}} = 0,025mol\\
{m_{Fe(1)}} = 0,025 \times 56 = 1,4g\\
{m_{Fe(2)}} = 2{m_{Fe(1)}} = 2,8g\\
{n_{Fe(2)}} = \dfrac{{2,8}}{{56}} = 0,05mol\\
{n_{Cu}} = {n_{Fe(2)}} = 0,05mol\\
{m_{Cu}} = 0,05 \times 64 = 3,2g\\
b)\\
FeS{O_4} + BaC{l_2} \to BaS{O_4} + FeC{l_2}\\
{n_{FeS{O_4}(1)}} = {n_{Fe}} = 0,025mol\\
{n_{BaS{O_4}}} = {n_{FeS{O_4}(1)}} = 0,025mol\\
{m_{BaS{O_4}}} = 0,025 \times 233 = 5,825g
\end{array}\)
