Em xem lại đề nha:
\(\begin{array}{l}
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
2Al + 2NaOH + 2{H_2}O \to 2NaAl{O_2} + 3{H_2}\\
Fe + CuS{O_4} \to FeS{O_4} + Cu\\
{n_{Cu}} = \dfrac{{3,2}}{{64}} = 0,05\,mol\\
{n_{Fe}} = {n_{Cu}} = 0,05\,mol\\
{m_{Fe}} = 0,05 \times 56 = 2,8g
\end{array}\)