Em tham khảo nha :
\(\begin{array}{l}
3)\\
Fe + HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,2mol\\
{m_{Fe}} = n \times M = 56 \times 0,2 = 11,2g\\
\% Fe = \dfrac{{11,2}}{{20}} \times 100\% = 56\% \\
\% Cu = 100 - 56 = 44\% \\
4)\\
F{e_x}{O_y} + yCO \to xFe + yC{O_2}\\
{n_{Fe}} = \dfrac{m}{M} = \dfrac{{33,6}}{{56}} = 0,6\\
{n_{CO}} = \dfrac{V}{{22,4}} = \dfrac{{17,92}}{{22,4}} = 0,8mol\\
{n_{CO}} = \dfrac{y}{x} \times {n_{Fe}} \Leftrightarrow 0,8 = \dfrac{y}{x} \times 0,6\\
\Rightarrow \dfrac{y}{x} = \dfrac{{0,8}}{{0,6}} = \frac{4}{3}\\
\Rightarrow CTHH:F{e_3}{O_4}
\end{array}\)
