Giải thích các bước giải:

1,

\(\begin{array}{l}
3Fe + 2{O_2} \to F{e_3}{O_4}\\
{n_{Fe}} = 0,033mol \to {n_{F{e_3}{O_4}}} = \dfrac{1}{3}{n_{Fe}} = 0,011mol\\
{m_{F{e_3}{O_4}}} = 2,552g\\
{n_{{O_2}}} = \dfrac{2}{3}{n_{Fe}} = 0,022mol \to {V_{{O_2}}} = 0,4928l\\
{V_{KK}} = 5{V_{{O_2}}} = 2,464l
\end{array}\)

2,

\(\begin{array}{l}
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{Zn}} = 0,2mol = {n_{{H_2}}} \to {V_{{H_2}}} = 4,48l\\
CuO + {H_2} \to Cu + {H_2}O\\
{n_{CuO}} = 0,15mol\\
{n_{{H_2}}} > {n_{CuO}}
\end{array}\)

Hiđro dư

\({n_{Cu}} = {n_{CuO}} = 0,15mol \to {m_{Cu}} = 9,6g\)

3,

\(\begin{array}{l}
Zn + HCl \to ZnC{l_2} + {H_2}\\
{n_{Zn}} = 0,3mol = {n_{{H_2}}} \to {V_{{H_2}}} = 6,72l\\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
{n_{F{e_2}{O_3}}} = 0,12mol\\
\dfrac{{{n_{{H_2}}}}}{3} < {n_{F{e_2}{O_3}}}
\end{array}\)

\(F{e_2}{O_3}\) dư

\({n_{Fe}} = \dfrac{2}{3}{n_{{H_2}}} = 0,2mol \to {m_{Fe}} = 11,2g\)

4,

\(N{a_2}O,{K_2}O,CaO,S{O_2},BaO,MgO,FeO,CuO,ZnO,{P_2}{O_5},HgO,NO\)