Em tham khảo nha:
\(\begin{array}{l}
1)\\
2KMn{O_4} \xrightarrow{t^0} {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
PbO + {H_2} \xrightarrow{t^0} Pb + {H_2}O\\
S + {O_2} \xrightarrow{t^0} S{O_2}\\
4Al + 3{O_2} \xrightarrow{t^0} 2A{l_2}{O_3}\\
2Mg + {O_2} \xrightarrow{t^0} 2MgO\\
2)\\
RO + {H_2} \xrightarrow{t^0} R + {H_2}O\\
{n_{RO}} = {n_R} \Leftrightarrow \dfrac{4}{{{M_R} + 16}} = \dfrac{{3,2}}{{{M_R}}}\\
\Rightarrow {M_R} = 64g/mol \Rightarrow R:Cu\\
3)\\
a)\\
{n_{F{e_2}{O_3}}} = \dfrac{{16 \times 75\% }}{{160}} = 0,075\,mol\\
{n_{CuO}} = \dfrac{{16 \times 25\% }}{{80}} = 0,05\,mol\\
F{e_2}{O_3} + 3{H_2} \xrightarrow{t^0} 2Fe + 3{H_2}O\\
CuO + {H_2} \xrightarrow{t^0} Cu + {H_2}O\\
{n_{Fe}} = 2{n_{F{e_2}{O_3}}} = 0,15\,mol\\
{n_{Cu}} = {n_{CuO}} = 0,05\,mol\\
{m_{Fe}} = 0,15 \times 56 = 8,4g\\
{m_{Cu}} = 0,05 \times 64 = 3,2g\\
b)\\
{n_{{H_2}}} = 0,075 \times 3 + 0,05 = 0,275\,mol\\
{V_{{H_2}}} = 0,275 \times 22,4 = 6,16l
\end{array}\)
