Em tham khảo nha:
\(\begin{array}{l}
8)\\
2M + 2{H_2}O \to 2MOH + {H_2}(1)\\
{M_2}O + {H_2}O \to 2MOH(2)\\
2MOH + {H_2}S{O_4} \to {M_2}S{O_4} + 2{H_2}O(3)\\
{n_{MOH(3)}} = 2{n_{{H_2}S{O_4}}} = 2 \times 0,1 \times 2 = 0,4\,mol\\
{n_{{H_2}}} = \dfrac{{1,12}}{{22,4}} = 0,05mol \Rightarrow {n_{MOH(1)}} = 2{n_{{H_2}}} = 0,1\,mol\\
{n_{MOH(2)}} = 0,4 - 0,1 = 0,3\,mol\\
{n_M} = {n_{MOH(1)}} = 0,1mol\\
{n_{{M_2}O}} = \dfrac{{0,3}}{2} = 0,15\,mol\\
{m_{hh}} = 18g \to 0,1 \times {M_M} + 0,15 \times (2{M_M} + 16) = 18\\
\Rightarrow {M_M} = 39g/mol \Rightarrow M:Kali(K)\\
9)\\
CuO + {H_2} \to Cu + {H_2}O\\
F{e_3}{O_4} + 4{H_2} \to 3Fe + 4{H_2}O\\
hh:CuO(a\,mol),F{e_3}{O_4}(b\,mol)\\
\left\{ \begin{array}{l}
80a + 232b = 29,6\\
3b \times 56 - 64a = 4
\end{array} \right.\\
\Rightarrow a = 0,25;b = 0,12\\
m = 0,25 \times 80 + 0,12 \times 232 = 47,84g\\
10)\\
CTHH:F{e_x}{O_y}\\
x:y = \dfrac{{{m_{Fe}}}}{{56}}:\dfrac{{{m_O}}}{{16}} = \dfrac{7}{{56}}:\dfrac{3}{{16}} = 2:3\\
CTHH:F{e_2}{O_3}
\end{array}\)
