27/
[tex]n_{H_2}=2,24:22,4=0,1[/tex]
[tex]2\overline{M}+2H_2O\rightarrow 2\overline{M}OH+H_2[/tex]
[tex]0,2<----------0,1[/tex]
[tex]\overline{M}=\frac{6,2}{0,2}=31\Rightarrow Na;K[/tex]
[tex]\left\{\begin{matrix} x+y=0,2 & \\ 23x+39y=6,2& \end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=0,1 & \\ y=0,1& \end{matrix}\right.[/tex]
%[tex]m_{Na}=\frac{0,1.23}{6,2}=37[/tex] (đáp án đề sai)
%[tex]m_{K}=100-37=63[/tex]
28/
[tex]n_M=\frac{5,4}{M};n_{M_2O_3}=\frac{10,2}{2M+48}[/tex]
[tex]4M+3O_2\rightarrow 2M_2O_3[/tex]
Theo PT ta có:
[tex]\frac{10,2}{2M+48}.2=\frac{5,4}{M}[/tex]
[tex]\Rightarrow M=27(AL)[/tex]
[tex]V_{O_2}=0,29.V_{kk}\Rightarrow V_{kk}=\frac{V_{O_2}}{0,29}=\frac{0,15.22,4}{0,29}=11,59(l)[/tex]