2Na+2H2O--->2NaOH+H2
x----------------------------0,5x
2K+2H2O---->2KOH+H2
y--------------------------0,5y
a) n\(_{H2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
Theo bài ra ta có
\(\left\{{}\begin{matrix}23x+39y=6,2\\0,5x+0,5y=0,1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
m\(_{Na}=0,1.23=2,3\left(g\right)\)
m\(_K=0,1.39=3,9\left(g\right)\)
b) Theo pthh
n\(_{kiềm}=n_{KL}=0,2\left(mol\right)\)
m\(_{kiềm}=0,2\left(40+56\right)=19,2\left(g\right)\)
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