`2Na+2H_2O->2NaOH+H_2`
x-----------------------------`1/2`x mol
`2K+2H_2O->2KOH+H_2`
y---------------------------`1/2` y mol
`n_(H_2)=(6,72)/(22,4)=0,3 mol`
Ta có phương trình :
\(\left\{{}\begin{matrix}23x+39y=9,3\\\dfrac{1}{2}x+\dfrac{1}{2}y=0,3\end{matrix}\right.\)
-> nghiệm vô lí
`#YBTran~`