2Na + 2H2O -> 2NaOH + H2 (1)
2K + 2H2O -> 2KOH + H2 (2)
nH2=0,15(mol)
Đặt nNa=a
nK=b
Ta có hệ:
\(\left\{{}\begin{matrix}23a+39b=7,7\\0,5\left(a+b\right)=0,15\end{matrix}\right.\)
=>a=0,25;b=0,05
mNa=0,25.23=5,75(g)
%mNa=\(\dfrac{5,75}{7,7}.100\%=74,675\%\)
%mK=100-74,675=25,325%