nK = 7,8 : 39 = 0,2 (mol)
pthh : 2K + 2H2O ---> 2KOH + H2
0,2--------------------------> 0,1 (mol)
=> VO2 = 0,1 . 22,4 = 2,24 (l)
a) \(PTHH:2K+2H_2O\rightarrow2KOH+H_2\)
b) \(n_K=\dfrac{m}{M}=\dfrac{7,8}{39}=0,2\left(mol\right)\)
Theo pthh: \(n_{H_2}=\dfrac{1}{2}n_K=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)
\(V_{H_2}=n.22,4=0,1.22,4=2,24\left(l\right)\)