a,
$2K+2H_2O\to 2KOH+H_2$
$2K+2C_2H_5OH\to 2C_2H_5OK+H_2$
b,
$V_{C_2H_5OH}=40,25.50\%=20,125(ml)$
$D_{C_2H_5OH}=0,8g/ml\to to m_{C_2H_5OH}=D.V=20,125.0,8=16,1g$
c,
$n_{C_2H_5OH}=\dfrac{16,1}{46}=0,35(mol)$
$V_{H_2O}=40,25-20,125=20,125(ml)$
$D_{H_2O}=1g/ml\to m_{H_2O}=DV=20,125g$
$\to n_{H_2O}=\dfrac{20,125}{18}=\dfrac{161}{144}(mol)$
Ta có:
$2n_{H_2}=n_{C_2H_5OH}+n_{H_2O}$
$\to n_{H_2}=\dfrac{1057}{1440}(mol)$
$\to V_{H_2}=22,4n_{H_2}=16,44l$