a,
$2K+2H_2O\to 2KOH+H_2$
$2K+2C_2H_5OH\to 2C_2H_5OK+H_2$
b,
$V_{C_2H_5OH}=56,25.46\%=25,875(ml)$
$D_{C_2H_5OH}=0,8g/ml\to to m_{C_2H_5OH}=D.V=25,875.0,8=20,7g$
c,
$n_{C_2H_5OH}=\dfrac{20,7}{46}=0,45(mol)$
$V_{H_2O}=56,25-25,875=30,375(ml)$
$D_{H_2O}=1g/ml\to m_{H_2O}=DV=30,375g$
$\to n_{H_2O}=\dfrac{30,375}{18}=1,6875(mol)$
Ta có:
$2n_{H_2}=n_{C_2H_5OH}+n_{H_2O}$
$\to n_{H_2}=\dfrac{1,6875+0,45}{2}=1,06875(mol)$
$\to V_{H_2}=1,06875.22,4=23,94l$