a)
nCO2 = 0,3mol => nC=0,3 mol
nH2O = 0,45 mol => nH = 0,9 mol
=> mO (A) = 6,9-0,3.12-0,9=2,4 g
=> nO (A) = 2,4:16=0,15 mol
A: CxHyOz
x:y:z=0,3:0,9:0,15=2:6:1
=> (12.2+6+16)n=23.2
=> n =1
=> A là C2H6O
b) CH3-CH2-OH
nC2H5OH = nH2O - nCO2 = 0,15 mol
C2H5OH + Na = C2H5ONa + 1/2 H2
0,15..............................................0,075
=> V H2 = 0,075.22,4=1,68 lít