a.C2H5OH + Mg ----/----> ko xay ra
2CH3COOH + Mg --------> ( CH3COO)2Mg + H2
b. Ta co:
n H2=1,12/22,4=0,05 mol
Theo PTHH : n CH3COOH= 2 n H2 =0,1 mol
=> m CH3COOH= 0,1.60=6g
=> m C2H5OH=8,3-6=2,3g
=> %m C2H5OH=2,3/8,3.100%=27,7%
=> %m CH3COOH=100%-27,7%=72,3%
c.Theo PTHH : n Mg = n H2=0,05 mol
=> m Mg =0,05.24=1,2g