pt 2 CH3COOH+Mg\(\rightarrow\)(CH3COO)2Mg+H2
nMg=7,2:24=0,3 mol
mCH3COOH=120*20:100=24g suy ra nCH3COOH=0,4 mol
theo pt thì Mg dư
theo pt n(CH3COO)2Mg=1/2nCH3COOH=0,2 mol suy ra n(CH3COO)2Mg=28,4g
mdd sau phản ứng =7,2+120-0,4=126,8 g
suy ra C%(CH3COO)Mg=22,3%