a, 2CH3COOH + Mg --> (CH3COO)2Mg + H2
nCH3COOH= 0,1 .2 = 0,2 (mol)
n(CH3COO)2Mg=\(\dfrac{1}{2}\)nCH3COOH= \(\dfrac{1}{2}.0,2=0,1\left(mol\right)\)
--> m(CH3COO)2Mg= 0,1. 142=14,2 (g)
c,nH2=\(\dfrac{1}{2}\)nCH3COOH= \(\dfrac{1}{2}.0,2=0,1\left(mol\right)\)
--> VH2= 0,1.22,4=2,24 (l)