n H2 = \(\dfrac{3,36}{22,4}\)= 0,15 mol
ptpu : Mg + 2 CH3COOH \(\overrightarrow{ }\)(CH3COO)2Mg + H2
0,15 0,3 0,15
theopu nMg= 0,15mol \(\Rightarrow\) mMg = 0,15 * 24= 3,6 g
b.
thep pu n CH3COOH ct = 0.3 * 60=18g
C% CH3COOH= \(\dfrac{18}{400}\) *100= 4,5%
c.
CH3COOH + C2H5OH\(\rightarrow\) CH3COOC2H5 + H2O
0,3 \(\rightarrow\) 0,3
metyaxetat lt = 0,3*88= 26,4g
m etyaxetat tt= 26,4 * 80=%=21,12g