nCH3COOH=12/60=0,2moli
nC2H5OH=9,2/46=0,2moli
CH3COOH+C2H5OH=CH3COOC2H5+H2O
Initial 0,2moli 0,2moli - -
transformt x x 0 0
echilibru 0,2-x 0,2-x x x
Kc=x²/(0,2-x)(0,2-x)
4=x²/(0,04-0,4x+x²)
4x²-1,6x+0,16=x²
3x²-1,6x+0,16=0
Δ=√b²-4ac=√2,56-1,92=√0,64=0,8
x1=1,6+0,8/6=2,4/6=0,4moli variantă incorectă
x2=1,6-0,8/6=0,8/6=0,13moli
deci la echilibru vor fi 0,1333moli de ester iar masa este=0,1333x88=11,733 g