100% vin............................ 11,5% etanol
100 kg vin ......................... m = 11,5 kg etanol
a)
CH3CH2OH --bacterium aceticum--> CH3COOH + H2O
b)
11,5 kg n kg
CH3CH2OH --bacterium aceticum--> CH3COOH + H2O
46 60
=> n = 11,5x60/46 = 15 kg acid acetic