acid + alcool ⇄ ester + H2O
initial : a moli b moli 0 0
au reactionat: x x = 0,8b
la echilibru : (a-x) (b-x) = 0,2b x x
" "" " ' ' ' ' ' 2 moli
ester : CH3 - COO-CH2-C6H5 M = 150g/mol⇒ n ester = 1 = x
⇒ 0,8b = 1 b = 1,25moli a - 1 = 2 a = 3 moli
K = 1·1/(2·0,25) = 2 C acid = 1·100/3 = 33,33%
A.: adevarat B: fals C: adevarat D: adevarat E : adevarat