4) Ethanol (6.9g) and ethanoic acid (9.0g) were kept at a constant temperature until equilibrium was reached. At this time analysis showed that 3.0g ethanoic acid remained. Determine the value of the equilibrium constant for the reaction:
CH3COOH + CH3CH20H = CH3COOCH2CH3 + H2O
5) Calculate Kc for the equilibrium below if [A] = 0.02 mol dm-3 [B2] = 0.10 mol dm-3 and [AB] = 0.40 mol dm-3
2A + B2 = 2AB
My answer for question 5 is [AB]/[A]2[B2] = [0.16] / [0.0004] [0.10] = 4000 4X105
Please can someone answer and explain asap thankyou so much, am really struggling :-(