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Quick Equilibria Help

When 4.50 g of ethanoic acid and 10.0 g of ethanol were allowed to reach equilibrium at60°C, the equilibrium mixture was found to contain 4.40 g of ethyl ethanoate. Calculate thepercentage yield of ethyl ethanoate, based on ethanoic acid. Show your working wherenecessary.

My method
Theoretical
CH3COOH + CH3CH3OH ===== CH3COOCH2CH3 + H2O
mol CH3COOH - 0.075
mol CH3CH2OH - 5/23
mol CH3COOCH2CH3 formed - 5/23
mass CH3COOCH2CH3 formed - 88 x 5/23

Therefore % yield is 4.4/mass above x 100 = 23%

What have I done wrong?
Reply 1
0.075 < 5/23

n(ethylethanoate)theory = n(ethanol)

n(ethylethanoate)achieved = 0.05

% = 0.05/0.075 x100 = 67%
Reply 2
Original post by Pigster
0.075 < 5/23

n(ethylethanoate)theory = n(ethanol)

n(ethylethanoate)achieved = 0.05

% = 0.05/0.075 x100 = 67%


Thanks!!!!!!!!!!!!

Could you help me with this thread please? http://www.thestudentroom.co.uk/showthread.php?t=3284233&p=55296221#post55296221

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