HOW DO I DO THIS QUESTION???? 6 marker Watch

brokegirl2.0
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Ethanoic acid and ethanol react together to form the ester ethyl ethanoate,
CH3COOC2H5, and water.
CH3COOH(l) + CH3CH2OH(l) U CH3COOCH2CH3(l) + H2O(l) ǻH = –3.5 kJ mol–1

*(iii) An experiment was carried out to determine the value of Kc for this reaction.
Ɣ 0.120 mol of ethanoic acid was added to 0.220 mol of ethanol.
Ɣ 5.00 cm3
of 1.00 mol dm–3 hydrochloric acid was added as a catalyst. This
contains 0.278 mol of water.
Ɣ The mixture was left to reach equilibrium.
Ɣ The mixture was titrated with 1.00 mol dm–3 sodium hydroxide, which
reacted with both of the acids.
Ɣ The titre was 45.0 cm3
.
Use these data to determine the value for Kc.

Thanksssss
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icequeenTM
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You can’t ask someone to do a question for you. Have you even attempted it yourself?
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brokegirl2.0
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I've been stuck on it for a while, I wouldn't have asked otherwise.
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icequeenTM
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Fair enough
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brokegirl2.0
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(Original post by icequeenTM)
Fair enough
soooooooooo, any help coming my way or you only here for moral support
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icequeenTM
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(Original post by brokegirl2.0)
soooooooooo, any help coming my way or you only here for moral support
No but I hope you find your answer. Have you tried typing it all into the search bar?
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dip0
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[QUOTE=brokegirl2.0;77919678]Ethanoic acid and ethanol react together to form the ester ethyl ethanoate,
CH3COOC2H5, and water.
CH3COOH(l) + CH3CH2OH(l) U CH3COOCH2CH3(l) + H2O(l) ǻH = –3.5 kJ mol–1

*(iii) An experiment was carried out to determine the value of Kc for this reaction.
Ɣ 0.120 mol of ethanoic acid was added to 0.220 mol of ethanol.
Ɣ 5.00 cm3
of 1.00 mol dm–3 hydrochloric acid was added as a catalyst. This
contains 0.278 mol of water.
Ɣ The mixture was left to reach equilibrium.
Ɣ The mixture was titrated with 1.00 mol dm–3 sodium hydroxide, which
reacted with both of the acids.
Ɣ The titre was 45.0 cm3
.
Use these data to determine the value for Kc.

Thanksssss[/QUOTE/] this should answer your question. Just realised mol of ester is 0.08 not 0.14
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MikeyA
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^^^^^^^^^^^^^ and Kc the whole thing
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brokegirl2.0
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[QUOTE=dip0;77920160]
(Original post by brokegirl2.0)
Ethanoic acid and ethanol react together to form the ester ethyl ethanoate,
CH3COOC2H5, and water.
CH3COOH(l) + CH3CH2OH(l) U CH3COOCH2CH3(l) + H2O(l) ǻH = –3.5 kJ mol–1

*(iii) An experiment was carried out to determine the value of Kc for this reaction.
Ɣ 0.120 mol of ethanoic acid was added to 0.220 mol of ethanol.
Ɣ 5.00 cm3
of 1.00 mol dm–3 hydrochloric acid was added as a catalyst. This
contains 0.278 mol of water.
Ɣ The mixture was left to reach equilibrium.
Ɣ The mixture was titrated with 1.00 mol dm–3 sodium hydroxide, which
reacted with both of the acids.
Ɣ The titre was 45.0 cm3
.
Use these data to determine the value for Kc.

Thanksssss[/QUOTE/] this should answer your question. Just realised mol of ester is 0.08 not 0.14
Thank you soooooooo much it makes sense now. Thanks
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