# Edexcel ial chemistry - unit 04 - may/june 2019Watch

#1
I need help with this question(picture of question is in attachment). My answer seems to be wrong.
Please guide me on this question.
0
1 week ago
#2
n(NaOH) = 0.045 mol
n(HCl) = 5 x 10^-3 mol
n(CH3COOH) = 0.04 mol

CH3COOH + C2H5OH <-> CH3COOC2H5 + H2O
I 0.12, 0.22, 0, 0.278
E 0.04, 0.14, 0.08, 0.358
Kc = 5.1
Last edited by BobbJo; 1 week ago
0
#3
(Original post by BobbJo)
n(NaOH) = 0.045 mol
n(HCl) = 5 x 10^-3 mol
n(CH3COOH) = 0.04 mol

CH3COOH + C2H5OH <-> CH3COOC2H5 + H2O
I 0.12, 0.22, 0, 0.278
E 0.04, 0.14, 0.08, 0.358
Kc = 5.1
Why did the number of moles of products increase by 0.08 mol? Could you explain it?
Was it because that n(CH3COOH)=n(NaOH)-n(HCl) ?
Last edited by Saman_B9; 1 week ago
0
1 week ago
#4
(Original post by Saman_B9)
Why did the number of moles of products increase by 0.08 mol? Could you explain it?
Was it because that n(CH3COOH)=n(NaOH)-n(HCl) ?
0.08 mol CH3COOH reacted
from stoichiometry
1:1 ratio so 0.08 mol of each product is formed
0
#5
(Original post by BobbJo)
0.08 mol CH3COOH reacted
from stoichiometry
1:1 ratio so 0.08 mol of each product is formed
Yeah, now I understood it
0
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