chemistry 2011 p2 variant 22

[Arslanjaved]

guys i am confused on this question here in paper 2011 mayjune variant 22
Q1
the part in which table has to be filled i am confused on the values we have to fill on the equilibrium stage. My answers to previous parts are correct.

 

[littlecloud11]

guys i am confused on this question here in paper 2011 mayjune variant 22
Q1
the part in which table has to be filled i am confused on the values we have to fill on the equilibrium stage. My answers to previous parts are correct.

no of moles of CH3CO2H = no of moles of ROH = .04 at equilibrium
no of moles of CH3 CO2R = no of moles of H2O = .06 at equilibrium

you found the number of moles that reacted with NaOH ryt? and tht was .04
after the reaction with the alcohol was complete the remaining acid was titrated with NaOH. If .04 moles reacted with NaOH and initially .1 moles were present, then .06 moles of acid reacted with the alcohol. so the acid moles present at equilibrium were .1-.06= .04
and as the acid to ester and acid to water ration was 1:1, if .06 moles of acid reacted, .06 moles of ester and water was formed.

 

[biba]

i was stuck at the same part. thanx ​