Quote:
Originally Posted by Hype en Ecosse
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This one's simple enough. Keep in mind here that we can't omit water as a factor like we could if the solutions were aqueous.
0.2 moles of ethanol and methanoic acid have been mixed to give 0.15 moles of ethyl methanoate (therefore, 0.15 moles of water) at equilibrium. Meaning we've got 0.05 moles of ethanol and methanoic acid left. When calculating your equilibrium concentration of stuff you have to make sure that you ratio correctly - dead easy here because everything's 1:1
![K = \frac{[HCOOC_2H_5][H_2O]}{[C_2H_5OH][HCOOH]}]( "K = \frac{[HCOOC_2H_5][H_2O]}{[C_2H_5OH][HCOOH]}")
Plug the numbers in, there's your answer. ;)
I get 9.
This ones a bit harder because it's not equimolar and different numbers of moles of reactants are required for the reaction, but if you can understand what I've done above, you might be able to work it out from here.
0.2 moles of ethanol and methanoic acid have been mixed to give 0.15 moles of ethyl methanoate (therefore, 0.15 moles of water) at equilibrium. Meaning we've got 0.05 moles of ethanol and methanoic acid left. When calculating your equilibrium concentration of stuff you have to make sure that you ratio correctly - dead easy here because everything's 1:1
![K = \frac{[HCOOC_2H_5][H_2O]}{[C_2H_5OH][HCOOH]}](http://www.thestudentroom.co.uk/latexrender/pictures/1f/1f11c59e4dc6c5340422d3450720c2d1.png "K = \frac{[HCOOC_2H_5][H_2O]}{[C_2H_5OH][HCOOH]}")
Plug the numbers in, there's your answer. ;)
I get 9.
This ones a bit harder because it's not equimolar and different numbers of moles of reactants are required for the reaction, but if you can understand what I've done above, you might be able to work it out from here.