HW 11.41

Moderators: Chem_Mod, Chem_Admin

Pauline Tze 3B
Posts: 57
Joined: Sat Jul 09, 2016 3:00 am
Been upvoted: 1 time

HW 11.41

Postby Pauline Tze 3B » Sun Nov 13, 2016 8:58 pm

Screen Shot 2016-11-13 at 8.48.42 PM.png


I understand how the solution manual answers this question (though is the first step where it converts 25.0g of the reactant to moles seems irrelevant), but I'm not understanding why my approach didn't work.
I tried to find the concentration of CO2 and NH3 separately, then plug that back into the Kc equation.
For CO2, the concentration was correct, but for NH3, I ended up with a much larger concentration than the answer.
Why doesn't it work to subtract 17.4 mg from 25.0g, convert that to moles of NH3 then divide by 0.250 L?

Is it that in decomposition chemical equations, the mass isn't preserved on both sides of the equation?

Thank you in advance!

Hao 1I
Posts: 26
Joined: Wed Sep 21, 2016 3:00 pm
Been upvoted: 1 time

Re: HW 11.41

Postby Hao 1I » Sun Nov 13, 2016 10:55 pm

Subtracting 17.4g of CO2 from 25.0g of ammonium carbonate does not work. 25.0g of ammonium carbonate is present initially (before any reaction happened) but only 17.4g CO2 are present in equilibrium. Because 17.4g of CO2 are present in equilibrium, there are still still some amount of ammonium carbonate left from the left hand side of the equation but not all of 25.0g.


Return to “Equilibrium Constants & Calculating Concentrations”

Who is online

Users browsing this forum: No registered users and 28 guests