Homework 11.41

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Angela_Pu_3C
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Joined: Fri Jul 15, 2016 3:00 am

Homework 11.41

Postby Angela_Pu_3C » Mon Nov 28, 2016 7:37 pm

Why do we plug in 1.581x10-3? The x (or molar concentration of CO2) is 1.581M (resulting in Kc=1.58) so I'm not sure how/why we use 1.581x10-3 instead. Thank you!

Joyce Wu 3E
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Joined: Fri Jul 22, 2016 3:00 am
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Re: Homework 11.41

Postby Joyce Wu 3E » Mon Nov 28, 2016 7:54 pm

We plug in 1.58x10^-3 because that is the molarity (mols/Liter) of CO2. The problem says that at equilibrium, there are 0.0174 grams of CO2. Convert that to mols, it is 3.95x10^-4 mols. We are told that this is within 0.250 liters; thus, the molarity of CO2 would be (3.95x10^-4 mol)/(.250 L) = 1.58x10^-3

I hope this helps!

Angela_Pu_3C
Posts: 22
Joined: Fri Jul 15, 2016 3:00 am

Re: Homework 11.41

Postby Angela_Pu_3C » Mon Nov 28, 2016 7:58 pm

Yeah I realized I forgot a conversion thank you!

Matt_Coopersmith_4H
Posts: 20
Joined: Wed Sep 21, 2016 2:57 pm

Re: Homework 11.41

Postby Matt_Coopersmith_4H » Mon Nov 28, 2016 7:59 pm

Hi! When looking at the problem, make sure to play close attention to the units. For the NH4(NH2CO2), the question gives you 25.0 g, so you find the molarity by dividing by the molar mass to get moles, then by diving by liters (.250).

However, the problem also gives you 17.4 mg. 17.4 mg translates into .0174 g, which you can then use to find the Molarity of CO2. This is how you get a Molarity of .00158 or 1.58 x 10^-3.

From here, you would use a rice table and use the equilibrium constant equation to find that Kc = 1.58 x 10^(-8)

Hope this helps!


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