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Topic: How does one solve a volume - volume limiting reagent problem?  (Read 13515 times)

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Offline BookwormElena

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How does one solve a volume - volume limiting reagent problem?
« on: October 09, 2010, 05:56:12 PM »
I know how to find limiting reagent problems when given to masses, but I am having difficulty trying to figure out how to find the limiting reagent when given volumes and two different coefficients. ex:
   2 N2(g) + 3 O2(g)  :rarrow: 2 N2O3

The nitrogen gas has a volume of 4.2 L and the oxygen has a volume of 6.3 L.
   Therefore would you divide the L of oxygen by 3 to see if there is enough nitrogen? Or vice versa? Since you can't find moles by dividing by the molar mass I am at a loss.

Any explanation would be highly appreciated! =)

Offline Borek

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Re: How does one solve a volume - volume limiting reagent problem?
« Reply #1 on: October 09, 2010, 06:47:30 PM »
Do you know ideal gas equation? Do you know Avogadro's law? What volume does 1 mole of gas occupy at STP?
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Offline BookwormElena

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Re: How does one solve a volume - volume limiting reagent problem?
« Reply #2 on: October 09, 2010, 06:53:59 PM »
Yes, the ideal gas equation is PV=nRT, and Avagadro's number is 6.022 x 1023 molecules/atoms per element/compound. One mol of liquid occupies 22.4 L at STP, but I'm not sure what mol of gas would occupy. The problem is set in STP so no need for any conversions right?

Offline BookwormElena

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Re: How does one solve a volume - volume limiting reagent problem?
« Reply #3 on: October 09, 2010, 07:09:34 PM »
If I divide the 6.3L of O2 by 2, and divide the 4.2L of N2 by 2 also, I should have approximately 2 times more L of O2 than N2 right?

Offline josharoon

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Re: How does one solve a volume - volume limiting reagent problem?
« Reply #4 on: October 09, 2010, 09:45:01 PM »
One mol of liquid occupies 22.4 L at STP, but I'm not sure what mol of gas would occupy.
Its actually one mole of gas which occupies 22.4L at STP.

Now I'm not 100% on how to answer this question but here it goes,
If you know that one mole = 22.4L you can find the amount of moles of each gas:
4.2/22.4=0.18750 <-- (moles of N2)
6.3/22.4=0.28125 <-- (moles of O2)

So as we know the ratio in the equation is 2N:3O we will have to divide the moles by their ratio so its 1:1 hence giving us the restricting reagent.
0.18750/2=0.9375
0.28125/3=0.9375
Therefore given your amounts of each gas there shouldn't be any excess of either  ;)

Offline Borek

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Re: How does one solve a volume - volume limiting reagent problem?
« Reply #5 on: October 10, 2010, 03:59:08 AM »
Avagadro's number is 6.022 x 1023 molecules/atoms per element/compound.

Avogadro's law says something about number of molecules in identical volume of gases.

As you were already told, 22.4L is a volume of gas, not liquid.
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Offline BookwormElena

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Re: How does one solve a volume - volume limiting reagent problem?
« Reply #6 on: October 10, 2010, 10:57:58 AM »

Yay! Thank you so much! That makes sense now. I was always tripped up on limiting reagent problems with high coefficients, but now I understand. Thank you!

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