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Topic: ideal laws & decomposition  (Read 6668 times)

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Offline jwxie

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ideal laws & decomposition
« on: October 26, 2009, 10:39:50 PM »
Quote
Ammonium nitrite undergoes decomposition to produce only gases:
NH4NO2 (s) ->  N2(g) + H2O (g)
How many liters of gas will be produced by the decomposition of 32.0g of NH4NO2 at 525C and 1.5 atm?

My approach is this:

finds the mole (n) of one of the gases (let say N2)

32g NH4NO2 x  (1 mole / molar mass of NH4NO2) x (1 mole N2 / 1 mole NH4NO2)
This stioche will gives us the mole of N2 presents

Now I will use the gase law forumla  PV = nRT to find V (N2)

So N2 = x amount in liters

but since there are more than one gases, so I have to do similar setup for H2O (g)

My question is, am I correct about the approach, that the answer (liter) is equal to N2 (v) + H2O (v).

In general,
2 NaN3(s) -->  2Na(l) + 3N2(g) only one gas so I only find one
but in that question, I have two gases, H2O and N2.

So the total volume is = N2+H2O. am I correct? or am I wrong?

Thanks

Offline orgoclear

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Re: ideal laws & decomposition
« Reply #1 on: October 27, 2009, 01:24:24 AM »
NH4NO2 (s) ->  N2(g) + H2O (g)

You have to balance this equation first to get the stoichiometric coefficients. I dont know how you can calculate the moles of N2 and H2O produced without knowing the stoichiometric ratios

NH4NO2 -> N2 + 2H2O

mol. mass of NH4NO2 = 64 g. So moles of NH4NO2 present is 0.5

So moles of N2 produced = 0.5 and moles of H2O produced = 1

Now you can use PV = nRT
where P=1.5 atm V=liters of gas produced, n=1.5 R= Gas Constant and T=525+273 K


Coming back to your method,
How will you calculate V(N2) without knowing P(N2) ?
Remember when more than one gas is present each gas will contribute some amount of pressure to the total

From Dalton's Law of Partial Pressures,
P(N2) = P(total)*X(N2) where X(N2) = mole fraction of N2
and P(H2O) = P(total)*X(H2O) where X(H2O)= mole fraction of H2O

Now you can use PV = nRT for each case and add the volumes obtained

But it remains to see that the first method is far more easier to calculate and that both methods are equivalent

Offline jwxie

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Re: ideal laws & decomposition
« Reply #2 on: October 27, 2009, 09:14:03 AM »
oh right... i didn't see it was H4, instead I saw H2
thanks for the correction and great reply.


Just a question. If only one gas is present, then the given 1.5atm is the pressure of that gas???

and here, 1.5atm is the total pressure of the N2 and H2O??

Offline BetaAmyloid

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Re: ideal laws & decomposition
« Reply #3 on: October 27, 2009, 10:34:36 AM »
Just a question. If only one gas is present, then the given 1.5atm is the pressure of that gas???

and here, 1.5atm is the total pressure of the N2 and H2O??

1) Yes, the gas given would be at 1.5 atm and the other would be assumed to be at that same pressure unless you were calculating for the new pressure of the ungiven gas.

2) No, it is saying that both gases are at 1.5 atm unless it states otherwise. 
Discovery consists of seeing what everybody has seen and thinking what nobody has thought - Albert Szent-Györgyi

Offline jwxie

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Re: ideal laws & decomposition
« Reply #4 on: October 27, 2009, 12:56:54 PM »
then since it's a mixture of gases, then the P(total) = P(N2) + P(H2O)???

I mean I see how each gas contribute partial pressure...
« Last Edit: October 27, 2009, 01:14:59 PM by jwxie »

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