# Homework Help: Finding mole percent

1. Oct 9, 2011

### Saitama

1. The problem statement, all variables and given/known data
A mixture of NH3 and N2H4 is placed in a sealed container at 300K. The total pressure is 0.5atm. The container is heated to 1200K, when both substances decompose completely according to the equations,
2NH3--->N2+3H2
N2H4--->N2+2H2
After decomposition is complete, the total pressure at 1200K is found to be 4.5atm. Find the mole percent of N2H4 in the original mixture.

2. Relevant equations

3. The attempt at a solution
I am not getting any idea on how to start.
Please give me some ideas on beginning with it.

Thanks! :)

2. Oct 10, 2011

### Staff: Mentor

Assume the initial partial pressures of both were p0 and p1.

Can you express both initial and final pressures as functions of p0 and p1?

3. Oct 10, 2011

### Saitama

I can express initial pressure like this:-
p0= 0.5 x XNH3
p1= 0.5 x XN2H4

XNH3--> mole fraction of NH3
XN2H4---->mole fraction of N2H4.

But i don't have the moles. :(

4. Oct 10, 2011

### Staff: Mentor

Why not p0 + p1 = 0.5 atm?

You don't need moles - you can express molar fraction using just total and partial pressures.

5. Oct 10, 2011

### Saitama

Yes, we can do p0+p1=0.5 atm.

But why then we need molar fraction here?

6. Oct 10, 2011

### Staff: Mentor

Molar fraction, mole percent - same thing, just expressed in slightly different way. That's what you have to calculate, aren't you?

7. Oct 10, 2011

### Saitama

Yeah I have to calculate that.

I am having problems finding the final pressure. The gases has changed to N2 and H2. What should i do now?

8. Oct 10, 2011

### Staff: Mentor

You can find ratio between initial and final number of moles from the stoichiometry. A volume doesn't change, number of moles is directly related to the pressure.

9. Oct 19, 2011

### Saitama

Sorry Borek. I did not replied to this post for a long time. :)

I worked on your hints and they proved to be very useful. I was only stuck at the stoichiometric calculation which was the easiest part. :P

Here's how i solved it:-

I assumed moles of NH3 as 'a' and that of N2H4 as 'b'.

Then using stoichiometry i found the total moles of N2 as $\frac{a+2b}{2}$ and of H2 as $\frac{3a+4b}{2}$.

Before decomposition, using ideal gas equation:-
(0.5V)/(300R)=(a+b)

After decomposition, using ideal gas equation:-
(4.5V)/(1200R)=(2a+3b)

After solving, i get a=3b.

Mole fraction of N2H4= (b)/(a+b)=b/4b=0.25

Since we are asked to find mole percent, therefore 0.25*100=25%.

Thanks for the help Borek.

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