Chemistry- ideal gases

  • Thread starter Dell
  • Start date
  • #1
590
0
a container with a volume of 61cm3 is filled with gas F2 at a pressure of 250mmHg and a temp of -85 C. a small amount of solid P4 is put into the container and a reaction occurs according to

P4(s) + 6F2(g)--> 4PF3(g)

the container is again cooled to -85 C and the pressure is now measured at 203mmHg. find the mass of the PF3

V=0.061 L
T=188.15 K
Pi=0.329atm
Pf=0.267atm

now i dont know if all of the F reacted since i dont know how many mols of P there were so i cant just find the mols of F and multiply. but the most i could have is 2/3 the amount of mols of F2

using PV=nRT i can say that sincs , V,R,T=const
RT/V=P/n=const
now to find the amount of mols at the beginning, n=PV/RT=1.3*10-3mol F2

(P/n)i=(P/n)f and i know everything except for n (PF3) which i calculate to come to 1.055*10-3mol

from here i can work out the mass of the PF3

m=n*mw=1.055*10-3mol*(30.97+3*19)g/mol=0.093g

but if i look at the amount of mols that i have here, something doesnt add up since i was meant to have 2/3 the amount of mols of F2 but i have about 0.8 times the amount of mols of F2, meaning that i have more mols than i should according to the equation.

what is wrong here?
 

Answers and Replies

  • #2
Borek
Mentor
28,886
3,431
Not all F2 reacted.

Think what unknowns do you have and what equations can you write to describe vessel content after the reaction ended.
 
  • #3
590
0
pv=nrt, i know the pressure, volume, and temp, only n is missing,
n=pv/rt=(0.267*0.061)/(188.15*0.08205)=1.055*10^-3mol

is this right??
 
  • #4
Borek
Mentor
28,886
3,431
Yes, but thats sum of two different ns. You have a mixture.
 
  • #5
590
0
i think i see what you are saying
pv=[n(PF3)+n(F2)]rt

[n(PF3)+n(F2)]=1.055*10^-3mol

now i know that n(PF3)=(2/3)*n(initial F2)=(2/3)*1.3*10^-3 =8.667*10^-4

therefore 1.88333*10^-4 didnt react. is this right?
 
  • #6
Borek
Mentor
28,886
3,431
n(PF3)=(2/3)*n(initial F2)

That means everything reacted, doesn't it?

Try to write fluorine moles balance.
 
  • #7
590
0
then im not sure what to do...

as far as i can see they are blanaced, 6F2->4F3

how do i know how much reacted??
 
  • #8
590
0
is this right:

the 1.055*10-3mol is made up of (2/3)*1.055*10-3mol of PF3 and (1/3)*1.055*10-3mol of F2, therefore of the 1.3*10-3mol of F2, 3.5*10-4mol didnt react?
 
  • #9
Borek
Mentor
28,886
3,431
No.

You have two substances in the container after ther reaction.

Assume you have nF2 moles of fluorine after the reaction, and nPF3 moles of phosphorus trifluoride.

Obviously

nF2 + nPF3 = 1.055x10-3

How many moles of fluorine is in the mixture of nF2 and nPF3? Write equation that will take into account fact that 1 mole of nPF3 contains 1.5 mole of F2.

You know number of moles of fluorine that was present in the container before the reaction. Has it changed?

This will give you two equations in two unknowns.
 
  • #10
590
0
i see what you are saying, but i seem to get stuck with it every time,
can i not say

X=number of moles of F2 that reacted, therefore


(4/6)X+[(1.3*10^-3)-x]=1.055*10^-3

[moles of PF3] + [moles of P2 remaining] = 1.055*10^-3

then i get X=moles F2 that reacted=7.35*10^-4moles
 
  • #11
Borek
Mentor
28,886
3,431
X=number of moles of F2 that reacted, therefore

(4/6)X+[(1.3*10^-3)-x]=1.055*10^-3

That's equivalent. I was aiming at:

nF2+nPF3=1.055*10-3
nF2+1.5nPF3=1.3*10-3

Note, that [(1.3*10^-3)-x] from your equation is just amount of fluorine left as calculated from the second equation.

7.35*10^-4moles

That's correct :smile:
 

Related Threads on Chemistry- ideal gases

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
6K
Replies
5
Views
4K
Replies
4
Views
2K
  • Last Post
Replies
8
Views
5K
  • Last Post
Replies
1
Views
6K
Top