Chemistry- ideal gases

[Dell]

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

P_4(s) + 6F_2(g)--> 4PF_3(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
P_i=0.329atm
P_f=0.267atm

now i don't know if all of the F reacted since i don't know how many mols of P there were so i can't 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 doesn't 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?

 

[Borek]

Not all F₂ reacted.

Think what unknowns do you have and what equations can you write to describe vessel content after the reaction ended.

 

[Dell]

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??

 

[Borek]

Yes, but that's sum of two different ns. You have a mixture.

 

[Dell]

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?

 

[Borek]

n(PF3)=(2/3)*n(initial F2)

That means everything reacted, doesn't it?

Try to write fluorine moles balance.

 

[Dell]

then I am not sure what to do...

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

how do i know how much reacted??

 

[Dell]

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?

 

[Borek]

No.

You have two substances in the container after ther reaction.

Assume you have n_F2 moles of fluorine after the reaction, and n_PF3 moles of phosphorus trifluoride.

Obviously

n_F2 + n_PF3 = 1.055x10^-3

How many moles of fluorine is in the mixture of n_F2 and n_PF3? Write equation that will take into account fact that 1 mole of n_PF3 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.

 

[Dell]

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

 

[Borek]

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:

n_F2+n_PF3=1.055*10^-3
n_F2+1.5n_PF3=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