Solve C2H2 Mol Fraction Problem: STP & Ideal Gas Eq.

[AsuraSky]

So I've been thinking about this problem for quite some time now any help will be appreciated.

A sample of $$C_{2}H_{2}(g)$$ has a pressure of 7.4 kPa. After some time a portion of it reacts to form $${C}_6{H}_6 (g)$$. The total pressure of the mixture of gases is then 3.6 kPa. Assume the volume and the temperature do not change. Find the mol fraction of $$C_{2}H_{2}(g)$$ that has undergone reaction.

The mol fraction, X, is given by

$$X = P_{1}/P_{total} = n_{1}/n_{total}$$

where P= pressure and n= number of moles

I think that $$3C_{2}H_{2}(g)->{C}_6{H}_6 (g)$$. So I assumed STP and thought that if i can find the number of moles of $${C}_6{H}_6 (g)$$ then I can use stoichiometry to find the moles of $$C_{2}H_{2}(g)$$ and subtract that from the number of moles found by using the ideal gas equation and then use the mol fraction formula to find it. However, this results in a negative number (3.25858*10^-3-4.754*10^-3) and I can't seem to figure out what I did wrong. Like I said, any help would be appreciated.

 

[chemisttree]

You shouldn't need to use the ideal gas equation.

Consider that N_total = N₁ + N₂
where N₁ = C2H2 and N₂ = C6H6

Initially N₂ = 0 and later N₂ = N₁ - N_total.

Now we can express N₂ in terms of pressure and N₁... right?

There are other things to work out but this understanding is critical to solving the problem.

 

[AsuraSky]

ok, thanks i'll try to work it out from there,
also, am i on the right track with 3C₂H₂-->C₆H₆ or is this chemical equation irrelevant to solving the problem?

 

[Borek]

It is very important - combined with Avogadro's hypothesis it explains why pressure went down. Think about stochiometry of the process.

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[AsuraSky]

ok, so so far i have
3C₂H₂-->C₆H₆
n_total=n_{C2H2 final} + n_C6H6
$$\frac{P_{C_{2}H_{2 initial}}}{n_{C_{2}H_{2 initial}}}=\frac{P_{C_{2}H_{2 final}}}{n_{C_{2}H_{2 final}}}$$ from avogadro's law
P_{C2H2 final}=P_total-P_C6H6
n_C6H6=n_{C2H2 final}-n_total=n_{C2H2 final}-(P_totaln_{C2H2 final})/P_{C2H2 final}
n_{C2H2 initial} - x moles of C₂H₂ = n_{C2H2 final}
x moles of C₂H₂/3 = n_{C6H6 final}
n_{C2H2 initial}-3*n_C6H6=n_{C2H2 final}

i'm not sure how to proceed from here, is there a way to find the change in moles based on the change of pressure and use that to find the mol fraction?

 

[Borek]

is there a way to find the change in moles based on the change of pressure and use that to find the mol fraction?

If you have no other ideas, assume some volume - say 1L - and do caculations on real numbers. Or assume volume V and do calculations on symbols - in the end V will cancel out. The simplest approach is to use pV=nRT and realize it can be written as n=kp, where k is some constant - that in the end will cancel out as well.

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[AsuraSky]

Ok, so I used your way and did P=nk which means that P/n = k, so P₁/n₁=P₂/n₂=k. After substitution that gives me 7.4/n₁=3.6/n₂. I then found the ratio of n₁ to n₂ to be n₁/n₂= 7.4/3.6 = 37/18 so I interpreted that to mean that we started out with 37 mols of C₂H₂ initially and ended up with a mixture of C₂H₂ and C₆H₆ that totaled in 18 mols.

so Let w = mols of C₂H₂
y = mols of C₆H₆

37 - x mols = w
x mols/3 = y from the stoichiometric ratio
w+y = 18
so that means that
37-3y = w
37-3(18-w) = w
w = mols of C₂H₂ = 17/2
since the total amount of C₂H₂ and C₆H₆ is 18 the mol fraction should be (17/2)/18 = .47222...since this is online homework, I get instant feedback which resulted in a wrong answer. Where did I go wrong?