Find out the molar fractions of all the involved gases

[TheSodesa]

Homework Statement

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5,00 moles of graphite and 5,00 moles of oxygen gas are stored in a metal bottle. The temperature is raised until the graphite starts burning. The burning produces a mixture of CO and CO2 gases in the vessel. After the temperature had returned to it's original value (after the reaction), the pressure inside the bottle had increased by 17%. Find out the molar fractions of each of the gases inside the bottle when all of the graphite is used up in the reaction.

Correct answers: x_O2 = 0,145; x_CO = 0,291; x_CO2 = 0,564

Homework Equations

x_i = p_i/p_total = n_i/n_total, where x = molar fraction.

Dalton's law: Sum(p_i) = p_total

pV = nRT

The Attempt at a Solution

I started out by forming the chemical equation:

3C + 2O₂ -> 2CO + CO₂

From this I calculated that n_CO = (2/3)n_C = 10/3 mol and n_CO2 = (1/3)n_C = 5/3 mol ( since we know that C is the limiting reactant). We can also see that the amount of used up oxygen is n_O2 = n_CO = 10/3 mol, so the amount oxygen gas remaining in the bottle after the reaction is n_O2g = 5mol - n_O2 = 5/3 mol, so the total amount of gas in the bottle (after the reaction) would be 20/3mol.

Next I assumed that the volume of C is negligible, so V is constant. T is also constant between the initial and final stages.This means that:

V₀ = V₁ <=> n₀/p₀ = n₁/p₁
= n₁/1,17p₀ (From pV = nRT).

=> n_1gas = 1,17n_0gas = 1,17 * 5 mol = 5,85 mol. However, the value of n_1gas does not equal the total amount of moles I got from adding up the individual amounts of the different gases, and I'm not sure what to do next. I don't see how I could use Dalton's law to my advantage here, since i can't actually calculate the pressures because I know nothing about the temperature or the volume of the bottle.

Can anybody point me in the right direction? I am aware that I might not even need the actual values of the pressures or moles since I'm supposed to find the percentages, but I just don't know how to proceed.

 

[Chestermiller]

There are 2 reactions occurring in parallel:

$C + O_2 ⇒ CO_2$
$C+\frac{1}{2}O_2⇒CO$

If x moles of C convert by the first reaction, then 5-x moles of C convert by the second reaction.

Chet

 

[TheSodesa]

There are 2 reactions occurring in parallel:

$C + O_2 ⇒ CO_2$
$C+\frac{1}{2}O_2⇒CO$

If x moles of C convert by the first reaction, then 5-x moles of C convert by the second reaction.

Chet

Alright, I'm going to try this after I'm done cooking (and eating) dinner. I'll report back when I'm done/if I get stuck again.

 

[TheSodesa]

There are 2 reactions occurring in parallel:

$C + O_2 ⇒ CO_2$
$C+\frac{1}{2}O_2⇒CO$

If x moles of C convert by the first reaction, then 5-x moles of C convert by the second reaction.

Chet

Alright, I'm back!
Turns out I was more tired than I though since I passed out in bed right after dinner so I couldn't get back to the problem before this morning. Anyways, your hint helped and I got the right answer, so thank you for that.

For future reference, should I always, always, always write separate chemical equations if I know that there are multiple reactions going on at the same time, even if they are drawing from the same pool of chemical(s) like C and O₂ in this problem?

 

[Chestermiller]

Alright, I'm back!
Turns out I was more tired than I though since I passed out in bed right after dinner so I couldn't get back to the problem before this morning. Anyways, your hint helped and I got the right answer, so thank you for that.

For future reference, should I always, always, always write separate chemical equations if I know that there are multiple reactions going on at the same time, even if they are drawing from the same pool of chemical(s) like C and O₂ in this problem?

I hate to answer "always" questions. Partly, it depends on the context of the problem statement. I would just use my judgement on the specific problem.

 

[TheSodesa]

I hate to answer "always" questions. Partly, it depends on the context of the problem statement. I would just use my judgement on the specific problem.

Got it. Thanks for the help.