What is the partial pressure of co2(product) given the v,t of 2 reacta

[ryu1]

Homework Statement

Hi

In a container of 1.5L there are 2 gases O2 and C3H8.
their partial pressures are O2 = 5atm, C3H8=0.1atm.
the temperature is 20C.
the reaction produces H2O and CO2.
Temperature is constant.

What is the partial pressure of CO2 after the reaction? (need to find the reaction formula)
(please show calculations)
Thanks!

Homework Equations

C3H8 + 5O2 --> 3CO2 + 4H2O

The Attempt at a Solution

Have found the moles of O2 (0.312) and moles of C3H8 (0.006) are these correct?, from that I assume I need to use Stoichiometry to find the moles of CO2 and from that to use PV=nRT to find the pressure of CO2.
But I can't figure it out.
Please help, Thanks!

 

[Sunil Simha]

Have found the moles of O2 (0.312) and moles of C3H8 (0.006) are these correct?, from that I assume I need to use Stoichiometry to find the moles of CO2 and from that to use PV=nRT to find the pressure of CO2.
But I can't figure it out.

You have got the right values of number moles of O₂ and C₃H₈ and the right idea to tackle the problem.

Start by finding out how many moles of O₂ will be consumed in the reaction if all of C₃H₈ is burnt. Then if 1 mole of C₃H₈ gives 3 moles of CO₂ then how much does 0.006 mole give?

Be careful while dealing with the water formed in the reaction. Think whether it will contribute to the total pressure after the reaction.

 

[ryu1]

You have got the right values of number moles of O₂ and C₃H₈ and the right idea to tackle the problem.

Start by finding out how many moles of O₂ will be consumed in the reaction if all of C₃H₈ is burnt. Then if 1 mole of C₃H₈ gives 3 moles of CO₂ then how much does 0.006 mole give?

Be careful while dealing with the water formed in the reaction. Think whether it will contribute to the total pressure after the reaction.

I think that if all C3H8 is consumed than according to the equation, 5 times that O2 will be consumed? (so it's 0.03 moles?)

If 1 mole of C3H8 gives 3 moles of CO2 then I do
3/1 * 0.006 = 0.018 moles ? (=3 times the number)

will it be correct to use the ideal gas law now?
P= (0.018mol*(0.082L*atm * K^-1 * mol^-1 )* 293.15K) / 1.5L

I don't see how water fits in.
Also I don't understand how people in Yahoo answers just figured out that the partial pressure is 0.3 atm for CO2 (just because for 1 C3H8 we get 3CO2 ?)

Thanks!

 

[Borek]

I don't see how water fits in.

It doesn't, as question doesn't ask about it - only about partial pressure of CO₂. But in other problems it can be important - think, what form water takes at 20 °C?

Also I don't understand how people in Yahoo answers just figured out that the partial pressure is 0.3 atm for CO2 (just because for 1 C3H8 we get 3CO2 ?)

As there are 3 moles of CO₂ produced for each mole of the hydrocarbon, partial pressure of CO₂ after reaction went to completion must be three times higher than the initial pressure of hydrocarbon. No need for any more elaborate calculations. No magic, just a direct application of Avogadro's hypothesis.

 

[ryu1]

It doesn't, as question doesn't ask about it - only about partial pressure of CO₂. But in other problems it can be important - think, what form water takes at 20 °C?




As there are 3 moles of CO₂ produced for each mole of the hydrocarbon, partial pressure of CO₂ after reaction went to completion must be three times higher than the initial pressure of hydrocarbon. No need for any more elaborate calculations. No magic, just a direct application of Avogadro's hypothesis.

Water takes the form of liquid so it's important because then I can't apply the Ideal gas law to it?

from the calculation I wrote I got 0.288 , it's almost 0.3, it is wrong?
If a reactant (gases) to product mole ratio is for example 1:6 so always the partial pressure of the product is 6 times the pressure of the reactant? (only in gases?)

Thanks.

 

[Borek]

Water takes the form of liquid so it's important because then I can't apply the Ideal gas law to it?

yes. It doesn't count as a gas.

from the calculation I wrote I got 0.288 , it's almost 0.3, it is wrong?

No idea how you got this result, so it is hard to comment. But it should be 0.3, not any other number.

If a reactant (gases) to product mole ratio is for example 1:6 so always the partial pressure of the product is 6 times the pressure of the reactant? (only in gases?)

Yes, but remember we are assuming constant volume.