Complete Combustion - which hydrocarbon will consume more O2

[RoboNerd]

Homework Statement

one gram of each of the following gases is introduced into a 10 L container at 25 degrees C

a) propane
b) ethane
c) methane
d) pentane

which gas will consume the greatest mass of oxygen upon complete combustion?

The solutions says that the right answer is D. I do not know why

Homework Equations

no equations. just chemical ones.

The Attempt at a Solution

My logic is that if I have greater number of moles of hydrocarbon, I will need more oxygen to consume.

I thus thought that since methane has a lesser molar mass, we have greater moles, so we will need greater moles [which leads to a higher mass] of O2 gas needed.

Could anyone please advise on how to solve this?

Thanks!

 

[Ygggdrasil]

Start with writing out balanced chemical equations for the complete combustion of each of these hydrocarbons.

 

[RoboNerd]

OK.

My equations are as following

• propane: C3H8 + 5O2 --> 3CO2 + 4H2O
• ethane: 2C2H6 + 7O2 ----> 4CO2 + 6H2O
• CH4 + 2O2 ---> Co2 + 2H2O
• C5H12 + 8O2 ----> 5CO2 + 6H2O

I believe these equations are correct. right?

I just did some stoichiometry with approximate values, and I got 0.111 moles of O2 needed to combust the hydrocarbon in answer D.

My other calculated numbers are:
a) 0.1136 mol O2 needed
b) 0.1167 mol o2 needed
c) 0.125 mol o2 needed
d) 0.111 mol o2 needed.

So far, it seems that the hydrocarbon in answer d does not have the maximum need for O2 moles.

Could I have made an algebraic error?

For example, a sample calculation for my number for part A is:

(1 g C3H8)/(44 g C3H8) * ( 5 mol O2 / 1 mol C3H8) = 0.1136.

I could be wrong.

Please advise and thanks for your help!

 

[Ygggdrasil]

I did the calculations as well and got the same numbers as you. The answer should be C (methane), and the book is probably wrong.

 

[RoboNerd]

Wow. This seems really strange.

Thanks so much for the help!