Jump to content

Recommended Posts

Posted

Hello everyone,
I would like a helping hand for this exercise:

"A mixture of carbon monoxide, methane and nitrogen occupies a volume of 40 mL. Burning this mixture with an excess of oxygen, a volume of 42 mL is observed after cooling at room temperature and together the formation of 36 mL of dioxide of carbon. Determine the volume of each component of the mixture. "

I think this is the reaction: CO + CH4 + N2 + O2 --> CO2 + N2 + H2O, balanced: 2CO + 2CH4 + 5O2 --> 4CO2 + 4H2O removing N2 because acts as a spectator molecule.

Now, how do I proceed??

Posted (edited)

He tells me that I have 42 mL of final mixture and that's it; also I have the molecular mass.

Edited by Gio_5
Posted
2 hours ago, Gio_5 said:

He tells me that I have 42 mL of final mixture and that's it; also I have the molecular mass.

The question you quoted says "the formation of 36 mL of dioxide of carbon"

 

Posted

Ok, I have the volume of CO2, but how can I determine the volume of mixture CO+CH4+N2? This is the question. I don’t have nothing.

Posted
12 hours ago, Gio_5 said:

Ok, I have the volume of CO2, but how can I determine the volume of mixture CO+CH4+N2? This is the question. I don’t have nothing.

Does the volume of a gas depend on the number of moles? 

Posted
On 9/11/2019 at 11:53 AM, swansont said:

Does the volume of a gas depend on the number of moles? 

Obviously, with the ideal gas law PV=nRT

Posted
33 minutes ago, Gio_5 said:

Obviously, with the ideal gas law PV=nRT

OK, so can you figure out the # of moles of CO2 that were formed? 

Posted
56 minutes ago, Gio_5 said:

Ok, yes n=PV/RT=1,5*10-3 mol

You have a balanced chemical equation, telling you how many moles of each reactant relates to moles of product. And you can convert from moles to volume.

Posted
21 hours ago, swansont said:

You have a balanced chemical equation, telling you how many moles of each reactant relates to moles of product. And you can convert from moles to volume.

Ok swansont, really thank you very much.

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.