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Gio_5

Combustion

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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??

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You know how much CO2 was formed.

What does that tell you about the rest?

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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

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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"

 

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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.

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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? 

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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

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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? 

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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.

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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.

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