[Vultux]

Methane is a very powerful greenhouse gas. One pound of methane traps 25 times more heat in the atmosphere than a pound of carbon dioxide. Methane is also the main ingredient in natural gas. Because methane can be captured from landfills, it can be burned to produce electricity, heat buildings, or power garbage trucks. What volume of oxygen is necessary to completely react with 8.46 x 10¹⁹ molecules of methane gas, CH₄ during combustion?



So I started with balancing the equation, which is CH₄ + 2O₂    CO₂ + 2H₂O

I think next I'm supposed to go from 8.46 x 10¹⁹ particles of methane gas (CH₄) to moles of (CH₄) to moles of oxygen to volume of oxygen. What I'm confused about is how I'm supposed to set the equation up and I would appreciate any help in doing so.p

[billnotgatez]

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[AWK]

Start from balanced reaction

[AWK]

Now change number molecules to moles and calculate moles of oxygen in relation to that of methane.
Hint - Avogadro number.

[Vultux]

So I divide particles by Avagadros which becomes 1.40*10^-4 moles. I'm confused on how I do moles to moles. I know it requires the molar mass of both of them but I'm confused on how to do so.

[Borek]

No, you don't need molar masses.

Do you know how to read the reaction equation?

[Vultux]

Not really, my teacher is horrible at teaching so I have to learn everything for Chemistry online by myself. Read it and I'm guessing I have to set a ratio of Oxygen to Methane gas which would be 2/4 then do something?

[AWK]

Now use Avogadro Law nad take molar volume at STP.

[Borek]

Not really, my teacher is horrible at teaching so I have to learn everything for Chemistry online by myself. Read it and I'm guessing I have to set a ratio of Oxygen to Methane gas which would be 2/4 then do something?

Use the ratio to find number of moles of oxygen needed to burn the methane. Then follow AWK's advice.

[Vultux]

So it would be (1.40*10^-4)/2 which makes 0.00007 moles of Oxygen then what? Do I multiply by Avagadros to get volume?

[AWK]

(1.40*10-4)/2

Do I multiply by Avagadros to get volume?

[Borek]

So it would be (1.40*10^-4)/2

No. Look at the reaction equation - every mole of methane needs TWO moles of oxygen. If so, 1.40*10^-4 moles of methane need how many?

then what? Do I multiply by Avagadros to get volume?

No. Avogadro's number tells you how many molecules are in the mole, not what volume they occupy.

Do you know how to deal with units? Avogadro's number has units of

[tex]\frac{number~of~molecules}{mole}[/tex]
Number of moles has - not surprisingly - units of just [itex]mole[/itex]. When you multiply the two, you get
[tex]mole\times\frac{number~of~molecules}{mole}[/tex]
Mole cancels out and the result is [itex]number~of~molecules[/itex].

Do you know what is a volume occupied by 1 mole of a gas at STP? Or do you know the ideal gas equation? If not, google these things.

[Vultux]

Would it be 2.8*10^-4 moles of Oxygen? and then multiply that 22.4? Sorry if I'm being retarded and dumb, It's just that my teacher taught us conversions the first week during school and it's been a while.

[Borek]

Would it be 2.8*10^-4 moles of Oxygen? and then multiply that 22.4?

Yes.

Sorry if I'm being retarded and dumb, It's just that my teacher taught us conversions the first week during school and it's been a while.

I strongly suggest you get back to conversions then, as you will need them all the time.

[Vultux]

So the answer would be 6.3*10^-3 liters of Oxygen?

I strongly suggest you get back to conversions then, as you will need them all the time.

Yeah I know, I'll be learning it after I finish this today.