How many liters of O2 (g) are needed to react completely with 56.0 L of CH4 (g) at STP to produce CO2 (g) and H2O (g)? Given: CH4 (g) + 2O2 (g) → CO2 (g) + H2O (g)

I'm not really sure how to approach this problem. I'm thinking the conversion factor for STP (22.4L/1mol) has something to do with it, but I'm not sure where I'd apply it.

1 Answer
Nov 25, 2015


#"112 L O"_2#


The trick here is that you don't actually need to use mole ratios, you can go by volume ratios.

That happens because when a reaction deals with gases at the same conditions for pressure and temperature, the mole ratios that exist between the chemical species that take part in that reaction are equivalent to the volume ratios.

In your case, you know that the reaction takes place at STP, Standard Temperature and Pressure, so that all the chemical species share the same conditions for temperature and pressure.

If you use the ideal gas law equation to write the number of moles of methane and the number of moles of carbon dioxide - remember, both gases have the same #P# and #T#- you will get

#P * V_(CH_4) = n_(CH_4) * R * T#


#P * V_(CO_2) = n_(CO_2) * R * T#

Divide these two equations to get

#( color(red)(cancel(color(black)(P))) * V_(CH_4))/(color(red)(cancel(color(black)(P))) * V_(CO_2)) = (n_(CH_4) * color(red)(cancel(color(black)(R * T))))/(n_(CO_2) * color(red)(cancel(color(black)(R * T))))#

This confirms that the mole ratio that exists between methane and carbon dioxide is equal to

#overbrace(n_(CH_4)/n_(CO_2))^(color(green)("mole ratio")) = overbrace(V_(CH_4)/V_(CO_2))^(color(blue)("volume ratio"))#

So, the balanced chemical equation for this combustion reaction looks like this

#"CH"_text(4(g]) + color(red)(2)"O"_text(2(g]) -> "CO"_text(2(g]) + "H"_2"O"_text((g])#

You have a #1:color(red)(2)# mole ratio between methane and oxygen, which means that you will also have a #1:2# volume ratio between the two gases.

In other words, if the reaction needs #color(red)(2)# times more moles of oxygen than of methane, then it will also need a volume of oxygen that's #color(red)(2)# times bigger than that of methane.

This means that you have

#56.0color(red)(cancel(color(black)("L CH"_4))) * (color(red)(2)" L O"_2)/(1color(red)(cancel(color(black)("L CH"_4)))) = color(green)("112 L O"_2)#