What volume of oxygen is needed to completely combust 3.57 L of methane gas (CH4)?

First, let's look at stoichiometry.

CH4 + 2O2 --> CO2 + 2H2O

We need 2 Os for CO2 and 2 for our 2 H2O, so we need a total of 4 O or 2 O2. So, we need 2 moles of O2 per mole of CH4.

Assuming ideal gas law at constant pressure and temperature, he have:

Vc/Vo = nc/no

where "c" = CH4 and "o" = O2

rearranging, Vo = Vc*no/nc

since we have 2 moles of O2 per CH4, no = 2 nc, so no/nc =2

Vo = 2*Vc

Vo = 2*3.57L

Vo=7.14L