Dalton's Law of partial pressures......... states that in a gaseous mixture, the partial pressure exerted by a component gas is the same as the pressure it would exert if it ALONE occupied the container. The total pressure is the sum of the individual partial pressures........
And so we works out.....
nCH4=MassMolar Mass=8⋅g16⋅g⋅mol−1=12⋅mol
nH2=MassMolar Mass=2⋅g2.0⋅g⋅mol−1=1⋅mol
And thus we have 32 moles of gas, THAT WE ASSUME to BEHAVE IDEALLY. Since we know (or should know) that 1 mole of Ideal Gas occupies 22.4⋅L under standard conditions.....(and this should be given as supplementary material on any exam paper).
Vgaseous mixture=32⋅mol×22.4⋅L⋅mol−1=33.6⋅L
Oh, and by the way, 1⋅L≡1⋅dm−3. Just to expand a bit on this, I thought it might be useful to explain the equivalence of 1⋅L and 1⋅dm3. Now chemists commonly use litres as they are convenient units, and of course they also use cubic centimetres which are equivalent to millilitres; 1⋅cm3≡1⋅mL.
Now 1⋅m3 is an impossibly large volume, and this is equal to 1000⋅L, which is also equal to 1000⋅dm3. If you ever shift 4−5⋅m3 of concrete, this is a hot and hard day's work.
Now 1⋅dm3=1×(10−1⋅m)3 (i.e. the d means deci, i.e. ×10−1), and so 1⋅dm3=1×(10−1⋅m)3=1×10−3⋅m3 ≡1⋅L as required.
