Chemistry Question on burning

The equation for the combustion of methanol is the following:
2CH3OH + 3O2 -> 2CO2 + 4H20
What volume of oxygen at STP is required to completely burn 35.0g of methanol?

2 CH3OH + 3 O2 → 2 CO2 + 4 H2O

(35.0 g CH3OH) / (32.0420 g/mol) x (3/2) x (22.4 L/mol) = 36.7 L O2 at STP

35g 1mol CH3OH 3mol O2 22.414L
---------- x --------------------- x ------------------- x --------------- = 36.7g O2
1 32.05g 2mol CH3OH 1mol O2

It takes away all the spaces I have in between the proportions so use your imagination.