What mass of oxygen is needed for the complete combustion of 6.20 × 10^-3 g of methane?

First, let's write the balanced equation, so we'll know what we're dealing with

CH₄ + 2 O₂  ⇒  2 H₂O + CO₂

Now, we'll convert g to mole  ↔  mole to mole  ↔  mole to g

6.20 × 10-3 g CH4  (1 mole CH4) / (16.05 g CH4)   (2 moles O2) / (1 mole CH4)   (32.00 g O2) / (1 mole O2) = 0.024723 g O2

This combustion rxn will produce 2.47 x 10^-2 g O₂