How many grams of oxygen are required to burn 4.8 mol of butane?

The balanced equation for the reaction is 2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O. This shows that 13 moles of diatomic oxygen are required to burn 2 moles of butane. By proportionality, (4.8/2)13 or 31.2 moles of oxygen are required to burn 4.8 moles of butane. This corresponds to 31.2(32) or 1.0 X 103 grams of oxygen.