At STP, 1 mole of any gas occupies 22.4 liters. The molar mass of methane (CH₄) is 16 g/mol, so 27.8 g of methane is 27.8 g / 16 g/mol = 1.7375 mol.

The balanced chemical equation for the combustion of methane is:

CH₄ + 2O₂ → CO₂ + 2H₂O

From the balanced chemical equation, we can see that 1 mole of methane reacts with 2 moles of oxygen. Therefore, 1.7375 mol of methane will react with 2 × 1.7375 mol = 3.475 mol of oxygen.

At STP, 1 mole of oxygen occupies 22.4 liters, so 3.475 mol of oxygen will occupy 3.475 mol × 22.4 L/mol = 77.76 liters.

Therefore, 77.76 liters of oxygen are needed to exactly react with 27.8 g methane at STP.