How many milligrams of oxygen are required to create 8.72 liters of carbon dioxide gas at STP from the complete combustion of methane (CH4)? Answer needs to be in scientific notation.

Answer 1

Balanced reaction is CH₄ + 2 O₂ --> CO₂ + 2 H₂O. At STP, each mol of gas occupies 22.4 liters.

(8.72 L CO₂) * (1 mol gas / 22.4 L) * (2 mol O₂ /1 mol CO₂) * (32.0g O₂ /1 mol O₂) = 24.9g = 2.49 x 10⁴ mg of O₂.