f 6.9L of CO2 is produced in a reaction following the complete combustion of CH4 with oxygen at STP. What is the minimum volume of oxygen that could yield this amount of product?

Answer 1

First, balance the reaction: CH₄ + 2 O₂ --> CO₂ + 2 H₂O. At STP each mole of gas occupies 22.4L. You could divide the 6.9 mol CO₂ by 22.4 to get moles, then multiply required moles of O₂ by 22.4 -- or just look at the mole ratio 2:1 of O₂ to CO₂ and get the answer of 2*6.9 = 13.8 L of O₂.