According to this eq, 22.4 L (1 moles) of Oxygen requires 44.8 L (2 mole) CH₄ for complete reaction. So, the volume of CH₄ required to consume 0.66 L of O₂ is calculated as,
22.4 L O₂ required to consume = 44.8 L CH₄ 0.660 L O₂ will require = X L of CH₄
Solving for X, X = (44.8 L × 0.660 L) ÷ 22.4 L
X = 1.320 L of CH₄
Result: 1.320 L of CH₄<span>gas is needed to react completely with 0.660 L of O</span>₂<span> gas to form methanol (CH</span>₃OH<span>).</span>