Determine the volume of methane (ch 4) gas needed to react completely with 0.660 l of o 2 gas to form methanol (ch 3 oh).

Balance Chemical Equation for this reaction is,

2 CH₄ + O₂ → 2CH₃OH

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 gas is needed to react completely with 0.660 L of O ₂ gas to form methanol (CH ₃OH).