What volume of gas is generated when 58.0 l of oxygen gas reacts at stp?

The question is incomplete. This is the complete question:

What volume of gas is generated when 58.0 l of oxygen gas reacts at STP according to the following balanced equation?

CH₃CH₂OH (l) + 3O₂ → (g) 2CO₂ (g) + 3H₂O (l)

Answer: Volume of CO₂ = 58 l * 2 / 3 = 38.7 l

Explanation

1) STP means standard temperature and pressure. That is 0°C and 1 atm.

2) Also, you can use the known fact that at STP 1 mole of gas occupies a volume of 22.4 liters.

3) With that you may calculate the number of moles of oxygen gas that reacts, but it is not necessary since the conditions of pressure and temperature do not change.

That permits you to use the same mole ratio of the equation.

4) The balanced equation is CH₃CH₂OH (l) + 3O₂ → (g) 2CO₂ (g) + 3H₂O (l), so the mole ratio is

3 mol O₂ : 2 mol CO₂

⇒ Volume of CO₂ = 58 l * 2 / 3 = 38.7 l