If 22.5 liters of oxygen reacted with excess of hydrogen, how many liters of water vapor could be produced?

45 L

Explanation:

Assuming normal temperature and pressure (NTP) and ideal gas behaviour, 22.5 L are equivalent to:

PV = nRT

n = PV/RT

n = (1*22.5)/(0.082*273.15)

n = 1 mol

Oxygen and hydrogen react to form water as  follows:

2 H2 (g) + O2 (g) -> 2 H2O (g)

From the balanced equation, we know that 1 mol of Oxygen produces 2 moles of water. The volume occupied by these 2 moles is:

PV = nRT

V = nRT/P

V =  2*0.082*273.15/1

V = 45 L

Know we can see the assumption of NTP conditions was unnecessary, the ratio between moles of oxygen consumed and moles of water produced is 1:2, so the ratio between volumes.