For the reaction 2H₂ (g) + O₂ (g) → 2H₂O (g), how many liters of water can be made from 7.5 L of oxygen gas and an excess of hydrogen at STP?

The amount of 1 mole of a gas at STP is calculated to be 22.4L. We first calculate the amount of oxygen gas in moles.

oxygen gas in moles = 7.5 L / (22.4 L/mol) = 0.335 moles O2

Then, we calculate the moles of H2O produced based on the given balanced chemical reaction.

(0.335 moles O2) (2 moles H2O / 1 mole O2) = 0.67 moles H2O

Then, converting again the calculated moles to volume,

volume of water = (0.67 moles) x (22.4 L/mole) = 15 L of H2O

Thus, the amount of gaseous water produced is 15 L.