At STP, how many liters of oxygen are required to react completely with 3.6 liters of hydrogen to form water, 2H 2 (g) + O 2 (g) - --> 2H 2O (g) ?

1.80L

Explanation:

Firstly, we need to calculate the number of moles of hydrogen reacted.

At STP, 1 mole of a gas occupies a volume of 22.4L, hence to get the number of moles that occupied a volume of 3.6L, we simply divide 3.6 by 22.4

This gives : 0.161 mole

Now from the balance equation of chemical reaction, we can see that 2 moles of hydrogen reacted with one mole of oxygen, hence, 0.161 mole of hydrogen would react with 0.161/2 mole of oxygen = 0.0805 mole

As said earlier, a single mole at STP would occupy a volume of 22.4, hence, 0.0805 mole of oxygen will occupy a volume of 0.0805 * 22.4 = 1.80L