HELP PLEASE Assuming all volume measurements are made at the same temperature and pressure, how many liters of water vapor can be produced when 8.23 liters of oxygen gas react with excess hydrogen gas? Show all of the steps taken to solve this problem.

Balanced equation: 2 H2 (g) + O2 (g) yields 2 H2 O(g)

Question

HELP PLEASE Assuming all volume measurements are made at the same temperature and pressure, how many liters of water vapor can be produced when 8.23 liters of oxygen gas react with excess hydrogen gas? Show all of the steps taken to solve this problem. 


Balanced equation: 2 H2 (g) + O2 (g) yields 2 H2 O(g)

anonymous · Answer

Finding the mol of oxygen used:
8.23 liters = $$8.23dm ^{3}$$
1mol of any gas at room temperature and pressure takes up 24dm3 of space.

Hence mol of O2 used = 8.23/24 = (approx)0.3429 mol

By molar ratio, 1 O2 reacts with H2 to yield 2 H2O (g)

Hence 0.3429 = approx 0.686 (3.sf) H2O gas produced