When 90 g of water is split by electrolysis (2H2O → 2H2 + O2) at STP, how many liters of oxygen gas will be collected?

Question 6 options:

44.8 L

22.4 L

56.0 L

67.2 L

2H2O - > 2H2 + O2

Stoichiometric ratio:

2 : 2 : 1

H2O : O2

2 : 1

Mass of 1 mol of H2O = [ (2 x 1) + 16] g = 18g

18g of H2O = 1 mol of H2O

90g of H2O = 90g x 1 mol / 18g

= 5 mol of H2O

2 mol of H2O : 1 mol of O2

5 mol of H2O : 5/2 mol of O2

1 mol of O2 = 22.4 L of O2

5/2 mol of O2 = 5/2 mol x 22.4 L / 1 mol

= 56 L of O2