Question

1. How many moles of oxygen gas are needed to form 21.8 liters of water vapor?

Answer 1

0.781 moles

Explanation:

We begin by balancing the chemical equation;

O₂ (g) + 2H₂ (g) → 2H₂O (g)

21.8 Liters = 21.8 Kgs

To find how many moles are in 28.1 Kg H₂O;

Molar mass of H₂O = 18 g/mol

28.1/18

= 1.56 moles

The mole ratio between water vapor and oxygen is;

1 : 2

x : 1.56

2x = 1.56

x = 1.56 / 2

x =  0.781

0.781 moles