Question

Potassium chlorate (KClO3) decomposes to form potassium chloride (KCl) and pure oxygen (O2) gas as shown below.

(mc022-1.jpg)

What amount of KClO3 is required to supply 28.0 L of oxygen gas at STP? (The molar mass of KClO3 is 122.55 g/mol.)
A. 102 g
B. 102.12 g
C. 306.36 g
D. 306.4 g

Answer 1

The balanced chemical reaction is:

2KClO3 ----> 2KCl + 3O2

We are given the amount of oxygen gas to be produced from potassium chlorate. This will be the starting point for the calculations.

28.0 L O2 ( 1 mol / 22.4 L) (2 mol KClO3 / 3 mol O2) ( 122.55 g / 1 mol) = 102.13 g KClO3 is needed