Question

Potassium chlorate is the prime ingredient in most Chinese fireworks. If I have 25 grams of

potassium chlorate, how many moles of oxygen can I produce?
(2 step mole problem)

Answer 1

To determine how many moles of oxygen can be produced from 25 grams of potassium chlorate, we first need to calculate the number of moles of potassium chlorate in 25 grams.

The molar mass of potassium chlorate (KClO3) is 122.55 g/mol. Therefore, the number of moles of KClO3 in 25 grams is:

25 g / 122.55 g/mol = 0.204 moles of KClO3

Next, we can use the stoichiometry of the chemical equation to determine the number of moles of oxygen produced. According to the balanced equation, for every 1 mole of KClO3, 3 moles of oxygen are produced. Therefore, the number of moles of oxygen produced from 0.204 moles of KClO3 is:

0.204 moles KClO3 x 3 moles O2 / 1 mole KClO3 = 0.612 moles O2

Therefore, 25 grams of potassium chlorate can produce 0.612 moles of oxygen.