Question

What volume of oxygen gas, at STP, can be formed from the decomposition of 61.0g of solid potassium chlorate to form solid potassium chloride and oxygen gas?

Answer 1

The balanced equation is

2 KClO₃ ----> 2 KCl + 3 O₂

Number of moles of potassium chlorate = 61.0 g / 122.55 g/mol = 0.498 mole

from the balanced equation we can say that

2 mole of potassium chlorate produces 3 mole of O2 so

0.498 mole of potassium chlorate will produce

= 0.498 mole of potassium chlorate *(3 mole of O2 / 2 mole of potassium chlorate)

= 0.747 mole of O2

Therefore, the number of mole of oxygen = 0.747

at STP, 1 mole of O2 = 22.4 L so

0.747 mole of O2 = 16.7 L

Therefore, the volume of oxygen gas produced would be 16.7 L