Answer to Question #168723 in Chemistry for Angelina

Question

If 5.0 g of KClO₃ are decomposed, what volume of O₂ is produced at STP?

Answer 1

Solution:

The balanced chemical equation:

2KClO₃ → 2KCl + 3O₂

According to the equation above: n(KClO₃) / 2 = n(O₂) / 3

Moles of KClO₃ = Mass of KClO₃ / Molar mass of KClO₃

The molar mass of KClO₃ is 122.55 g mol^-1.

Hence,

n(KClO₃) = (5.0 g) / (122.55 g mol^-1) = 0.0408 mol

n(O₂) = 3 × n(KClO₃) / 2 = (3 × 0.0408 mol) / 2 = 0.0612 mol (according to the equation)

At STP, one mole of any gas occupies a volume of 22.4 L.

Thus 0.0612 mol of O₂ occupies:

(0.0612 mol × 22.4 L) / 1 mol = 1.37088 L of O₂ = 1.37 L of O₂

At STP, 0.0612 mol of O₂ occupies a volume equal to 1.37 L.

OR (according to the equation):

(5.0 g KClO₃)×(1 mol KClO₃ / 122.55 g KClO₃)×(3 mol O₂ / 2 mol KClO₃)×(22.4 L O₂ / 1 mol O₂) = 1.37 L

1.37 L O₂

Answer: 1.37 liters of O₂ is produced at STP.