Answer to Question #146486 in General Chemistry for Madison Dupnik

Question

In the reaction 2KClO3(s)= 2KCl(s)+ 3O2(g)

If the 50.0 L of oxygen at STP are needed, how many grams of KclO3 must be decomposed 

Answer 1

Solution:

The balanced chemical equation:

2KClO₃(s) = 2KCl(s)+ 3O₂(g)

At STP, 1 mole of an ideal gas = 22.4 liters.

Thus:

50.0 L O₂ × (1 mole/22.4 L) = 2.232 moles of O₂

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

n(KClO₃) = 2 × n(O₂) / 3

n(KClO₃) = (2 × 2.232 mol ) / 3 = 1.488 mol

Moles of KClO₃ = 1.488 mol.

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

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

Thus:

Mass of KClO₃ = Moles of KClO₃ × Molar mass of KClO₃

Mass of KClO₃ = 1.488 mol × 122.55 g mol^-1 = 182.3544 g = 182.35 g

Mass of KClO₃ is 182.35 grams = 182 grams (to 3 significant figures).

Answer: 182 grams of KClO₃ must be decomposed.