Answer to Question #270397 in Chemistry for Brie

Question

If this reaction produced 51.4 g KCl how many grams of O2 were produced

Answer 1

Solution:

Calculate the moles of KCl:

The molar mass of KCl is 74.55 g/mol

Therefore,

(51.4 g KCl) × (1 mol KCl / 74.55 g KCl) = 0.6895 mol KCl

Balanced chemical equation:

2KClO₃(s) → 2KCl(s) + 3O₂(g)

According to stoichiometry:

2 mol of KClO₃ produces 2 mol of KCl and 3 mol of O₂

Thus, if 0.6895 mol of KCl is produced, then:

(0.6895 mol KCl) × (3 mol O₂ / 2 mol KCl) = 1.03425 mol O₂

1.03425 mol O₂ is produced

Calculate the mass of O₂:

The molar mass of O₂ is 32.0 g/mol

Therefore,

(1.03425 mol O₂) × (32.0 g O₂ / 1 mol O₂) = 33.096 g O₂ = 33.1 g O₂

Answer: 33.1 grams of O₂ were produced