Question

Oxygen is prepared by catalytic decomposition of potassium chlorate (KClO3). Decomposition of potassium chlorate gives potassium chloride (KCl) and oxygen (O2).If 3.6 mol of oxygen is needed for an experiment, how many grams of potassium chlorate must be decomposed?
(At. mass of K = 39, Cl=35.5, O = 16)

Answer 1

Answer :

★ Balanced chemical reaction :

$\underline{\boxed{\bf{\blue{2KClO_3+\Delta\longrightarrow 2KCl+3O_2}}}}$

• KClO₃ : Potassium chlorate
• KCl : Potassium chloride
• O₂ : Oxygen gas

➤ Molar mass of KClO₃ :

• K + Cl + 3(O)
• 39 + 35.5 + 3(16)
• 122.5 g/mol

It is clear from the chemical reaction that,

3 moles of oxygen gas requires 2 moles of potassium chlorate.

Therefore, 3.6 moles of oxygen gas requires

• 3.6×2/3 = 2.4 moles

⧪ Mass of potassium chlorate :

➙ Mass = Moles × Molar mass

➙ Mass = 2.4 × 122.5

➙ Mass = 294 g

Cheers!