(i) What is limiting reactant? (ii) Oxygen is prepared by catalytic d

1.	(i) What is limiting reactant? (ii) Oxygen is prepared by catalytic decomposition of potassium chlorate (KCIO3). Decomposition of potassium chlorate gives potassium chloride (KCI) and oxygen (O2). If 2.45 mol of oxygen is needed for an experiment, how many grams of potassium chlorate must be decomposed?
Answer» (i) It is the reactant which is entirely consumed when reaction goes to completion. (ii) 2KClO₃(s) → 2KCl(s) + 3O₂(g) Molecular weight of KClO₃ = 39 + 35.5 + 3 × 16 = 122.5 For 3 moles of O₂ we need = 2 × 122.5 g of KClO₃ For 2.4 moles of O₂ we need = \(\frac{2\times122.5}{3}\) x 2.4 = 196 g of KClO₃.

(i) What is limiting reactant? (ii) Oxygen is prepared by catalytic decomposition of potassium chlorate (KCIO3). Decomposition of potassium chlorate gives potassium chloride (KCI) and oxygen (O2). If 2.45 mol of oxygen is needed for an experiment, how many grams of potassium chlorate must be decomposed?

Answer»

(i) It is the reactant which is entirely consumed when reaction goes to completion.

(ii) 2KClO₃(s) → 2KCl(s) + 3O₂(g)

Molecular weight of KClO₃ = 39 + 35.5 + 3 × 16

= 122.5

For 3 moles of O₂ we need

= 2 × 122.5 g of KClO₃

For 2.4 moles of O₂ we need

= \(\frac{2\times122.5}{3}\) x 2.4 = 196 g of KClO₃.