Answer to Question #198971 in Chemistry for Hsn

Question

2KCLO₃→ 2KCl + 3O_2(g)

1.5 mole of KCLO₃ was decompose during an experiment in laboratory 40 dm³

of O₂ was produced find out 

percentage yield of O_2.

Answer 1

Solution:

Balanced chemical equation:

2KClO₃(aq) → 2KCl(aq) + 3O₂(aq)

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

Therefore,

n(O₂) = 3 × n(KClO₃) / 2 = (3 × 1.5 mol) / 2 = 2.25 mol

At Standard Temperature and Pressure (STP), one mole of any gas occupies a volume of 22.4 L.

Thus 2.25 mol of O₂ occupies:

(2.25 mol) × (22.4 L / 1 mol) = 50.4 L O₂

50.4 L O₂ - theoretical yield

40 dm³ O₂ = 40 L O₂ - practical yield

Therefore,

%O₂ = (40 L / 50.4 L) × 100% = 79.365% = 79.37%

%O₂ = 79.37%

Answer: The percentage yield of O₂ is 79.37%