Question

Oxygen gas can be prepared by heating potassium chlorate according to the following equation:

17.2KClO3(s)2KCl(s) + 3O2(g)

The product gas, O2, is collected over water at a temperature of 25 °C and a pressure of 758 mm Hg. If the wet O2 gas formed occupies a volume of 9.07 L, the number of moles of KClO3 reacted was ____
mol. The vapor pressure of water is 23.8 mm Hg at 25 °C.

Answer 1

Moles KClO₃ = 0.239

Further explanation

Given

Reaction

2KClO₃(s) ⇒2KCl(s) + 3O₂(g)

P water = 23.8 mmHg

P tot = 758 mmHg

V = 9.07 L

T = 25 + 273 = 298 K

Required

moles of KClO₃

Solution

P tot = P O₂ + P water

P O₂ = P tot - P water

P O₂ = 758 - 23.8

P O₂ = 734.2 mmHg = 0.966 atm

moles O₂ :

n = PV/RT

n = 0.966 x 9.07 / 0.082 x 298

n = 0.358

From equation, mol ratio KClO₃ : O₂ = 2 : 3, so mol KClO₃ :

= 2/3 x mol O₂

= 2/3 x 0.358

= 0.239