Question

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

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

The product gas, O2, is collected over water at a temperature of 25 °C and a pressure of 755 mm Hg. If the wet O2 gas formed occupies a volume of 6.22 L, the number of grams of O2 formed is ________
g. The vapor pressure of water is 23.8 mm Hg at 25 °C.

Answer 1

Mass of O₂ formed = 7.84 g

Further explanation

Given

Reaction

2KClO₃(s) ⇒2KCl(s) + 3O₂(g)

P water = 23.8 mmHg

P tot = 755 mmHg

V = 6.22 L

T = 25 + 273 = 298 K

Required

mass of O₂

Solution

P tot = P O₂ + P water

P O₂ = P tot - P water

P O₂ = 755 - 23.8

P O₂ = 731.2mmHg = 0.962 atm

• Moles O₂ :

Ideal gas law :

n = PV/RT

n = 0.962 x 6.22 / 0.082 x 298

n = 0.245

• Mass O₂ :

= mol x MW

= 0.245 x 32

= 7.84 g