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.
Mass of O₂ formed = 7.84 g
Further explanationGiven
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