Consider the thermal decomposition of potassium chlorate.
2 KClO3(s) ---> 2 KCl(s) + 3 O2(g)
A mixture containing KCl and KClO3 weighing 1.80 g was heated producing
1.40 x 102 mL of O2 gas at STP. What percent of the original mixture was KClO3?
To find the percent of KClO3 in the original mixture, we need to calculate the mass of KClO3 and the total mass of the mixture.
1. Calculate the moles of O2 gas produced:
Using the ideal gas law, PV = nRT, where P is the pressure (STP - 1 atm), V is the volume (1.40 x 102 mL = 0.140 L), n is the number of moles, R is the gas constant (0.0821 L·atm/K·mol), and T is the temperature (STP - 273 K):
(1 atm)(0.140 L) = n(0.0821 L·atm/K·mol)(273 K)
n = (1 atm x 0.140 L) / (0.0821 L·atm/K·mol x 273 K)
n ≈ 0.00612 mol
2. Use the stoichiometric ratio of the balanced equation to find the moles of KClO3:
From the balanced equation, we know that 2 moles of KClO3 produce 3 moles of O2. So, the moles of KClO3 = (0.00612 mol O2) x (2 mol KClO3 / 3 mol O2).
Moles of KClO3 ≈ 0.00408 mol
3. Calculate the mass of KClO3:
To find the mass of KClO3, we need to assume that the mixture is composed solely of KCl and KClO3, and assume their molar masses:
Mass of KClO3 = (moles of KClO3)(molar mass of KClO3)
≈ (0.00408 mol)(122.55 g/mol) (the molar mass of KClO3 is 122.55 g/mol)
≈ 0.500 g
4. Calculate the total mass of the mixture:
The total mass of the mixture = mass of KCl + mass of KClO3
= 1.80 g - mass of KClO3 (since we're given the total mass of the mixture, 1.80 g)
= 1.80 g - 0.500 g
= 1.30 g
5. Calculate the percent of KClO3 in the mixture:
Percent of KClO3 = (mass of KClO3 / total mass of mixture) x 100%
= (0.500 g / 1.30 g) x 100%
≈ 38.5%
Therefore, approximately 38.5% of the original mixture was KClO3.