You are given 1.382 g of a mixture of KClO3 and KCl. When heated, the KClO3 decomposes to KCl

and O2,

2 KClO3 (s) → 2 KCl (s) + 3 O2 (g),

and 279 mL of O2 is collected over water at 22 °C. The total pressure of the gases in the collection flask is 742 torr. What is the weight percentage of KClO3 in the sample?

The formula weight of KClO3 is 122.55 g/mol. The vapor pressure of water at 22 °C is 19.8 torr.

% KClO₃ = 64.7%

We can use the ideal gas law to solve for the number of moles that were produced and then use stoichiometry to relate this back to the number of grams of KClO₃ that would be required to produce this amount of oxygen. When using the ideal gas law we need to subtract the vapor pressure of water from the total pressure of the gases because we need to work with the pressure of only oxygen, not the pressure of oxygen and water vapor. Also, when using the ideal gas law the volume must be in liters (so divide by 1,000), the temperature in Kelvin (so add 273), and the pressure in atm (so divide by 760).

P_Total = P_H2O + P_O2

742 torr = 19.8 torr + P_O2

P_O2 = 722.2 torr

722.2 torr x (1 atm/760 torr) = 0.950... atm

PV = nRT

n (PV)/(RT)

n = (0.950... atm x 0.279 L)/(0.0821 L^.atm/mol^.K x 295 K)

n = 0.0109... mol O₂

0.0109... mol O₂ x (2 mol KClO₃/3 mol O₂) x (122.55 g KClO₃/1 mol KClO₃) = 0.894... g KClO₃

% KClO₃ = (Mass of KClO₃/Mass of Mixture) x 100

% KClO₃ = (0.894... g/1.382 g) x 100

% KClO₃ = 64.7%