First, you have to determine the reaction involved between potassium chlorate to oxygen. When potassium chlorate, or KClO₃, is heated, it is decomposed into potassium chloride, KCl, and oxygen gas. The reaction is written below:
2 KClO₃ ⇒ 2 KCl + 3 O₂
So, the for every 2 moles of KClO₃ heated, it yields 3 moles of O₂. Let us first convert 110 grams to moles using the molar mass of KClO₃ equal to 122.55 g/mol.
110 g (1 mol/122.55 g) = 0.8976 mol KClO₃
Then, we use the stoichiometric ratios in the reaction:
Amount O₂ produced = 0.8976 mol KClO₃ (3 moles O₂/ 2 moles KClO₃)
Amount O₂ produced = 1.3464 moles
Now, we assume the oxygen is ideal gas to be able to use the ideal gas equation: PV =nRT. At STP (standard temperature and pressure), the pressure is equal to 101.325 Pa while the temperature is 273.15 K. Substituting the values:
(101.325)(V) = (1.3464)(8.314)(273.15)
V = 30.18 m³
Since 1 m³ = 1000 L, the amount of volume in liters of oxygen produced is 30,176 L.
