How many litres of oxygen at stp can be obtain from 110g of potasium chlorade?

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.