How many grams of oxygen are formed when 58.6 g of kno3 decomposes?

The formula for potassium nitrate is: 2KNO₃ → 2KNO₂ + O₂

If moles = mass : molar mass

then moles of KNO₃ = 58.6 g : [ (39 * 1) + (14 * 1) + (16 * 3) ] g/mol

= 58.6 g : (101 g) g/mol

= 0.5802 mol

Now the mole ratio of KNO₃ : O₂ is 2 : 1

∴ if moles of KNO₃ = 0.5802 mol

then moles of O₂ = 0.5802 mol : 2

= 0.2901 mol

Now since mass = moles * molar mass

∴ the mass of oxygen produced = 0.2901 mol * (16 * 2) g/mol

= 9.283 g

∴ the grams of oxygen gas produced when 58.6 grams of potassium nitrate is decomposed is ~ 9.28 g