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

User IslandCow
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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
User JArgente
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