When potassium chlorate is heated, it decomposes into potassium chloride and oxygen in the reaction 2KCIO3 -> 2KCI + 3O2. How much oxygen is liberated when 50 g of potassium chlorate is heated?
figure the moles of potassium chlorate in 50 grams. Then you will get 3/2 moles of oxygen times the moles of KClO3
To determine the amount of oxygen liberated when 50 grams of potassium chlorate (KCIO3) is heated, we need to calculate the molar mass of potassium chlorate and use stoichiometry to find the amount of oxygen produced.
1. Calculate the molar mass of potassium chlorate (KCIO3):
- Molar mass of K = 39.10 g/mol (potassium)
- Molar mass of Cl = 35.45 g/mol (chlorine)
- Molar mass of O = 16.00 g/mol (oxygen)
- Molar mass of KCIO3 = Molar mass of K + (Molar mass of Cl * 1) + (Molar mass of O * 3)
= 39.10 + (35.45*1) + (16.00*3)
= 39.10 + 35.45 + 48.00
= 122.55 g/mol
2. Determine the moles of potassium chlorate in 50 grams:
- Moles = Mass / Molar mass
- Moles of KCIO3 = 50 g / 122.55 g/mol
= 0.408 moles
3. Use the balanced chemical equation to determine the moles of oxygen produced:
- According to the balanced equation: 2KCIO3 -> 2KCI + 3O2
- For every 2 moles of KCIO3, 3 moles of O2 are produced.
- Therefore, the moles of O2 = (Moles of KCIO3) * (3 moles of O2 / 2 moles of KCIO3)
= 0.408 moles * (3/2)
= 0.612 moles
4. Convert moles of oxygen to grams:
- Moles = Mass / Molar mass
- Mass of O2 = Moles * Molar mass of O
= 0.612 moles * 16.00 g/mol
= 9.79 grams
Therefore, when 50 grams of potassium chlorate is heated, approximately 9.79 grams of oxygen is liberated.