The amount of oxygen formed when $12.26 \mathrm{g} \mathrm{of} {\mathrm{KClO}}_3$ is heated is

Step 1: Decomposition Reaction

The decomposition reaction of KClO₃ is:

        2KClO₃ → 2KCl + 3O₂
    

From the reaction, 2 moles of KClO₃ produce 3 moles of O₂.

Step 2: Molar Mass of KClO₃

The molar mass of KClO₃ is calculated as:

• K = 39.1
• Cl = 35.45
• O₃ = 3 × 16 = 48

Molar mass of KClO₃ = 39.1 + 35.45 + 48 = 122.55 g/mol

Step 3: Moles of KClO₃

The number of moles of KClO₃ in 12.26 g is:

Moles of KClO₃ = Mass of KClO₃ / Molar mass of KClO₃

Moles = 12.26 / 122.55 ≈ 0.1 mol

Step 4: Moles of O₂ Produced

From the reaction, 2 mol of KClO₃ produce 3 mol of O₂. Therefore, 0.1 mol of KClO₃ will produce:

Moles of O₂ = 0.1 × (3 / 2) = 0.15 mol

Step 5: Mass of O₂

The molar mass of O₂ is 32 g/mol. The mass of O₂ produced is:

Mass of O₂ = Moles of O₂ × Molar mass of O₂

Mass = 0.15 × 32 = 4.8 g

Final Answer:

The amount of oxygen (O₂) formed is 4.8 g.