1. Write the balanced chemical equation:

2 KClO₃(s) → 2 KCl(s) + 3 O₂(g)

2. Calculate the moles of KClO₃:

* Molar mass of KClO₃ = 39.10 g/mol + 35.45 g/mol + 3 * 16.00 g/mol = 122.55 g/mol

* Moles of KClO₃ = (8.15 g) / (122.55 g/mol) = 0.0665 mol

3. Calculate the moles of O₂ produced:

* From the balanced equation, 2 moles of KClO₃ produce 3 moles of O₂.

* Moles of O₂ = (0.0665 mol KClO₃) * (3 mol O₂ / 2 mol KClO₃) = 0.0998 mol O₂

4. Convert the temperature to Kelvin:

* T (K) = T (°C) + 273.15

* T (K) = 22.0 °C + 273.15 = 295.15 K

5. Convert the pressure to atmospheres:

* 1 atm = 760 mm Hg

* P (atm) = (728 mm Hg) / (760 mm Hg/atm) = 0.958 atm

6. Use the ideal gas law to calculate the volume:

* PV = nRT

* V = (nRT) / P

* Where:

* V = volume (L)

* n = moles (mol)

* R = ideal gas constant (0.0821 L·atm/mol·K)

* T = temperature (K)

* P = pressure (atm)

* V = (0.0998 mol * 0.0821 L·atm/mol·K * 295.15 K) / 0.958 atm

* V = 2.54 L

Therefore, 2.54 L of O₂ would be produced at 22.0 °C and 728 mm Hg.