Answer:
2.74 L
Explanation:
The reaction that takes place is:
2KClO₃(s) + Δ → 2KCl (s) + 3O₂ (g)
To calculate the volume of oxygen gas released, we calculate the moles of O₂ produced, using the mass of reactant given by the problem:
10.0 g KClO₃ * [tex]\frac{1molKClO_{3}}{122.55gKClO_{3}} *\frac{3molO_{2}}{2molKClO_{3}} =[/tex] 0.1224 mol O₂
At STP, 1 mol of gas occupies 22.4 L, so the amount occupied by 0.1224 mol is:
0.1224 mol * 22.4 L·mol⁻¹= 2.74 L
