Final answer:
The volume of O₂ gas formed at STP from the decomposition of 50.0 g of KClO₃ is approximately 13.71 liters.
Step-by-step explanation:
The question asks to determine the volume of O₂ gas formed at Standard Temperature and Pressure (STP) when 50.0 g of KClO₃ decomposes. The decomposition reaction for potassium chlorate (KClO₃) is:
2KClO₃ (s) → 2KCl (s) + 3O₂ (g).
First, we calculate the number of moles of KClO₃ that 50.0 g represents:
Number of moles = mass (g) / molar mass (g/mol) = 50.0 g / 122.55 g/mol = 0.408 mol KClO₃.
From the balanced equation, 2 moles of KClO₃ produce 3 moles of O₂. Therefore, 0.408 mol of KClO₃ will produce:
(0.408 mol KClO₃) × (3 mol O₂ / 2 mol KClO₃) = 0.612 mol O₂.
At STP, 1 mol of any gas occupies 22.4 L. Thus, the volume of O₂ produced is:
Volume O₂ = moles × molar volume at STP = 0.612 mol × 22.4 L/mol = 13.71 L.
So, the volume of O₂ produced at STP is approximately 13.71 liters.