Final answer:
To determine the volume of O₂ formed when 50.0 g of KClO₃ decomposes, calculate the number of moles of KClO₃, use the stoichiometry to find the number of moles of O₂, then convert moles to volume at STP.
Step-by-step explanation:
To determine the volume of O₂ formed when 50.0 g of KClO₃ decomposes, we first need to calculate the number of moles of KClO₃. The molar mass of KClO₃ is 122.55 g/mol, so we can divide the mass by the molar mass to find the number of moles.
50.0 g / 122.55 g/mol = 0.408 mol KClO₃
The balanced equation shows that the reaction produces 3 moles of O₂ for every 2 moles of KClO₃. Using this ratio, we can calculate the number of moles of O₂ produced.
0.408 mol KClO₃ * (3 mol O₂ / 2 mol KClO₃) = 0.612 mol O₂
Finally, we can convert the number of moles of O₂ to volume at STP (standard temperature and pressure). 1 mole of gas occupies 22.4 L at STP, so we can multiply the number of moles by the molar volume.
0.612 mol O₂ * 22.4 L/mol = 13.7 L O₂ (at STP)