Use that equation to determine the number of liters of oxygen gas that could be produced from 10.0 g of KClO3 at STP. Show your work. 2KClO3---> 2KCl + 3O2

Question

g

g/mol

mol

L

22.4= L/mol

mol

Answer 1

2KClO₃ ==> 2KCl + 3O₂

At STP, 1 mol of any ideal gas occupies 22.4 L (memorize this value)

So, how many moles of O₂ are formed from 10.0 g of KClO₃?

molar mass KClO₃ = 122.5 g / mol

molar mass O₂ = 32 g / mol

mols KClO₃ used = 10.0 g x 1 mol / 122.5 g = 0.0816 mols KClO₃

mols O₂ produced = 0.0816 mols KClO₃ x 3 mols O₂ / 2 mols KClO₃ = 0.1224 mols O₂ produced

Liters of O₂ = 0.1224 mols x 22.4 L / mol = 2.74 L O₂ @ STP