Answer:
342.5 L of O₂ will be produced at STP
Explanation:
This is the reaction of decomposition:
2KClO₃ → 2KCl + 3O₂
Let's determine the moles of the reactant salt.
Molar mass KClO₃ = 122.55 g/m
1.25 kg = 1250 g
1250 g / 122.55 g/m = 10.2 moles
Ratio is 2:3
2 moles of chlorate decompose in 3 mol of oxygen
10.2 moles of chlorate will decompose in (10.2 .3)/2 = 15.3 moles
Let's apply the Ideal gases Law equation
STP = 273K of T° and 1 atm of pressure.
1 atm . V = 15.3 mol . 0.082 L.atm/mol.K . 273K
V = (15.3 mol . 0.082 L.atm/mol.K . 273K) / 1 atm → 342.5 L
