how many liters of oxygen will be produced at STP if 1.25 kg potassium chlorate decomposes completely?

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