Question

If a 0.500-g sample of a potassium nitrate mixture is decomposed by heating, what is the milliliter volume of oxygen gas released at STP?

Answer 1

At STP conditions :

pressure = 1 atm

temperature = 273 K

moles of KNO3 = 0.5 / 101.1 = 4.95 x 10^-3 mol

2 KNO₃ (s) → 2 KNO₂ (s) + O₂ (g)

2 mol KNO3   --------------> 1 mol O2

4.95 x 10^-3 KNO3 ----------> ??

moles of O2 = 4.95 x 10^-3 / 2 = 2.47 x 10^-3 mol O2

P V = n R T

1 x V = 2.47 x 10^-3 x 0.0821 x 273.15

V = 0.0554 L

volume of oxygen gas = 55.4 mL