Question

Potassium chlorate(s) can be decomposed to O2(g) and potassium chloride(s) by heating. Calculate the mass of potassium chlorate that must react to produce 187.1 mL of dry O2, measured at STP.

Answer 1

The balanced equation is

2 KClO₃ ----> 2 KCl + 3 O₂

The volume of potassium chlorate = 187.1 mL = 0.1871 L

At STP,

22.4 L of O2 = 1 mole so

0.1871 L of O2 = 0.00835 mole of O2

Therefore, the number of moles of O2 = 0.00835

From the balanced equation we can say that

3 mole of O2 is produced by 2 mole of KClO3 so

0.00835 mole of O2 will be produced by

= 0.00835 mole of O2 *(2 mole of KClO3 / 3 mole of O2)

= 0.005567 mole of KClO3

mass of 1 mole of KClO3 = 122.5495 g

the mass of 0.005567 mole of KClO3 = 0.682 g

Therefore, the mass of potassium chlorate reacts would be 0.682 g