Question

Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO3(s).

The equation for the reaction is

2KClO3⟶2KCl+3O2

Calculate how many grams of O2(g) can be produced from heating 98.6 g KClO3(s).

mass:

Answer 1

Number of moles of KClO3 = 98.6 g / 122.55 g/mol = 0.805 mole

From the balanced equation we can say that

2 mole of KClO3 produces 3 mole of O2 so

0.805 mole of KClO3 will produce

= 0.805 mole of KClO3 *(3 mole of O2 / 2 mole of KClO3)

= 1.21 mole of O2

mass of 1 mole of O2 = 32.0 g so

the mass of 1.21 mole of O2 = 38.7 g

Therefore, the mass of O2 produced would be 38.7 g