For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.

2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)

2.72 g KClO3

0.361 g KClO3

83.6 kg KClO3

22.5 mg KClO3

The balanced chemical reaction is given as follows:

2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)

The starting amount of the reactant are given above. These values would be used for the calculations. We do as follows:

2.72 g KClO3 (1 mol / 122.50g) (3 mol O2 / 2 mol KClO3) (32 g O2 / 1 mol O2) = 1.06 g O2

0.361 g KClO3 (1 mol / 122.50g) (3 mol O2 / 2 mol KClO3) (32 g O2 / 1 mol O2) = 0.14 g O2

83.6 kg KClO3 (1000g / 1kg) (1 mol / 122.50g) (3 mol O2 / 2 mol KClO3) (32 g O2 / 1 mol O2) = 3275.76 g O2

22.5 mg KClO3 (1 g / 1000 mg) (1 mol / 122.50g) (3 mol O2 / 2 mol KClO3) (32 g O2 / 1 mol O2) = 0.009 g O2