Question

For the reaction shown, calculate how many grams of oxygen form when each of the following completely reacts. 2KClO3(s)→2KCl(s)+3O2(g)

2.73g KClO3

0.336g KClO3

89.6g KClO3

20.5g KClO3

Answer 1

Solution :

For the given reaction,

2KClO3 ---> 2KCl + 3O2

A) 2.73 g KClO3

Moles of KClO3 = Mass /Molar mass = 2.73 g / 122.55 g mol-1 = 0.02228 mol

Since, 2 mol of KClO3 forms 3 mol of O2

Hence, 0.02228 mol of KClO3 will form = 0.02228 mol x 3/2 = 0.03342 mol

Thus,

Mass of O2 = Moles x Molar mass of O2 = 0.03342 mol x 32.0 g mol-1

Mass of O2 = 1.069 g

B) 0.336 g KClO3

Moles of KClO3 = 0.336 g / 122.55 g mol-1 = 0.00274 mol

Moles of O2 = 0.00274 mol x 3/2 = 0.00411 mol

Mass of O2 = 0.00411 g x 32.0 g mol-1 = 0.132 g

C) 89.6 g KClO3

Moles of KClO3 = 89.6 g / 122.55 g mol-1 = 0.731 mol

Moles of O2 = 0.731 mol x 3/2 = 1.0965 mol

Mass of O2 = 1.0965 mol x 32.0 g mol-1 = 35.088 g

D) 20.5 g KClO3

Moles of KClO3 = 20.5 g / 122.55 g mol-1 = 0.1673 mol

Moles of O2 = 0.1673 mol x 3/2 = 0.251 mol

Mass of O2 = 0.251 mol x 32.0 g mol-1 = 8.032 g