Question

When heated, KClO3, decomposes into KCl and O2, if this reaction produced 65.0g of KCl, how much O2 was produced in grams?

Answer 1

Given:

Mass of KCl = 65.0 g

Solution:

Decomposition reaction:

KClO3 --- > KCl + O2

Mol ratio of KCl : KClO3 is 1:1

Calculation of moles of KCl

Mol KCl = mass of it in g / molar mass

Molar mass of KCl = 74.5513 g /mol

n KCl = 65.0 g x 1 mol / 74.5513 g = 0.87188 mol

Moles of KClO3 = moles of KCl x 1 mol KClO3 / 1 mol KCl

= 0.87188285 mol KCl x 1 mol KClO3 / 1 KCl

= 0.87188285 mol KClO3

Now we use mole ratio of KClO3: O2

Mol ratio of KClO3 : O2 is 1 : 1

Therefore moles of O2 produced = mol of KClO3 x 1 mol O2 / 1 mol KClO3

=0.87188 mol O2.

Mass of O2 = mole O2 x molar mas

= 0.87188 mol O2 x 31.998 g /mol

= 27.898 g

Mass of O2 produced

= 27.898 g