How to calculate this chemical equation?

Question

You are given a mixture of KCl and KClO3. 

You add MnO2 as a catalyst to the mixture in a test tube and heat the test tube to achieve constant mass.

Determine the percent composition of KClO3 in the mixture.

Given: K: 39.10 g/mole, Cl: 35.45 g/mole, O: 16.00 g/mole

          heat

 2KClO3(s) ---------→ 2KCl(s)+ 3O2(g)

Mass of Test tube + MnO2: 32.0118 g 

Test tube + MnO2 + KClO3 + KCl: 33.2282 g 

Test tube + MnO2 + reaction product: 33.0251 g after 1st heating 

                      33.0233 g after 2nd heating

1. Calculate the mass of O2 released.

2. Calculate the moles of O2 released.

3. Calculate the moles of KClO3 required by the moles of O2 released. 

4. Calculate the mass of KClO3 required.

5. Calculate the mass of KClO3 + KCl mixture. 

6. Calculate the percent of KClO3 in KClO3 + KCl mixture. 

Answer 1

 2KClO₃(s) → 2KCl(s)+ 3O₂(g) ... balanced equation

To find % composition of KClO3 in an impure sample, we need to find the mass of KClO3 present. To do this, we must first find moles KClO3 present. To do this, we will find moles of O2 produced and using the 2:3 mol ratio of the balanced equation, convert to moles KClO3.

mass of sample = 33.2282 g - 32.0118 g = 1.2164 g

mass of sample after heating = 33.0233 g

mass lost due to heating = 33.2282 g - 33.0233 g = 0.2049 g lost

What is lost during heating? It would be the O2 gas that is formed and has escaped.

(1). mass O2 released = 0.2049 g

(2). moles O2 released = 0.2049 g O2 x 1 mol O2/31.9988 g = 0.00640337 moles O2

(3). moles KClO3 needed = 0.00640337 moles O2 x 2 mol KClO3 / 3 mol O2 = 0.0042689 moles KClO3

(4). mass KClO3 = 0.0042689 moles KClO3 x 122.55 g/mol = 0.52315 g KClO3

(5). mass of KCl + KClO3 sample = 1.2164 g (see above)

(6). % KClO3 = 0.52315 g / 1.2164 g (x100) = 43.008%