A 7.11g sample of potassium chlorate was decomposed according to the following equation: 2KCLO3 -->2KCL + 3O2 How many moles of oxygen are formed?

A 7.11g sample of potassium chlorate was decomposed according to the following equation: 2KCLO3 -->2KCL + 3O2, the moles of oxygen are formed - 0.0870

Given:

sample of potassium chlorate =  7.11g

reaction:

2KClO₃ -->2KCl + 3O₂

This means 2 moles of KClO₃ produces 2 moles of KCl and 3 moles of O₂

Solution:

KClO₃ = (1 x 39.1) + (1 x 35.45) + (3 x 16)  

          = 39.1 + 35.45 + 48  

          = 122.55 g

So, 1 mole of KClO₃ = 122.55 g

Then, 7.11 g of KClO₃

=

= 0.0580 moles

We know,

2 moles of KClO₃ = 3 moles of O₂

Then,

0.0580 moles of KClO₃ = × 0.0580 moles

= 0.0870 moles of oxygen

Thus, In this reaction, the moles of oxygen are formed - 0.0870.

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The answer to your question is 0.087 moles of Oxygen

Explanation:

Data

mass of KClO₃ = 7.11 g

moles of O₂ = ?

Balanced chemical reaction

               2KClO₃  ⇒   2KCl  +  3O₂

Process

1.- Calculate the molar mass of KClO₃

KClO₃ = (1 x 39.1) + (1 x 35.45) + (3 x 16)

           = 39.1 + 35.45 + 48

           = 122.55 g

2.- Convert the mass to KClO₃ to moles

                  122.55 g 1 mol

                      7.11 g    x

                       x = (7.11 x 1)/122.55

                       x = 0.058 moles

3.- Calculate the moles of Oxygen

                  2 moles of KClO₃ 3 moles of Oxygen

                  0.058 moles          x

                  x = (0.058 x 3) / 2

                  x = 0.087 moles of Oxygen

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