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.

Learn more:

link

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

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.