# The decomposition of potassium chlorate (KClO3) is used as a source of oxygen in the laboratory. How much potassium chlorate is needed to produce 10.2 mol of oxygen? Answer in units of mol

## 15.3 mol

## Laila is El wrongo! Used the factor upside down to get 15.3.

2KClO3 ==> 2KCl + 3O2

mols KClO3 = mols O2 x (2 mols KClO3/3 mols O2) = 10.2 x (2/3) = ?

## 2KClO3 ==> 2KCl + 3O2

mols KClO3 = mols O2 x (2 mols KClO3/3 mols O2) = 10.2 x (2/3) = ?

## To find out how much potassium chlorate (KClO3) is needed to produce a certain amount of oxygen, we need to use the ratio of moles between KClO3 and O2 from the balanced chemical equation.

The balanced chemical equation for the decomposition of potassium chlorate is:

2 KClO3 -> 2 KCl + 3 O2

From the equation, we can see that 2 moles of KClO3 produce 3 moles of O2.

To find out how much KClO3 is needed to produce 10.2 moles of O2, we can set up a proportion:

2 moles KClO3 / 3 moles O2 = x moles KClO3 / 10.2 moles O2

Cross-multiplying, we get:

2 moles KClO3 * 10.2 moles O2 = 3 moles O2 * x moles KClO3

Simplifying, we have:

20.4 moles KClO3 = 3 moles O2 * x moles KClO3

Dividing both sides of the equation by 3 moles KClO3, we get:

20.4 moles KClO3 / 3 moles KClO3 = x moles KClO3

Simplifying further:

x = 20.4 moles KClO3 / 3 moles KClO3

x = 6.8 mol

Therefore, you would need 6.8 moles of potassium chlorate (KClO3) to produce 10.2 moles of oxygen (O2).