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**Hint:**When potassium chlorate decomposes, potassium chloride and oxygen gas are formed. Then write the balanced reaction and find the number of moles and then weight of oxygen produced from the given weight of potassium chlorate.

**Complete step by step solution:**

We will need to find the balanced decomposition reaction of potassium chlorate. Then we can find the amount of oxygen gas generated.

- In the decomposition of potassium chlorate, potassium chloride and oxygen gas are the products. The balanced reaction can be written as below.

\[2KCl{O_3} \to 2KCl + 3{O_2}\]

Now, we will find the molecular weight of potassium chlorate.

Molecular weight of $KCl{O_3}$ = Atomic weight of K + Atomic weight of Cl + 3(Atomic weight of O)

Molecular weight of $KCl{O_3}$ = 39 + 35.5 + 3(16) = 122.5 $gmo{l^{ - 1}}$

We know that molecular weight of oxygen gas is 32 $gmo{l^{ - 1}}$

Now, we know that ${\text{Number of moles = }}{{{\text{Weight}}}}{{{\text{Molecular weight}}}}$

So, for $KCl{O_3}$ , we can write that ${\text{Number of moles = }}{5}{{122.5}} = 0.0408$ moles

We can see that from two moles of potassium chlorate, three moles of oxygen is produced.

So, from 0.0408 moles of potassium chlorate, the number of moles of oxygen generated will be ${{0.0408 \times 3}}{2} = 0.0612$ moles

We know that Mass = Number of moles $ \times $ Molecular weight

So, for oxygen, we can write that

Mass = 0.0612 $ \times $ 32 = 1.9584 g

**Thus, we can conclude that the mass of oxygen produced in the complete decomposition of 5 g of $KCl{O_3}$ will be 1.9684 g.**

**Note:**Note that here it is given that complete decomposition of $KCl{O_3}$ (Potassium chlorate) occurs. If the decomposition is not complete, then we need the % decomposition in order to predict the weight of oxygen gas produced.

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