The molar mass of potassium chlorate(KClO3) is 122.55g/mol.

$$Moles \space of \space KClO_3=\frac{Mass}{Molar \space Mass}$$

$$Moles \space of \space KClO_3=\frac{6.02g}{122.55g/mol}=0.0491 mol$$

Step 2: Determine the mole ratio between KClO3 and O2

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

$$2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)$$

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

Therefore, the mole ratio between KClO3 and O2 is 2:3.

Step 3: Use the mole ratio to calculate the moles of O2

Using the mole ratio, we can calculate the moles of O2 produced from 0.0491 mol of KClO3.

$$Moles \space of \space O_2=\frac{3 \space moles \space O_2}{2\space moles \space KClO_3}\times0.0491 \space mole \space KClO_3=0.07365 mol$$

Step 4: Convert the moles of O2 to grams

The molar mass of O2 is 32.00 g/mol.

$$Mass \space of \space O_2=Moles \times Molar \space Mass$$

$$Mass \space of \space O_2= 0.07365 mol \times 32.00g/mol=\boxed{2.360g}$$

Therefore, 6.02 g of potassium chlorate produces 2.360 grams of oxygen gas.