$$2KClO_3 + O_2 \rightarrow 2KClO_4$$
From the balanced equation, we can see that 2 moles of potassium chlorate (KClO3) react with 1 mole of oxygen gas (O2) to produce 2 moles of potassium perchlorate (KClO4).
So, we need to calculate the number of moles of oxygen gas used in the reaction. We are given that 500 g of oxygen gas is used. The molar mass of oxygen gas is 32 g/mol. Therefore, the number of moles of oxygen gas used is:
$$500 \text{ g O}_2 \times \frac{1 \text{ mol O}_2}{32 \text{ g O}_2} = 15.625 \text{ mol O}_2$$
According to the balanced chemical equation, 2 moles of potassium chlorate are produced for every 1 mole of oxygen gas consumed. So, the number of moles of potassium chlorate produced is:
$$15.625 \text{ mol O}_2 \times \frac{2 \text{ mol KClO}_4}{1 \text{ mol O}_2} = 31.25 \text{ mol KClO}_4$$
Finally, we convert the moles of potassium chlorate to grams using the molar mass of potassium chlorate, which is 138.55 g/mol:
$$31.25 \text{ mol KClO}_4 \times \frac{138.55 \text{ g KClO}_4}{1 \text{ mol KClO}_4} = 4337.31 \text{ g KClO}_4$$
Therefore, 4337.31 grams of potassium chlorate would be produced by reacting 500 g of oxygen gas with potassium chloride.
