Oxygen Gas can be produced by decomposing potassium chlorate using the reaction below. If 30.6 g of KCIO3 is heated and decomposes completely, how many liters of oxygen gas is produced at STP?

Answer 1

So to figure this out, first we need to write the decomposition reaction

2KClO₃ --> 2KCl + 3O₂

From this equation we see that it takes 2 moles of KClO₃ to make 3 moles of O₂The molar mass of KClO₃ is 122.5g.

So now let's turn grams of KClO₃ into moles of O_2.

30.6g x 1mol KClO₃ / 122.5g x 3 moles O₂ / 2 mol KClO₃ = 0.375 moles O₂

At STP, we know 1 mole of any gas has a volume of 22.4L. So, we just multiply the moles of oxygen by 22.4L and we get 8.4L