How many grams of potassium chlorate must be heated to obtain the oxygen required to completely burn 6 grams of carbon

To solve this problem, we need to use stoichiometry, which is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

The balanced chemical equation for the reaction of potassium chlorate (KClO3) decomposing into potassium chloride (KCl) and oxygen (O2) is:

2 KClO3 → 2 KCl + 3 O2

And the balanced chemical equation for the reaction of carbon (C) burning in oxygen (O2) to produce carbon dioxide (CO2) is:

C + O2 → CO2

From the second equation, we can see that 1 mole of carbon requires 1 mole of oxygen to completely burn.

The molar mass of carbon is approximately 12 grams/mole. So, 6 grams of carbon is 6/12 = 0.5 moles.

Therefore, we need 0.5 moles of oxygen to completely burn 6 grams of carbon.

From the first equation, we can see that 2 moles of potassium chlorate produce 3 moles of oxygen. So, 1 mole of potassium chlorate will produce 3/2 = 1.5 moles of oxygen.

Therefore, to produce 0.5 moles of oxygen, we need 0.5/1.5 = 0.33 moles of potassium chlorate.

The molar mass of potassium chlorate (KClO3) is approximately 122.55 grams/mole. So, 0.33 moles of potassium chlorate is 0.33 * 122.55 = 40.44 grams.

Therefore, approximately 40.44 grams of potassium chlorate must be heated to obtain the oxygen required to completely burn 6 grams of carbon.