Potassium chlorate (used in fireworks, flares and safety matches) forms oxygen and potassium chloride when heated.                KClO3(s) ⟶ KCl(s) + O2(g) [unbalanced]How many grams of oxygen are formed when 26.4 g of potassium chlorate is heated?

To solve this problem, we need to follow these steps:

1. Balance the chemical equation: The balanced equation for the reaction is 2KClO3(s) ⟶ 2KCl(s) + 3O2(g).

2. Calculate the molar mass of KClO3 and O2: The molar mass of KClO3 is approximately 122.55 g/mol (39.10 g/mol for K, 35.45 g/mol for Cl, and 16.00 g/mol for each of the three O atoms). The molar mass of O2 is approximately 32.00 g/mol (16.00 g/mol for each of the two O atoms).

3. Convert the mass of KClO3 to moles: Using the molar mass of KClO3, we can convert the given mass of KClO3 to moles. 26.4 g KClO3 * (1 mol KClO3 / 122.55 g KClO3) = 0.215 mol KClO3.

4. Use the stoichiometry of the reaction to find the moles of O2: From the balanced chemical equation, we know that 2 moles of KClO3 produce 3 moles of O2. So, 0.215 mol KClO3 * (3 mol O2 / 2 mol KClO3) = 0.3225 mol O2.

5. Convert the moles of O2 to grams: Using the molar mass of O2, we can convert the moles of O2 to grams. 0.3225 mol O2 * (32.00 g O2 / 1 mol O2) = 10.32 g O2.

So, when 26.4 g of potassium chlorate is heated, approximately 10.32 g of oxygen are formed.