2KClO3 → 2KCl + 3O2How many moles of KCl are produced when 67.43 g of KClO3 is decomposed?

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

Here are the steps:

1. First, we need to find the molar mass of KClO3. The molar mass of K (Potassium) is approximately 39.10 g/mol, Cl (Chlorine) is 35.45 g/mol, and O (Oxygen) is 16.00 g/mol. Since there are 3 Oxygen atoms, we multiply the molar mass of Oxygen by 3. So, the molar mass of KClO3 is 39.10 g/mol + 35.45 g/mol + (3 * 16.00 g/mol) = 122.55 g/mol.

2. Next, we need to convert the given mass of KClO3 to moles. We do this by dividing the given mass by the molar mass. So, 67.43 g / 122.55 g/mol = 0.55 moles of KClO3.

3. Now, we use the stoichiometric ratio from the balanced chemical equation. The equation tells us that 2 moles of KCl are produced for every 2 moles of KClO3 decomposed. So, the ratio is 1:1.

4. Finally, we multiply the moles of KClO3 by the stoichiometric ratio to find the moles of KCl produced. So, 0.55 moles * 1 = 0.55 moles of KCl.

So, when 67.43 g of KClO3 is decomposed, 0.55 moles of KCl are produced.