How many liters of oxygen will be released during the complete decomposition of 24.5

How many liters of oxygen will be released during the complete decomposition of 24.5 g of potassium chlorate (KCLO3) and how much potassium chloride is formed?

Let’s write the reaction equation:

2KClO3 → 2KCl + 3O2 (conditions: manganese (IV) oxide, temperature).

Let’s find the amount of potassium chlorate substance:

n (KClO3) = m (KClO3) / M (KClO3) = 24.5 g / 122.5 g / mol = 0.2 mol.

Let’s find the amount of oxygen substance according to the reaction equation:

n (O2) = 1.5 * n (KClO3) = 1.5 * 0.2 mol = 0.3 mol.

Let’s find the volume of oxygen:

V (O2) = n (O2) * Vm = 0.3 mol * 22.4 L / mol = 6.72 L.

Let’s find the amount of potassium chloride substance according to the reaction equation:

n (KCl) = n (KClO3) = 0.2 mol.

Let’s find the mass of potassium chloride:

m (KCl) = n (KCl) * M (KCl) = 0.2 mol * 74.5 g / mol = 14.9 g.

Answer: 6.72 L; 14.9 g