How many liters of oxygen can be obtained by decomposing 158 g of potassium permanganate?

First, you need to write down the reaction equation.

2KMnO4 = K2MnO4 + MnO2 + O2.

Let’s determine the amount of potassium permanganate substance. We use the following formula.

n = m / M.

M (KMnO4) = 158 g / mol.

n = 1/1 = 1 mol.

This means that the amount of substance is 1 mol.

Let’s make a proportion according to the reaction equation.

1 mole – x mole.

2 mol – 1 mol.

X = 1 * 1/2 = 0.5 mol.

Next, we determine the volume of oxygen.

V = nVm.

V = 0.5 * 22.4 = 11.2 liters.

This means that the volume of oxygen released as a result of the reaction is 11.2 liters.