Using the reaction equation Ba + O2 → BaO, calculate the volume of oxygen if 10 g of barium has entered into a reaction with oxygen.

2Ba + O2 = 2BaO
Let us find the amount of barium substance n = m / M = 10 g / 137 g / mol = 0.073 mol. By reaction with 2 mol of barium 1 mol of oxygen reacts, then 0.073 / 2 = n (O2) / 1. Then n (O2) = 0.073 / 2 = 0.0365 mol. Oxygen volume V = n * Vm = 0.0365 mol * 22.4 L / mol = 0.8176 L. (Vm is molar volume, constant for all gases under normal conditions).
Answer: the volume of oxygen is 0.8176 liters.