How much hydrogen and oxygen, measured under normal conditions, is required to form 7.2 g of water.

Reaction: 2H2 + O2 = 2H2O
According to the table. Periodic molar mass of water, M = 1 * 2 + 16 = 18 g / mol.
Find the amount of water: n = m / M = 7.2 g / 18 g / mol = 0.4 mol.
According to the reaction, from 2 moles of H2, 2 moles of water are obtained, that is, n (H2O) = n (H2) = 0.4 mol. To find the volume, we use the molar volume: Vm = 22.4 l / mol. The volume of hydrogen: V = n * Vm = 0.4 mol * 22.4 L / mol = 8.96 L.
According to the reaction from 1 mole of O2, 2 moles of water are formed, we find the amount of O2. n (O2) / 1 = 0.4 / 2. n (O2) = 0.2 mol. V (O2) = 0.2 mol * 22.4 L / mol = 4.48 L.