As a result of the combustion of Me II group m = 16g, 4.48 liters of oxygen were consumed. Define Me?

Combustion reaction of a metal of the 2nd group: 2Me + O2 = 2MeO
Find the amount of oxygen n = V / Vm = 4.48 l / 22.4 l / mol = 0.2 mol. Vm – molar volume, constant for all gases.
Let’s find the amount of metal substance. By reaction with 2 mol of metal, 1 mol of oxygen reacts, which means n (Me) / 2 = 0.2 / 1. n (Me) = 2 * 0.2 = 0.4 mol. Let us find the molar mass of this metal M = m / n = 16 g / 0.4 mol = 40 g / mol. In the second group of the periodic table, only one metal has a molar mass of 40 g / mol – calcium.