How much oxygen is required to completely oxidize 23 g of sodium?

Answer: V (O2) = 5.6 liters.
Explanation of the solution to the problem: to solve this problem, you first need to write down the equation of the chemical reaction: 4Na + O2 => 2Na2O.
Then you need to find the amount of sodium substance, for this you need to divide its mass by the molar mass of sodium:
n (Na) = m / M = 23/23 = 1 mol.
Now you need to find the amount of oxygen substance: n (O2) = n (Na) / 4, because the reaction shows that oxygen and sodium interact in a ratio of 1: 4, respectively.
n (O2) = 1/4 = 0.25 mol.
The last step is to find the volume of oxygen, for this we multiply the found amount of substance by the molar volume:
V (O2) = n (O2) * Vm = 0.25 * 22.4 = 5.6 liters.