What is the mass of the formed metal oxide, taking into account that 22.4 liters of oxygen

What is the mass of the formed metal oxide, taking into account that 22.4 liters of oxygen are consumed for the combustion of 48 g of metal?

Given:
m (Me) = 48 g
V (O2) = 22.4 l
Vm = 22.4 l / mol

Find:
m (MeO) -?

Solution:
1) Write a diagram of the chemical reaction:
Me + O2 -> MeO;
2) Calculate the amount of oxygen substance (under normal conditions):
n (O2) = V (O2) / Vm = 22.4 / 22.4 = 1 mol;
3) Calculate the mass of oxygen:
m (O2) = n (O2) * M (O2) = 1 * 32 = 32 g;
4) Calculate the mass of the metal oxide (according to the Lomonosov-Lavoisier mass conservation law):
m (MeO) = m (Me) + m (O2) = 48 + 32 = 80 g.

Answer: The mass of the metal oxide is 80 g.