Determine the mass of lithium that can be oxidized with 11.2 liters of oxygen.

1. Let’s write down the reaction of lithium oxidation with oxygen:

4Li + O2 = 2Li2O;

2.Calculate the chemical amount of oxygen:

n (O2) = V (O2): Vm;

n (O2) = 11.2: 22.4 = 0.5 mol;

3.It can be seen from the reaction equation that the amount of lithium is four times the amount of oxygen:

n (Li) = n (O2) * 4 = 0.5 * 4 = 2 mol;

4.determine the mass of lithium:

m (Li) = n (Li) * M (Li) = 2 * 7 = 14 g.

Answer: 14 g.