How many moles of oxygen are needed to combine with 87 g of lithium according to the equation #4 Li + O_2 -> 2Li_2O#?
Moles of Li? Approx. 12 mol. Therefore approx. 3 moles of oxygen gas are required.
You have the stoichiometric equation, and we ASSUME excess oxygen.
So moles of metal
Therefore, we require a stoichiometric quantity of oxygen gas, and you have kindly provided the equation:
Note that we deal with oxygen gas, the diatomic molecule.