If you mix 40 grams of copper with 60 grams of sulfur, how much copper(i) sulfide (cu2s) can be produced during the reaction? 2cu + s → cu2s

The balanced equation for the reaction is as follows;
2Cu + S ---> Cu₂S
stoichiometry of Cu to S is 2:1
we need to find the limiting reactant 
number of Copper moles - 40 g /63.5 g/mol = 0.63 mol 
number of S moles - 60 g / 32 g/mol = 1.88 mol
If Copper is the limiting reactant 
then 0.63 mol of Copper reacts with - 0.63/2 = 0.315 mol of S
but 1.88 mol of S present only 0.315 mol of S is required
and Copper is the limiting reactant therefore amount of product formed depends on amount of Cu present 
stoichiometry of Copper to Cu₂S is 2:1
then 0.63 mol of Copper forms - 0.63/2 = 0.315 mol of Cu₂S
0.315 mol of Cu₂S produced
mass of Cu₂S produced - 0.315 mol x 159 g/mol = 50.1 g of Cu₂S produced

Ido not understand.