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