How many grams of copper sulfide would be formed by the reaction of 1.35 g of copper with excess sulfur?
Copper metal + Sulphur ---> Copper (ii) sulphide which can be expressed by the balanced equation, Cu + S ---> CuS thus, 1 mole of copper metal reacts with one mole of sulphur to produce one mole of copper (II) sulphide The atomic weight of copper is 63.5, the atomic weight of sulphur is 32 1.35 grams of copper / 63.5 = 0.021 moles of copper metal 0.021 moles of copper metal will react with 0.021 moles of sulphur to produce 0.021 moles of copper (II) sulphide 0.021 x atomic weight of sulphur (which is 32) = 0.672 = 0.021 moles of sulphur 0.672 + 1.35 = 2.022 grams of copper (II) sulphide produced
