CuO(s) + H2SO4(aq) ---> CuSO4 + H2O(l) (a) How many moles of copper (II) sulphate would be formed from one mole of copper oxide?I've been trying to work out this question for a while now.
Looking at the stoichiometry of the balanced chemical reaction, it is apparent that one mole of copper oxide reacts with one mole of H2SO4 in aqueous solution to produce one mole of copper (II) sulphate and one mole of water. In terms of mass, (63.5+16)=79.5 g copper oxide reacts with (2x1+32+4x16)=98 g H2SO4 in aqueous solution to produce (63.5+32+4x16)=159.5 g copper (II) sulphate and (2x1+16)=18 g water. This proportionality shall be maintained for any amount of starting material (reactant). This means that starting with 10 g copper oxide, and sufficient H2SO4, one would get 159.5×10/79.5 or, approximately 20.063 g copper (II) sulphate, and so on.