how much would be produced if you mixed 3.0grams of iron with a solution containing 15.0 grams of copper (II) sulfate.
Since this specific reaction is one mol on one mol you can only get the amount of mol iron you have. That is 3.0/ 55.85=0.05371531 so this times the mass of a copper (II) ion (63.55) =3.41 grams of copper this is a redox reaction the Cu2+ reacts with the Iron giving you copper and Fe 2+. People correct immediately if I'm wrong please
First write the balanced equation Fe + CuSO4----------> FeSO4 + Cu Write the givens m Fe = 3.0 grams m CuSO4 = 15.0 grams. Now what do you need to find. The Copper that would be produced. Since your given 2 reactants here you would need to know which one is the limiting reactant. First find the moles of both the reactants For Fe m Fe = 3.0 grams ; Molar mass (Mm) = 55.9 grams/mol n Fe = 3.0 grams * 1 mol / 55.9 grams n Fe = 0.0537 moles of Fe For CuSO4 m CuSO4 = 15.0 grams ; Molar mass (Mm) = 159.6 grams / mol n CuSO4 = 15.0 grams / 159.6 grams / mol n CuSO4 = 0.09398 moles of CuSO4. Now calculate the limiting reactant. n CuSO4 needed for 0.0537 moles of Fe 0.0537 moles of Fe * 1 mole of CuSO4 / 1 mole of Fe 0.0537 moles of CuSO4 is needed. Since the given moles of CuSO4 is excess Fe is the limiting reactant. Now to calculate how much Copper would be produced n Cu = 0.0537 moles of Fe * 1 mole of Cu / 1 mole of Fe n Cu = 0.0537 moles of Cu would be produced. Now mass of the Cu Produced = moles * Mm(Molar mass) Molar mass of Cu = 0.0537 moles of Cu * 63.5 grams of Cu / 1 mole of Cu 3.41 grams of Copper are produced. So amount of Copper to be produced by mixing 3.0grams of iron with a solution containing 15.0 grams of copper (II) sulfate is 0.0537 moles or 3.41 grams of Copper are produced.
