If iron is oxidized to Fe2+ by a copper(II) sulfate solution, and 0.662 grams of

Question

If iron is oxidized to Fe2+ by a copper(II) sulfate solution, and 0.662 grams of iron and 14.6 mL of 0.437M copper(II) sulfate react to form as much product as possible, how many millimoles (mmol) of the non-limiting reactant will remain unused at the end of the reaction? Amount of non-limiting reactant remaining unused = mmol Copper(II) sulfate in solution is blue in color. Iron(II) sulfate is colorless. In the reaction described above, will the solution at the end of the experiment be blue or colorless? The solution will be

Explanation / Answer

work out moles of each reagent moles = mass / molar mass moles Fe = 0.283 g / 55.85 g/mol moles = 0.005067 moles Fe moles = molarity x litres moles CuSO4 = 0.595 M x 0.0209 L = 0.01244 moles CuSo4 Work out the limiting reagent, the limiting reagent is the one that will be all used up if the reaction goes to completion. CuSO4(aq) + Fe(s) --------> FeSO4(aq) + Cu(s) 1 mole Fe needs 1 mole CuSo4 to react So 0.005067 mol Fe needs 0.005067 mol CuSO4 You have 0.01244 mol CuSO4, which is more then enough, so Fe is the limitng reagent. Since it tells you that as much product as is possible is formed, it means the reaction went to completion and there is no Fe left. So you must have used 0.005076 moles of the CuSO4 (see work above) Left over CuSO4 = mole provided - moles used = 0.01244 mol - 0.005076 mol = 0.007634 moles CuSO4 left over 1 mole = 1000 mmole 0.007634 mol = (1000 x 0.007634) mmole = 7.63 mmol (3 sig figs)

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