If iron is oxidized to Fe2+ by a copper(II) sulfate solution, and 0.798 grams of

Question

If iron is oxidized to Fe2+ by a copper(II) sulfate solution, and 0.798 grams of iron and 14.1 mL of 0.470M 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_______ please answer both parts!

Explanation / Answer

Fe+Cu(II)SO4 -> Fe(II)SO4+Cu

Now,

moles of Fe = given mass/molar mass= 0.798 g/55.845 g/mol = 0.01429 mole

moles of CuSO4= Molarity x volume of solution in Litres

= 0.470M x 14.1/1000 = 0.00663 mole

Now, 1mole of Fe reacts with 1mole of CuSO4 to form 1 mole of FeSO4

So, moles of FeSO4 from each reactant is

0.01429 mole of Fe = 0.01429 mole of FeSO4

0.00663 mole of CuSO4 = 0.00663 mole of FeSO4

CuSO4 is limiting reactant and Fe is non limiting reactant

Now, 0.00663 mole of CUSO4 will react with 0.00663 mole of Fe

So, mole of Fe unused = 0.01429-0.00663 = 0.00766 mole = 0.00766 x 1000 = 7.66 millimoles

(Since, 1mole = 1000 millimoles)

At the end of experiment 0.00663 mole of CuSO4 is completely used up to form 0.00663 mole of FeSO4

So, The Solution will be colourless due to presence of FeSO4

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