If iron is oxidized to Fe2+ by a copper(II) sulfate solution, and 0.850 grams of

Question

If iron is oxidized to Fe2+ by a copper(II) sulfate solution, and 0.850 grams of iron and 18.8 mL of 0.525M 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

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

moles of Fe = given mass/molar mass = 0.850/55.845 = 0.01522 mole

moles of CuSO4 = Molarity x volume of solution in litres = 0.525 M x 18.8/1000 = 0.00987 mole

Now,

1 mole of Fe react with 1 mole of CuSO4 to produce 1 mole of FeSO4

So, 0.01522 mol of Fe = 0.01522 mol of FeSO4

0.00987 mol of CuSO4 = 0.00987 mol of FeSO4

So, CuSO4 is in limited quantity hence the Limiting reactant and Fe is non limiting reactant

Reaction goes into completion hence CuSO4 is completely used up to form FeSO4

Since, 0.00987 mol of CuSO4 react with 0.00987 mol of Fe to form 0.00987 mol of FeSO4

Hence, mol of Fe unused = 0.01522 - 0.00987 = 0.00535 mol = 0.00535 x 1000 = 5.35 mmol(since, 1mol=1000mmol).

At the end of experiment CuSO4 is completely used up to form FeSO4 hence the solution will be colourless due to presence of FeSO4.

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