If iron is oxidized to Fe^2+ by a copper(II) sulfate solution, and 0.490 grams o

Question

If iron is oxidized to Fe^2+ by a copper(II) sulfate solution, and 0.490 grams of iron and 18.0 mL of 0.580M 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 _____?

explain how you did it please it helps a lot

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

to find the amount of moles of copper sulphate, you multiply the concentration by the volume.
that's 0.531 x 0.0112

whichever has more moles represents the reagent in excess, which will dictate the colour of the solution.

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