If iron is oxidized to Fe 2+ by a copper(II) sulfate solution, and 0.308 grams o

Question

If iron is oxidized to Fe2+ by a copper(II) sulfate solution, and 0.308 grams of iron and 20.5 mL of 0.451M 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

Mass of Fe=0.308g

moles of Fe=0.308g/56g/mole=0.0055moles=5.5mmoles

mMoles of Cu2+=20.5mL*0.451moles/L=9.246mmoles

Fe + CuSO4 <==>FeSO4 + Cu2+

nonlimiting reagent is Cu2+

Amount of non-limiting reactant (Cu2+)remaining unused = 9.246-5.5=3.746mmolCu2+

since CuSO4 is still left the colour of the solution at the end of the experiment is blue

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