Answer to Question #166270 in General Chemistry for Felecia

Question

Hello, I need help with balancing the equation and finding the limited reactant and Excess. The equation says: Soild Iron (3) nail of 4.800 grams is placed in copper (2) sulfate solution of 9.883 grams yeild. For the equation I got 2Fe+ CuSo4 = Fe2(SO4)3 +3 Cu. Did I do this correct or not?

Answer 1

Iron (3) nail does not make any chemical sense because a pure metal cannot have any oxidation state other than zero.

There are two possible reactions of solid iron with CuSO₄ solution with this one to be more likely to occur:

Fe(s) + CuSO₄(aq) --> Cu(s) + FeSO₄(aq)

To find the limiting reactant, amounts of both Fe and CuSO₄ should be known. It is not very clear what "solution of 9.883 grams yield" means, but assuming that this is a mass of CuSO₄ in the solution, both amounts in moles can be calculated:

$n(Fe)=\frac{4.800g}{55.847g/mol}=0.08595mol$

$n(CuSO_4)=\frac{9.883g}{159.609g/mol}=0.06192mol$

Since they react in 1 : 1 molar ratio, the reactant with fewer moles (CuSO₄) is limiting, and Fe is in excess.

Similarly, for the second possible reaction

2Fe(s) + 3CuSO₄(aq) --> 3Cu(s) + Fe₂(SO₄)₃(aq),

the molar ratio of Fe to CuSO₄ is 2 : 3. Since there are more moles of Fe than CuSO₄ present, obviously CuSO₄ is also a limiting reactant, and Fe is in excess.