Gen Chem 2: coulombs

Question

8. How many Coulombs of charge are needed to reduce FeSO4 to yield 28.5g of Fe metal?

A) 157 C

B) 2.91 x 105 C

C) 2.46 x 104 C

D) 2.75 x 103 C

E) 3.62 x 104 C

F) 72.7 C

G) 5.01 x 104 C

H) 9.85 x 104 C

Answer 1

To reduce Fe²⁺(aq), which is what you have when you have FeSO₄ in solution, you need the following reaction:

Fe²⁺(aq) + 2e- ==> Fe(s)

So, for each mole of Fe(s) produced, you need TWO moles of electrons.

moles of Fe(s) desired = 28.5 g Fe x 1 mol Fe / 55.85 g = 0.5103 moles Fe

moles of electrons required = 0.5103 moles Fe x 2 mole e- / mol Fe = 1.021 moles of electrons

Recall that 1 mole of electrons = 96,485 coulombs (C). Thus ...

1.021 mols e- x 96,485 C / mol e- = 98,471 C = 9.85x10⁴ C (Answer H)