How many grams of molybdenum can be deposited from a solution containing aqueous Mo(NO3)3 by supplying 25.0 Amps of current for 12.0 hours?

Question

The molar mass of Mo is 95.96 g/mol. The Faraday constant is 96485 C/mol e-

Answer 1

Mo(NO₃)₃

Since each NO₃ is -1, the Mo must have an oxidation number of 3+. In order to go from Mo³⁺ ==> Mo, it requires 3 electrons / mole.

1 amp = 1 coulomb/sec

25 amps = 25 coulombs/sec

25 C/sec x 12.0 hours x 60 min/hour x 60 sec/min = 1,080,000 Coulombs

1,080,000 C x 1 mol e- / 96485 C = 11.19 moles of electrons

11.19 moles e- x 1 mol Mo / 3 mol e- = 3.73 moles Mo

grams Mo = 3.73 mol Mo x 95.96 g/mol = 358 g Mo deposited