How many faradays are required to deposit the vanadium

An aqueous solution of an unknown salt of vanadium is electrolyzed by a current of 2.50 amps for 1.50 hours. The electroplating is carried out with an efficiency of 95.0%, resulting in a deposit of 2.251 g of vanadium.

I'm stuck! Thanks for the help

2.50 amp x 1.50 hr x 3600 seconds/hr = 13500 amp seconds = 13500 coulombs
13500 C x 95% / 96,500 C/faraday = 0.1329 faraday

Vanadium forms salts with a +3 or +5 oxidation state.
2.251 g V / 50.95 g/mole = 0.0442 mole of V
0.133 faraday = 0.133 mole of electrons
0.133 mole of electrons / 0.0442 mole of V = 3 mole of electrons / mole of V
The vanadium in the unknown salt was +3 oxidation state.