Chemistry 104
posted by Janavi .
A current of 3.40 A is passed through a Ni(NO3)2 solution for 1.80 hours. How much nickel is plated out of the solution?

Current=charge/time (s)
Where
current=3.40A
Convert hr to seconds (s):
1.80hrs*(60 min/1 hr)*(1 min/60s)= time in s
Solve for charge:
seconds*current=charge (C)
1 mole of e^s=9.65 x 10 4 C
Solve for moles:
C*(1 mole/9.65 x 10 4 C)=moles of Ni(NO3)2
Solve for mass:
moles of Ni(NO3)2*(182.703 g/mole)= mass of Ni(NO3)2 
Note that there are two mols of electrons and you must take that into account. A minor problem is that you will not plate out Ni(NO3)2 but Ni metal.

Dr. Bob222 caught that for me. Look for **** for corrections.
Current=charge/time (s)
Where
current=3.40A
Convert hr to seconds (s):
1.80hrs*(60 min/1 hr)*(1 min/60s)= time in s
Solve for charge:
seconds*current=charge (C)
1 mole of e^s=9.65 x 10^4 C
Solve for moles:
**** C*(1 mole/9.65 x 10^4 C)=moles of e^s
**** The half reaction is the following:
Ni2+ + 2e > Ni
So, 2 moles of e^ is needed for 1 mole of Ni:
Solve for moles of Ni:
**** moles of e^s*(1 mole of Ni/2 mole of e^)= moles of Ni
Solve for mass:
**** moles of Ni*( 58.69 g/mole)= mass of Ni
Dr. Bob222 is correct: the question states Ni not Ni(NO3)2.