What mass of Al can be deposited by the passage of a constant current of 5.00 A through a Al(NO3)2 solution of 2.00 hours?

My work:
Q= I*t
= 5 * 7200sec = 36000

36000/96458= .3732
.3732* 1/3 * (26.98*3) = 10.06g

But this is wrong it's suppose to be 3.3.6g. why is that?

where did the three in the last parenthesis come from.

I'm sorry I made a typo it goes through a solution of Al(NO3)3 and that's why I mutliplied the m.w. of Al by 3 because you have 3 moles of Al per NO3

It isn't the nitrate that is plating out. It's the Al. Remove the 3. You have taken care of the 3e in the transfer of Al(III) to Al(s) by the 1/3.

To calculate the mass of aluminum (Al) deposited by the passage of a constant current through an Al(NO3)2 solution, we need to use Faraday's law of electrolysis. The formula for this law is:

mass = (charge * Molar mass of Al) / (3 * Faraday's constant)

Where:
- charge is the total charge passing through the solution, which can be calculated using the equation: charge (Q) = current (I) * time (t)
- Molar mass of Al is the atomic mass of aluminum, which is 26.98 g/mol
- 3 is the number of electrons involved in the reduction reaction of aluminum (each Al3+ ion gains 3 electrons to become Al)
- Faraday's constant is 96485.8 C/mol, which represents the amount of charge carried by 1 mole of electrons

Let's calculate it step by step:

1. Calculate the charge (Q) passing through the solution:
Q = I * t = 5.00 A * 2.00 hours * 3600 seconds/hour = 36,000 C

2. Calculate the mass of aluminum deposited:
mass = (charge * Molar mass of Al) / (3 * Faraday's constant)
mass = (36,000 C * 26.98 g/mol) / (3 * 96485.8 C/mol)
mass = 972,960 g / 289457.4
mass ≈ 3.36 g

Therefore, the correct mass of aluminum that can be deposited by the passage of a constant current of 5.00 A through an Al(NO3)2 solution for 2.00 hours is approximately 3.36 g.