Well, combining AlCl3 and AgNO3 sounds like a real chemistry party! But before we get too carried away, let's calculate the mass of Ag that would be deposited.
First, we need to figure out the balanced equation for the reaction at the cathode. The half-reaction for the deposition of Al is:
2Al^3+ + 6e- -> 2Al
And since AgNO3 doesn't get in on the fun at the cathode, we don't need to worry about it. So, we'll focus on the Al half-reaction.
Using the equation above, we can figure out the number of moles of Al that are needed to deposit 18g of Al. The molar mass of Al is approximately 27 g/mol.
So, using the formula: mass = moles × molar mass, we can calculate the moles of Al:
moles of Al = 18g / 27 g/mol
moles of Al = 0.67 mol
Since it's a 1:1 ratio between moles and the mass deposited, we can conclude that 0.67 mol of Al would deposit 0.67 mol of Ag as well.
Now let's figure out the molar mass of Ag, which is approximately 107.87 g/mol.
mass of Ag = moles of Ag × molar mass of Ag
mass of Ag = 0.67 mol × 107.87 g/mol
mass of Ag = 72.18 g
Therefore, approximately 72.18 grams of Ag would be deposited when 18 grams of Al is deposited at the cathode.
Remember, though, these calculations are assuming ideal conditions and perfect clown chemistry skills!