Two cells are connected in series . One contains AlCl3, and the other contains AgNO3 as the electrolytes. What mass of Ag is deposited when 18g of Al is deposited at cathode?

Answer 1

96,485 coulombs will deposit 27/3 = 9 g Al
This cell deposited 18 g Al so you must have had 96,485 x 18/2 = 48,242 C flowing through the cell.
96,485 coulombs will deposit 107.9/1 = about 108 but you need to do the math on all of these estimates.
So Ag deposited will be 107.9 x (48,242/96, 485) = about 107.9/2 = ? g Ag.
Post your work if you have further questions.

Answer 2

MAl/EAl=MAg/MAg

18g/9=MAg/108g
MAg=18gx108/9
=1994/9
=216g/mol
I think this is my answer for this question

Answer 3

Your answer of 216 grams Ag (not grams/mol) is correct and mine is wrong. I goofed in my first calculations. Here is what I should have written.
96,485 coulombs will deposit 27/3 = 9 g Al
This cell deposited 18 g Al so you must have had 96,485 x 18/9 = 192,970 C flowing through the cell.
96,485 coulombs will deposit 107.9/1 = about 108 g Ag but you need to do the math on all of these estimates.
So Ag deposited will be 107.9 x (192,970/96, 485) = about 107.9*2 = ? g which is 215.8 g Ag.