To solve the problem, I would use the following setup: O 0.25 x 299= V2 x 451
O =
0.25 x V2= 299 x 451
O V2= 0.25 x 299 x 451
0.25 x 451= V2 x 299

Since there is the same amount of agn03 in both silver chloride and silver nitrate they produce the same amount of moles. therefore silver chloride produces the same amount of silver nitrate, so your answer is 15.0 moles of silver nitrate. i hope this !

The correct answer is "atomic weight".