I would look at it this way.
Ag available from AgNO3 is 8.0 x (107.9/169.9) = about 5 g and that's the max either way.
From the battery 850 mAh = 0.850 Ah and that is 0.850 A/h x (60 min/hr) x (60 s/min) = about 3060 coulombs (C = A x sec).
You know that 96,485 C will deposit 1 equivalent of Ag and that is 107.9 g Ag. So 107.9 x (3060/96,485) = about 3.4 g. If you have about 5 g Ag in solution and the battery (technically that's a cell and not a battery) will allow you to collect only 3.4g then the battery is the limiting source. I assume you can do the rest. You should not only confirm the above but you should watch the number of significant figures AND you should do it more accurately than my about this and that. But my numbers are close.
seriously, any help would be much appreciated.
You conduct another experiment on silver condensation. You add 8.0 g of silver nitrate to 200 mL of pure water. You attach a single 850 mAh AAA battery (at 1.3 V) to the electrochemical cell. You can assume that the entire battery charge is useable for electroplating. You get 3.24 g of silver metal at the end of the experiment.
a) Was your silver recovery limited by the battery capacity or the available silver in solution?
b) What was your product yield (%) using the limiting resource?
2 answers
I'm not sure that I'm that comfortable with the "product" yield using the limiting resource. Although I understand what the problem is driving at for the % yield, I think the yield, by any standard is (3.24/5.08)*100 = ? no matter how you do it. I think the answer they're driving at is (3.24/3.42)*100 = ?
1 answer