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?

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.

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 = ?

To determine whether the silver recovery was limited by the battery capacity or the available silver in solution, we need to compare the amounts of silver metal obtained in the experiment (3.24 g) with the theoretical maximum amount of silver that could be obtained using both the battery capacity and the available silver in solution.

First, let's calculate the theoretical maximum amount of silver that could be obtained using the battery. The molar mass of silver (Ag) is 107.87 g/mol, so we can calculate the theoretical maximum amount of silver using the formula:

Mass of silver = (Battery capacity / Faraday's constant) * (molar mass of silver / 2)

Battery capacity = 850 mAh = 850 mA * 1 h = 0.85 A * 1 h = 0.85 C (Coulombs)
Faraday's constant = 96485 C/mol

Plugging in the values, we can calculate:

Mass of silver (battery-limited) = (0.85 C / 96485 C/mol) * (107.87 g/mol / 2) ≈ 0.0466 g

Now, let's calculate the theoretical maximum amount of silver that could be obtained using the available silver in solution. From the amount of silver nitrate added, we can calculate the number of moles of silver nitrate:

Number of moles of silver nitrate = mass of silver nitrate / molar mass of silver nitrate

mass of silver nitrate = 8.0 g
molar mass of silver nitrate = molar mass of silver (Ag) + molar mass of nitrate (NO3-) = 107.87 g/mol + 14.01 g/mol + 3 * 16.00 g/mol ≈ 169.87 g/mol

Number of moles of silver nitrate = 8.0 g / 169.87 g/mol ≈ 0.0471 mol

Since silver nitrate contains one mole of silver ions (Ag+) per mole of silver nitrate, the theoretical maximum amount of silver that could be obtained from the available silver in solution is approximately equal to the number of moles of silver nitrate:

Mass of silver (silver-limited) ≈ Number of moles of silver nitrate ≈ 0.0471 g

Comparing the two maximum amounts of silver, we can see that the battery-limited recovery is lower than the silver-limited recovery (0.0466 g < 0.0471 g). Therefore, the silver recovery in this experiment was limited by the battery capacity.

To calculate the product yield (%) using the limiting resource, we can divide the actual amount of silver obtained (3.24 g) by the maximum amount of silver using the limiting resource (0.0466 g) and multiply by 100:

Product yield (%) = (actual amount of silver / maximum amount of silver) * 100
= (3.24 g / 0.0466 g) * 100
≈ 6964.85 %

Therefore, the product yield using the limiting resource (battery-limited) is approximately 6964.85%.