Ok, For Cup C

Mass of cup,water,stirrer: 55.15
Mass of sodium bicarbonate: 2.02g
mass of citric acid: 0.77g
total mass: 57.94g
mass of cup,solution, stirrer after reaction: 57.64
difference (CO2): 0.30g

Part B: Limiting Reactants
Plastic Cup C
7. Determine which reactant is the limiting reactant in the plastic cup C. Describe your reasoning.

You know I know how to calculate the TY and AY and %Y now, but I'm confused on the finding limiting reactant? umm Let me see if I can figure it out and you let me know if I'm on the right track...
2.02g NaHCO3 mols = 2.02/84 = 0.0240 mols
the ratio is 3/1 for NaHCO3toH3C6H5O7 so do I do 0.0240/3 = citric acid? 0.008 moles citric acid? so since citric acid is smaller but it requires 3 of sodium bicarbonate, would that be the limiting reactant? because it takes more of them? am I figuring this out right?

8. Calculate the theoretical yield of carbon dioxide in the plastic cup C.
2.02/84=0.0240 *44 = 1.056 TY??

9. Calculate the percentage yield in the plastic cup C.

AY = 0.30/1.056 TY = 0.284*100 = 28.4% ?? is this right?
Chemistry Dr. Bob222 - DrBob222, Wednesday, March 4, 2015 at 12:18am
You know I know how to calculate the TY and AY and %Y now, but I'm confused on the finding limiting reactant? umm Let me see if I can figure it out and you let me know if I'm on the right track...
2.02g NaHCO3 mols = 2.02/84 = 0.0240 mols
the ratio is 3/1 for NaHCO3toH3C6H5O7 so do I do 0.0240/3 = citric acid? 0.008 moles citric acid? so since citric acid is smaller but it requires 3 of sodium bicarbonate, would that be the limiting reactant? because it takes more of them? am I figuring this out right?

8. Calculate the theoretical yield of carbon dioxide in the plastic cup C.
2.02/84=0.0240 *44 = 1.056 TY??
!!!!!!HERE IS WHERE I AM WRITING NEW AGAIN!!!
*** So I should be saying 0.008/84 *44? = 0.42% or am I to use the citric acid mass of 192.1? 0.008/192.1 = 0.00042 * 44 ==0.0018 I'm even more confused now... And before I didn't have to calculate the limiting reagent it's just now asking me to.. I don't know.. ughhhhhh..

If citric acid is the LR then you should use it to determine the theoretical yield. This calls into question some of the earlier answers since I didn't check them for the LR bit. But it stands to reason that if 1 g NaHCO3 is equivalent to 0.76 g citric acid then if you start with 2.02 g NaHCO3 it will take more than 0.76 g citric acid and there isn't enough there to do anything except with 1g NaHCO3.

9. Calculate the percentage yield in the plastic cup C.

AY = 0.30/1.056 TY = 0.284*100 = 28.4% ?? is this right?
TY needs to be recalculated.

I think I must have answered this elsewhere.

To determine the limiting reactant, you need to compare the moles of each reactant and their stoichiometric ratios. In your case, you correctly calculated the moles of sodium bicarbonate (NaHCO3) to be 0.0240 moles. However, your calculation for the moles of citric acid (H3C6H5O7) is incorrect.

To calculate the moles of citric acid, you need to use the given mass of citric acid (0.77g) and its molar mass (192.1 g/mol). So, moles of citric acid = mass of citric acid / molar mass of citric acid = 0.77g / 192.1 g/mol = 0.004 moles.

Now, compare the moles of sodium bicarbonate and citric acid using their stoichiometric ratio. The ratio is 3:1 for NaHCO3 to H3C6H5O7. So, divide the moles of citric acid by 3 to get the moles of NaHCO3 required for complete reaction: 0.004 moles / 3 = 0.00133 moles.

Since you have 0.0240 moles of sodium bicarbonate and only 0.00133 moles of citric acid, the limiting reactant is citric acid because it is present in a lesser amount and will fully react with the available sodium bicarbonate.

Moving on to calculating the theoretical yield of carbon dioxide (CO2):

The balanced reaction equation indicates that 1 mole of sodium bicarbonate reacts to produce 1 mole of carbon dioxide. Therefore, the theoretical yield (TY) of CO2 would be the same as the moles of sodium bicarbonate: 0.0240 moles.

To calculate the percentage yield (AY), divide the actual yield (AY) of CO2 (0.30 g) by the theoretical yield (TY) of CO2 (0.0240 moles) and multiply by 100:

AY = 0.30 g
TY = 0.0240 moles

AY = 0.30 g / 0.0240 moles = 12.5 g/mol

Now, calculate the percentage yield:
AY / TY = 12.5 g/mol / 0.0240 moles = 520.83 g/mol

The percentage yield is given by:
AY / TY * 100 = 520.83 g/mol * 100 = 52.08%

Therefore, the percentage yield of carbon dioxide in the plastic cup C is approximately 52.08%.