Plastic Cup B

Mass of cup, water and stirrer: 53.75g
mass of sodium bicarbonate: 2.02g
mass of citric acid: 1.52g
total mass of all: 57.29g
mass of cup, solution, stirrer after reaction: 56.59g
difference (CO2): 0.70 g

Now, Here's the problems it's basically just like before, however, I have a question about moles for 2 g not just 1.

Plastic Cup B
4. Show that the equivalence amount of citric acid for 2.00 g of sodium bicarbonate is 1.52 g.
*** 1 mol citric acid reacts with 3 mols sodium bicarbonate, molar mass NaHCO3=84g/mol
2 moles in 2 g? = 2/84? = 0.0238 mols 2g NaHCO3? then to find the citric acid? is it.. 0.0238/3 = 0.0079 then 0.0079*192.1 = 151.75 (152g?) is this correct??

5. Calculate the theoretical yield of carbon dioxide from 2.00 g of sodium bicarbonate.
Then, I use my own calculations of 2.02g NaHCO3
2.02/84=0.0240 mol NaHCO3 then 0.0240 2 mol NaHCO3*(2 mol CO2/2mol NaHCO3) *44 = 1.056 g CO2 TY

6. Calculate the percentage yield in the plastic cup B.
Actual Yield = 0.70/TY 1.056 =0.663 *100 = 66.3% Is this correct????

. Show that the equivalence amount of citric acid for 2.00 g of sodium bicarbonate is 1.52 g.

*** 1 mol citric acid reacts with 3 mols sodium bicarbonate, molar mass NaHCO3=84g/mol
2 moles in 2 g? = 2/84? = 0.0238 mols 2g NaHCO3? then to find the citric acid? is it.. 0.0238/3 = 0.0079 then 0.0079*192.1 = 151.75 (152g?) is this correct??
This looks ok except you threw away a digit on the 0.079 calculation and if 192.1 is the correct molar mass for citric acid. I can't get to my table right now.<?b>

5. Calculate the theoretical yield of carbon dioxide from 2.00 g of sodium bicarbonate.
Then, I use my own calculations of 2.02g NaHCO3
2.02/84=0.0240 mol NaHCO3 then 0.0240 2 mol NaHCO3*(2 mol CO2/2mol NaHCO3) *44 = 1.056 g CO2 TY
looks ok

6. Calculate the percentage yield in the plastic cup B.
Actual Yield = 0.70/TY 1.056 =0.663 *100 = 66.3% Is this correct????
It looks ok but 66.3 probably reports too many significant figures. You are allowed only two because of the 0.70.

Thanks!!!! I'm learning!!

To answer your question, let's go through each step and check the calculations:

1. Finding the number of moles of sodium bicarbonate:
You correctly calculated that 2 grams of sodium bicarbonate (NaHCO3) is equivalent to 0.0238 moles (2/84 = 0.0238 moles).

2. Finding the number of moles of citric acid:
Since the balanced equation shows that 1 mol of citric acid reacts with 3 moles of sodium bicarbonate, you need to divide the number of moles of sodium bicarbonate by 3:
0.0238 moles of NaHCO3 / 3 = 0.00793 moles of citric acid (rounded to four decimal places).

3. Calculating the mass of citric acid:
Next, you correctly multiplied the number of moles of citric acid by the molar mass of citric acid (192.1 g/mol):
0.00793 moles of citric acid * 192.1 g/mol = 1.52 g of citric acid.

This means that 1.52 grams of citric acid is the amount needed to react with 2.00 grams of sodium bicarbonate.

Moving on to the next question,

4. Calculating the theoretical yield of carbon dioxide (CO2) from 2.00 grams of sodium bicarbonate:
You correctly determined that 2.02 grams of sodium bicarbonate is equivalent to 0.0240 moles (2.02/84 = 0.0240 moles).

To calculate the theoretical yield of CO2, we need to use the stoichiometry of the balanced equation. Since the equation shows that 2 moles of NaHCO3 produce 2 moles of CO2, the calculation will be:
0.0240 moles of NaHCO3 * (2 moles CO2/2 moles NaHCO3) * 44 g/mol = 1.056 grams of CO2.

Therefore, the theoretical yield of carbon dioxide from 2.00 grams of sodium bicarbonate is 1.056 grams.

Moving on to the final question,

5. Calculating the percentage yield of the reaction in Plastic Cup B:
The percentage yield is calculated by dividing the actual yield by the theoretical yield, then multiplying by 100.

In this case, the actual yield is given as 0.70 grams of CO2.

Percentage yield = (0.70 g CO2 / 1.056 g CO2) * 100 = 66.3%

So, you are correct that the percentage yield in Plastic Cup B is 66.3%.