Plastic Cup B

Question

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????

Answer 1

. 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.

Answer 2

Thanks!!!! I'm learning!!