Dr. Bob,
I just did a home lab with citric acid and sodium bicarbonate and had to put them in cups and measure before and after.. I put 1.02g of sodium bicarbonate in a cup with water and stirrer (54.66g) then added .77g of citric acid and supposed to add up total mass (56.45g) and after the reaction it all weighed 56.21g. The difference is supposed to equal carbon dioxide.
Now my lab asks this question and I don't know if I'm supposed to figure it out using what they gave me or if I'm supposed to incorporate my personal lab results also? How do I approach and work this problem?
The reaction of citric acid and sodium bicarbonate is written as
H3C6H5O7(aq) + 3 NaHCO3(aq) → Na3C6H5O7(aq) + 3 H2O(l) + 3 CO2(g)
Show that the equivalence amount of citric acid for 1.00 g of sodium bicarbonate is 0.76 g.
Dr. Bob, I just found an old post of someone asking this question, and I figured it out.. 1 mole citric acid reacts with 3 moles sodium bicarbonate, molar mass of NaHCO3 = 84.0 g/mol
1 mole in 1.00 g = 1.00/84.0 = 0.0119 mole sodium bicarbonate
this will react with 0.0119/3 = 0.00397 mole citric acid
molar mass citric acid 192.1
0.00397 * 192.1 = 0.76g
Bam!!! I'm figuring this out!!! You are amazing! thanks for all you do!!
Now, I'm going to post how to figure out the theoretical yield of carbon dioxide from 1.00g sodium bicarbonate... Um.. I think it's the same mole ratio so would I just take the moles of NaHCO3 and divide by the moles of CO2? and then what? I found the moles of CO2 = 1/molar mass 44 = 0.227 so would it be 0.0119 NaHCO3/ 0.227 CO2 = 0.0524? but what is that? what is the 0.0524 number supposed to be?
Thanks Dr. Bob,
Jennie
mols in 1g NaHCO3 = 1/84 = ?
mols CO2 produced = same or 1/84
grams CO2 = mols x molar mass = theoretical yield.
You're right that mols CO2 = 1g/molar mass CO2 IF YOU HAD 1 g CO2 but you didn't. You said you had 1 g NaHCO3; furthermore, you calculated the mass CO2 as 0.24 g (56.45-56.21).
To determine the equivalence amount of citric acid for 1.00 g of sodium bicarbonate, you can use the balanced chemical equation and the stoichiometry of the reaction.
First, let's review the balanced chemical equation:
H3C6H5O7(aq) + 3 NaHCO3(aq) → Na3C6H5O7(aq) + 3 H2O(l) + 3 CO2(g)
According to the equation, 1 mole of citric acid (H3C6H5O7) reacts with 3 moles of sodium bicarbonate (NaHCO3) to produce 1 mole of sodium citrate (Na3C6H5O7), 3 moles of water (H2O), and 3 moles of carbon dioxide (CO2).
Now, let's calculate the molar masses of sodium bicarbonate (NaHCO3) and citric acid (H3C6H5O7):
- Sodium bicarbonate (NaHCO3):
Na: 22.99 g/mol
H: 1.01 g/mol
C: 12.01 g/mol
O: 16.00 g/mol
Adding them together:
NaHCO3 = 22.99 + 1.01 + (12.01 * 1) + (16.00 * 3) = 84.01 g/mol
- Citric acid (H3C6H5O7):
H: 1.01 g/mol
C: 12.01 g/mol
O: 16.00 g/mol
Adding them together:
H3C6H5O7 = (1.01 * 3) + (12.01 * 6) + (16.00 * 7) = 192.13 g/mol
Now, let's calculate how much citric acid is needed for 1.00 g of sodium bicarbonate:
First, we need to convert the mass of sodium bicarbonate to moles. Using the molar mass of sodium bicarbonate, we have:
1.00 g NaHCO3 * (1 mol NaHCO3 / 84.01 g NaHCO3) = 0.0119 mol NaHCO3
According to the balanced chemical equation, 1 mole of citric acid reacts with 3 moles of sodium bicarbonate. Therefore, the moles of citric acid needed would be:
0.0119 mol NaHCO3 * (1 mol H3C6H5O7 / 3 mol NaHCO3) = 0.00396 mol H3C6H5O7
Finally, to calculate the mass of citric acid needed, we can use the molar mass of citric acid:
0.00396 mol H3C6H5O7 * (192.13 g H3C6H5O7 / 1 mol H3C6H5O7) = 0.759 g H3C6H5O7
Therefore, the equivalence amount of citric acid for 1.00 g of sodium bicarbonate is approximately 0.76 g.