The fizz produced when an Alka-Seltzer® tablet is dissolved in water is due to the reaction between sodium bicarbonate (NaHCO3) and citric acid (H3C6H5O7): 3NaHCO3(aq)+H3C6H5O7(aq)→3CO2(g)+3H2O(l)+Na3C6H5O7(aq) In a certain experiment 1.20 g of sodium bicarbonate and 1.20 g of citric acid are allowed to react.
How many grams of carbon dioxide form?
How many grams of the excess reactant remain after the limiting reactant is completely consumed?
3NaHCO3(aq)+H3C6H5O7(aq)→3CO2(g)+3H2O(l)+Na3C6H5O7(aq)
I do these LR (limiting reagent) and ER (excess reagent) problems the long way. First convert grams to mols of each.
mols NaHCO3 = 1.2/84 = approx 0.014
mols citric acid = 1.2/192 = approx 0.0062
Now convert mols of EACH to mols CO2 formed IF YOU HAD EACH BY ITSELF AND ALL OF THE OTHER ONE YOU NEEDED.
For NaHCO3. 0.014 mols NaHCO3 x (3 mols CO2/3 mols NaHCO3) = 0.14 x 3/3 = 0.014.
For citric acid. 0.0062 mols citric acid x (3 CO2/1 mol citric acid) = 0.0062 x 3/1 = 0.019.
In LR problems you can't get more than the smaller amount so NaHCO3 is the LR and citric acid is the ER. How much CO2 is formed.
0.014 mols CO2 from the NaHCO3 x molar mass CO2 = grams CO2 formed.
How much citric acid is left? First, how much is used. Just another stoichiometry problem.
0.014 mols NaHCO3 x (1 mol citric acid/3 mols NaHCO3) = 0.014 mols NaHCO3 x 1/3 = 0.005 mols citric acid used. Amount left is 0.0062 initially - 0.005 used= ? left.
Convert to grams left. g citric acid left = mols citric acid left x molar mass citric acid = ?
Remember I've estimated ALL of these calculations so you should go through and repeat each but to more accuracy. Post your work if you get stuck.
To find the number of grams of carbon dioxide (CO2) formed in the reaction, we need to first determine the limiting reactant. The limiting reactant is the reactant that is entirely consumed during the reaction, thereby determining the quantity of product formed.
1. Calculate the moles of each reactant:
- Sodium bicarbonate (NaHCO3) = 1.20 g / molar mass of NaHCO3
- Citric acid (H3C6H5O7) = 1.20 g / molar mass of H3C6H5O7
2. Determine the stoichiometry of the reaction:
According to the balanced equation, 3 moles of sodium bicarbonate react with 1 mole of citric acid to produce 3 moles of carbon dioxide.
3. Compare the moles of each reactant to determine the limiting reactant:
Divide the moles of each reactant by their respective stoichiometric coefficients. The reactant resulting in a smaller mole ratio is the limiting reactant.
4. Once the limiting reactant is determined, calculate the moles of carbon dioxide formed:
Using the mole ratio mentioned earlier, the moles of the limiting reactant will also be the moles of carbon dioxide produced.
5. Finally, calculate the grams of carbon dioxide:
Multiply the moles of carbon dioxide by the molar mass of carbon dioxide to obtain the mass in grams.
To find the mass of the excess reactant remaining after the limiting reactant is consumed:
1. Determine the moles of the excess reactant using the mole ratio from the balanced equation.
2. Multiply the moles of the excess reactant by its molar mass to obtain the mass in grams.
To find out how many grams of carbon dioxide form, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in a chemical reaction, thus limiting the amount of product formed.
First, we need to find the number of moles of sodium bicarbonate (NaHCO3) and citric acid (H3C6H5O7) in the given masses.
Number of moles of sodium bicarbonate (NaHCO3):
moles = mass / molar mass
moles = 1.20 g / 84.01 g/mol (molar mass of NaHCO3)
moles = 0.01428 mol
Number of moles of citric acid (H3C6H5O7):
moles = mass / molar mass
moles = 1.20 g / 192.13 g/mol (molar mass of H3C6H5O7)
moles = 0.00624 mol
Next, we use the balanced equation to determine the stoichiometry of the reaction.
From the balanced equation: 3NaHCO3(aq) + H3C6H5O7(aq) → 3CO2(g) + 3H2O(l) + Na3C6H5O7(aq)
We can see that 3 moles of NaHCO3 produce 3 moles of CO2.
Based on the stoichiometry, we can calculate the moles of CO2 produced:
moles of CO2 = 3 moles of NaHCO3 × (0.01428 mol of NaHCO3 / 3 moles of NaHCO3)
moles of CO2 = 0.01428 mol
Now, we can calculate the mass of CO2 produced:
mass = moles × molar mass
mass = 0.01428 mol × 44.01 g/mol (molar mass of CO2)
mass = 0.6276 g
Therefore, approximately 0.6276 grams of carbon dioxide will form in the reaction.
To determine the mass of the excess reactant that remains after the limiting reactant is completely consumed, we need to calculate the moles of each reactant and compare them.
Since the stoichiometric ratio is 3:1 (NaHCO3:H3C6H5O7), there is an excess of citric acid.
moles of excess reactant (H3C6H5O7) = moles of starting reactant (H3C6H5O7) - moles used in reaction (0.01428 mol)
moles of excess reactant (H3C6H5O7) = 0.00624 mol - 0.01428 mol
moles of excess reactant (H3C6H5O7) = -0.00804 mol (negative value indicates it is completely consumed)
The negative value indicates that the citric acid is completely consumed. Therefore, no excess reactant remains.
In summary:
- The mass of carbon dioxide formed is approximately 0.6276 grams.
- There is no mass of excess reactant remaining after the citric acid is completely consumed.