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.
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?
1 answer