Question

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 1.25 g of sodium bicarbonate and 1.25 g of citric acid are allowed to react.

Part C How many grams of the excess reactant remain after the limiting reactant is completely consumed?

Express your answer using two significant figures.

Answer 1

3NaHCO3(aq)+H3C6H5O7(aq)→3CO2(g)+3H2O(l)+Na3C6H5O7(aq)

from this balanced equation it is clear that 3 moles of NaHCO3 required to rect with one mole of citric acid

now convert every thing in to moles using weight / molar mass formula

no of moles of sodium bicarbonate = weight of the sodium bicarbonate / molar mass of the sodium bicarbonate

= 1.25 g / 84.007 g/mol (molar mass of sodium bicarbonate = 84.007 grams/ mole taken from google)

= 0.0147 moles

now no of mles of citric acid = weight of citric acid / molar mass of citric acid

= 1.25 g / 192.124 g/mol (192.124 is molar mass of citric acid taken from google)

= 0.0065 moles

from balanced equation it is clear that one mole of citric acid required 3 moles of sodium bicarbonate

so 0.0065 moles of citric acid required 3 x 0.0065 moles of sodium bicarbonate = 0.0195 moles

so heretically dor 0.0065 moles required 0.0195 moles but there is only 0.147 so

limiting agent is bicarbonate

0.0147 moles of sodium bicarbonate required 1/3 x0.0147 moles citric acid = 0.0049 moles citric acid required

but we have 0.0065

so remaining amount is

0.0065 - 0.0049 = 0.0016 moles if citric acid left