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Discussion

Using the equation PV = nRT, one can calculate the expected volume of gas generated from 84.0 g (1.00 mol) of sodium bicarbonate. At room temperature (25 degrees Celsius, or 298 K) and 1 atm, and using the gas constant R = 0.08206 L^.atm/mol^.K, one would predict that 84.0 g of sodium bicarbonate would generate 24.5 L of carbon dioxide gas.

The reaction for the generation of carbon dioxide in this demonstration is

NaHCO₃(s) + HCl(aq) → NaCl(aq) + H₂O(l) + CO₂(g)

Narration

Approximately 1 mole of sodium bicarbonate is weighed. Excess hydrochloric acid in a measuring cylinder is placed in the flask, which is reweighed. The flask is connected to the gas measuring system. Note that the water level has not dropped, indicating a closed system with no leaks. Tilting the flask mixes the acid from the cylinder with the bicarbonate. The carbon dioxide gas generated by the reaction displaces the water from the cylinder. At zero degrees Celsius and one atmosphere pressure, the volume of one mole of gas should be 22.4 liters. Here under the conditions of this demonstration, one mole of carbon dioxide gas occupies 25 liters. The flask is reweighed, showing that there is a loss of mass of about 44 grams, which is the mass of the carbon dioxide collected. In other words, one mole of sodium bicarbonate produces one mole of carbon dioxide.