Question

You determine the volume of your plastic bag (simulated human stomach) is 1.09 L. How many grams of NaHCO₃ (s) are required to fill this container given a 49.8% CO₂ recovery, assuming the other contents in the bag take up a negligible volume compared to the gas. The temperature of the room is 24.0 °C and the atmospheric pressure is 753.5 mmHg.

Answer 1

Volume of gas = 1.09 L

Pressure of gas = 753.5 mm Hg

= (753.5 mm Hg/760)

= 0.991 atm

Temperature = 24.0 degree C

= 24.0 + 273.15

= 297.15 K

Calculate the moles of CO2 (n) as shown below:

n = PV/RT

= (0.991 atm)*(1.09 L)/[(0.0821 Latm/Kmol)*297.15 K]

= 0.0442 mol

Reaction involved:

2NaHCO3 -----------> Na2CO3 + CO2 + H2O

1 mol of CO2 = 2 mol of NaHCO3

0.0442 mol of CO2 = 2*0.0442 = 0.0884 mol of NaHCO3

% recovery of reaction = 49.8%

So, actual moles of NaHCO3 required = 0.0884/49.8%

= 0.1775 mol

Mass of NaHCO3 required = 0.1775 mol*(84 g/mol)

= 14.91 g (Answer)