Question

8. A student performs parts A and B of the experiment and attains an average percent yield from the Alka- Seltzer of 64.6%. Using this value and the following data perform the calculations for part B and complete the table below. Assume the pressure has not changed. As it states in the manual: You may apply the findings regarding percent-yield in Part A to Part B when calculating the moles of calcium carbonate Trial 2 Trial 1 Mass Antacid used (g) 0.4190 0.3978 Volume CO2 recovered (mL) 21.5 25.2 Water Temperature (C) 16.4 14.4 Moles CO2 recovered (moles) Theoretical moles CO2 (moles) Moles CaCO3 (moles) Mass CaCO3 (g) Mass Percent CaCO3 Average Percent Yield CaCO3 9. You determine the volume of your plastic bag (simulated human stomach) is 1.03 L. How many grams of NAHCO3 (s) are required to fill this container given a 64.6% CO2 recovery, assuming the other contents in the bag take up a negligible volume compared to the gas. The temperature of the room is 24.7 °C and the atmospheric pressure is 753.5 mmHg.

Answer 1

Question 8.

Trial 1.

According to the ideal gas equation:

PV = nRT

i.e. 1 atm * (21.5/1000) L = n * 0.0821 L.atm/mol.K * (14.4+273) K

i.e. The moles of CO2 (n) = 9.112*10^-4 mol

Moles of CaCO3 = 9.112*10^-4 mol

Mass of CaCO3 = 9.112*10^-4 mol * 100 g/mol = 0.09112 g

Mass percent of CaCO3 = (0.09112 g/0.3978 g)*100 = 22.91%

Trial 2.

According to the ideal gas equation:

PV = nRT

i.e. 1 atm * (25.2/1000) L = n * 0.0821 L.atm/mol.K * (16.4+273) K

i.e. The moles of CO2 (n) = 1.061*10^-3 mol

Moles of CaCO3 = 1.061*10^-3 mol

Mass of CaCO3 = 1.061*10^-3 mol * 100 g/mol = 0.1061 g

Mass percent of CaCO3 = (0.1061 g/0.4190 g)*100 = 25.32%

Therefore, the average percent yield of CaCO3 = (22.91+25.32)/2 = 24.12%