Atmospheric pressure, mm Hg 635 2 mma 3.0° a 1 · 1 mA Ha 614.1 mm Temperature of
ID: 637904 • Letter: A
Question
Atmospheric pressure, mm Hg 635 2 mma 3.0° a 1 · 1 mA Ha 614.1 mm Temperature of H,O, °C Vapor presure of H0 at this temperature, mm He All Pressure of CO2 in the system, mm Hg tg Na?ICO, + HCINaCI + CO,+ H,o (MM NaHCO, -84.01 g/mol) Trial 4 Mass of sample, g initials mL Vfinal, mL IS Moles of CO Moles of NaHCO, Mass of NaHCO, in sample, s Mass % of NaHCO3 in sample, % Average mass % NaHCO3 in Alka-Seltzer Show your calcrulations for Trial I on the next page, and submit them with your Data Sheets REPORT SHEET 129
Explanation / Answer
Trial 1
From the ideal gas equation
Moles of CO2 = pressure x volume / R x Temperature
= [(614.1/760) atm x 22.41 L] / [0.0821 L-atm/mol-K x (23+273)K]
= 0.7451 mol
From the stoichiometry of the reaction
Moles of NaHCO3 = moles of CO2 consumed
= 0.7451 mol
Mass of NaHCO3 = moles x molecular weight
= 0.7451 mol x 84.01 g/mol
= 62.598 g
Mass % NaHCO3 in sample
= mass of NaHCO3 x 100 / mass of sample
= 62.598 x 100 / 115
= 54.43 %
Trial 2
From the ideal gas equation
Moles of CO2 = pressure x volume / R x Temperature
= [(614.1/760) atm x 21.74 L] / [0.0821 L-atm/mol-K x (23+273)K]
= 0.7229 mol
From the stoichiometry of the reaction
Moles of NaHCO3 = moles of CO2 consumed
= 0.7229 mol
Mass of NaHCO3 = moles x molecular weight
= 0.7229 mol x 84.01 g/mol
= 60.727 g
Mass % NaHCO3 in sample
= mass of NaHCO3 x 100 / mass of sample
= 60.727 x 100 / 119
= 51.03 %
Trial 3
From the ideal gas equation
Moles of CO2 = pressure x volume / R x Temperature
= [(614.1/760) atm x 22.08 L] / [0.0821 L-atm/mol-K x (23+273)K]
= 0.7342 mol
From the stoichiometry of the reaction
Moles of NaHCO3 = moles of CO2 consumed
= 0.7342 mol
Mass of NaHCO3 = moles x molecular weight
= 0.7342 mol x 84.01 g/mol
= 61.677 g
Mass % NaHCO3 in sample
= mass of NaHCO3 x 100 / mass of sample
= 61.677 x 100 / 117
= 52.72 %
Average mass % NaHCO3 = (54.43 + 51.03 + 52.72)/3
= 52.727 %
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