Question

Atmospheric pressure, mm Hg 633.09 mm Hg Temperature of H,O, °C 23.3°C Vapor pressure of H.O at this temperature, mm Hg 1.1 mm Ha Cell. 99 mm Hg Pressure of CO, in the system, mm Hg 4. NaHCO, + HCI + NaCl + Co, + H,O (MM NaHCO, - 84.01 g/mol) Trial T2 4 Mass of sample, g 0.135g 0.1l6g 0.126g N/A Vinitial; ml 4.34mL 3.82mL 4. 14 mL N/A Vfinal mL 24.664mL 24,47 m2 24.60mL N/A Vect, mL (Vaal - Vinitial) 20.30m 20.65m2 20.46ml N/A Moles of CO, N/A Moles of NaHCO NA Mass of NaHCO, in sample,g IN/ Mass % of NaHCO, in sample, % Average mass % NaHCO, in Alka-Seltzer Show your calculations for Trial 1 on the next page, and submit them with your Data Sheets.

the lab is analysis of alka-stelzer via gas evolution

Answer 1

Using ideal gas equation

PV = nRT

or , number of moles (n) = $\frac{PV}{RT}$

given

Temperature = 23.3 + 273 = 296.3 K

calculated pressure of CO₂ gas = 611.99 mm Hg = 611.99 (mm Hg) $\times$$\frac{1\, atm}{760 \: mm \: Hg}$ = 0.805 atm

R = 0.082 L-atm/mol.K

moles of CO₂ = moles of NaHCO₃

mass of NaHCO₃ = moles$\times$ molar mass

molar mass of NaHCO₃ = 84.01 g/mol

,mass percentage in sample = { (mass of NaHCO₃$\times$ 100)/ total mass of the sample }

now

	Trial 1	Trial 2	Trial 3
volume of CO2	20.30 mL = 0.02030 L	20.65 mL = 0.02065 L	20.46 mL= 0.02046 L
moles of CO2	$\frac{PV}{RT}$ = $\small \frac{0.805\times 0.02030}{0.082\times296.3 }$ = 6.73 $\times$10-4	$\small \frac{0.805\times 0.02065}{0.082\times296.3 }$ = 6.84$\times$10-4	$\small \frac{0.805\times 0.02046}{0.082\times296.3 }$ = 6.78$\times$10-4
moles of NaHCO3	6.73 $\times$10-4	6.84$\times$10-4	6.78$\times$10-4
mass of NaHCO3	6.73$\times$10-4$\times$ 84.01 = 0.0565 g	6.84$\times$10-4$\times$ 84.01  = 0.0575 g	6.78$\times$10-4  $\times$ 84.01 = 0.0569 g
mass percentage of NaHCO3 in sample	(0.0565$\times$100)/0.135 = 41.85	(0.0575$\times$100/0.116 = 49.56	(0.0569$\times$100)/0.126 = 45.16

Average percentage mass of NaHCO₃

= (41.85+ 49.56 +45.16) /3

= 45.52 %