Question

Trial	Pressure( atm)	Temperature(K)	Reactant Mass	Water displaced (L)
Trial 1 NaHCO3	0.993	292.6K	1.0 g	0.163 L
Trial 2 NaHCO3	0.993	292.8 K	1.1 g	0.122L
Trial 3 Sodium Carbonate	0.993	292.7 K	0.9 g	0.125L
Trial 4 Sodium Carbonate	0.993	279.5K	1.0 g	0.112L

Beginning temp 293.15

Show your calculation for the molar volume of CO₂ from Trial 1. 2, 3, 4 according to your experimental data. Show all work, values, and units in your calculations in order to receive full credit.

Answer 1

Solution :-

Reaction equation

NaHCO3 + H+ ----- > H2O + CO2 + Na+

Using the mass of the NaHCO3 lets calculate the mols of the CO2 formed in each trial

In trial 1 ,2 and 4 mass of NaHCO3 used is 1.0 g

(1g NaHCO3 * 1 mol / 84.0066 g)*(1 mol CO2/1 mol NaHCO3) = 0.0119 mol CO2

in trial 3 mass of NaHCO3 used is 0.9 g

(0.9 g NaHCO3 * 1 mol / 84.0066 g)*(1 mol CO2/1 mol NaHCO3) = 0.0107 mol CO2

Now lets calculate the molar volume of the CO2 in each trial

Molar volume of CO2 = volume of water replaced / moles of CO2

Trial 1 Molar volume of CO2 = 0.163 L / 0.0119 mol = 13.7 mol /L

Trial 2 Molar volume of CO2 = 0.122 L / 0.0119 mol = 10.25 mol/L

Trial 3 Molar volume of CO2 = 0.125 L / 0.0107 mol = 11.68 mol /L

Trial 4 Molar volume of CO2 = 0.112 L / 0.0119 mol = 9.41 mol /L