From the following reaction between baking soda (NaHCO3) and vinegar (acetic aci

Question

From the following reaction between baking soda (NaHCO3) and vinegar (acetic acid, HC2H3O2), 0.500 L of gases are collected over water. Given the vapor pressure of water is 23.78 mmHg at 25°C, what is the mass of the CO2 gas if the total pressure of the gases collected is 721.0 mmHg (CO2 = 44.01 g/mol)? (1 atm = 760 mmHg (exact))

NaHCO3 (s) + HC2H3O2 (aq) --> H2O (l) + CO2 (g) + NaC2H3O2 (aq)

a. 0.557 g

b. 0.735 g

c. 0.826 g

d. 0.325 g

e. 0.912 g

I know the answer is C, but how do I get there?

Explanation / Answer

Given,

Vapor Pressure of Water at 25 degree C = 23.78 mm Hg

Total Pressure of Gas Collected = 721 mm Hg

We can see from the given reaction that the only gaseous product is CO2.

=> Pressure due to CO2 = 721 - 23.78 = 697.22 mm Hg

=> Presure in atm = 697.22 / 760 = 0.9174 atm ( 1 atm = 760 mm Hg)

Given, Volume = 0.5 L

Temperature = 25 degree C = 298.15 K

Using ideal gas law to calculate the number of moles of CO2, PV= nRT

=> 0.9174 x 0.5 = n x 0.0821 x 298.15

R(Gas constant) = 0.0821

=> n (moles of CO2) = 0.01874 moles

Mass = Moles x Molar Mass

=> Mass of CO2 = 0.01874 x 44.01 = 0.825 g

c) 0.826 g

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