Question

A scientist tries to generate a mixture of gases similar to a volcano by introducing 30.0 g of H20 vapor, 7.5 g of SO2, and 2.0 g of CO2 into a 30.0 L vessel held at 110.0˚C. (a)Calculate the total pressure of the CO2 in the mixture if the gasses don’t react. However, water and CO2 react to make liquid carbonic acid H2CO3. Write the balanced chemical reaction. Assuming the reaction is complete, how much carbonic acid is formed and how much of each molecule is in the gas phase. Finally, what is the pressure of the mixture if H2CO3 is not in the vapor phase?

Answer 1

Let us first calculate # of moles and mole fraction of each species from mixture.

# of moles = Given mass / Molar mass

# of moles of H2O (n1).0 / 18.0 = 1.67

# of moles of SO2 (n2) = 7.5 / 64 = 0.12

# of moles of CO2 (n3) = 2 / 44 = 0.05

Hence Total # of moles (n) = n1 + n2 + n3 = 1.67 + 0.12 + 0.05 = 1.84

Assuming H2OO = , SO2, and CO2 behave perfectly

Ideal gas equation holds true,

PV = nRT

P = ?, V = 30.0 L, n = 1.84 moles, R = 0.082 L.atm/mole.K, T = 110 ^oC = 383 K

On substituting all known values,

P x 30.0 = 1.84 x 0.082 x 383

P = 1.84 x 0.082 x 383 / 30.0

P = 1.93 atm

Pressure of CO2 = mole fraction of CO2 x Total pressure

= (n3/n) x P

= (0.05/1.84) x 1.93

= 0.052 atm

Partial pressure of CO2 in mixture is 0.052 atm.

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b) Reaction of CO2 and H2O to give H2CO3 (Carbonic acid) is,

H2O + CO2 ------------> H2CO3

mole equivalence is given as

1 mole of H2O 1 mole of CO2   1 mole of H2CO3

From above calculated # of moles it's clear that CO2 is limiting reagent and hence calculations needed to be done w.r.t. CO2

# of moles of CO2 = 0.05

Hence # of moles of H2O reacted = 0.05 and hence # of moles of H2O left = 1.67 - 0.05 = 1.62

Hence mole equivalence is given as,

0.05 mole of H2O    0.05 mole of H2CO3

Molar mass of H2CO3 = 62 g/mole

Mass of 0.05 mole H2CO3 = 0.05 x 62 = 3.1 g

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Now new number of moles of each species are,

n1 = 1.67 - 0.05 = 1.62 moles of H2O left in gas phase

n2 = 0.12 moles of SO2 left in gas phase

n3 = 0.05 - 0.05 = 0 i.e. no CO2 gas present in gas phase.

Hence new N = 1.62 + 0.12 = 1.74 moles

Using 1.74 moles let us again use Ideal gas equation to calculate P,

P = 1.74 x 0.082 x 383 / 30.0

P = 1.82 atm.

Pressure of mixture in vapour phase after reaction is 1.82 atm.

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