Question

the following reaction takes place


3 CuO(s) +2NH3(g)--->3 Cu(s) +N2(g)+ 3 H20(g)


50.0g of CuO(s) was placed in an 80.0 L vessel, and the vessel was evacuated. Ammonia gas was the gradually introduced into the vessel, slow enough for the reaction to proceed, until the total pressure in the vessel (after reaction) was 1.00atm. what are the partial pressures of all 3 gases in the vessel? (Temperature is 180degrees celcius) Please show how you arrived at the answers.

Answer 1

Task:

the following reaction takes place

3 CuO(s) +2NH3(g)-->3 Cu(s) +N2(g)+ 3 H2O(g)

50.0g of CuO(s) was placed in an 80.0 L vessel, and the vessel was evacuated. Ammonia gas was the gradually introduced into the vessel, slow enough for the reaction to proceed, until the total pressure in the vessel (after reaction) was 1.00atm. what are the partial pressures of all 3 gases in the vessel? (Temperature is 180degrees celcius) Please show how you arrived at the answers.

Solution:

The chemical equation for the reaction is

3 CuO(s) +2NH3(g) → 3 Cu(s) +N2(g)+ 3 H2O(g)

The total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual gases

p = p(N2) + p(H2O) + p(NH3)

MW(CuO) = 80 g/mol

v(CuO) = m(g) / MW(g/mol) = 50.0 / 80 = 0.625 mol

According to the chemical equation

v(H2O) = v(CuO) = 0.625 mol

v(N2) = v(CuO)/ 3 = 0.625 / 3 = 0.208 mol

According to the ideal gas law

pV = nRT

p = 1 atm

V = 80.0 L

The total number of moles is

n = pV/RT = 1.00·80.0/ 0.082·(273+180) = 2.15 mol

the mole fraction of each component is

χ(H2O) = n(H2O) / n = 0.625 / 2.15 = 0.291

χ(N2) = n(N2) / n = 0.208 / 2.15 = 0.097

χ(NH3) = n(NH3) / n = (2.15-0.625-0.208)=0.612

The partial pressures are

p(H2O) = p · χ(H2O) = 1.00 · 0.291 = 0.291 atm

p(N2) = p · χ(N2) = 1.00 · 0.0967 = 0.097 atm

p(NH3) = p · χ(NH3) = 1.00 · 0.612 = 0.612 atm

Answer: p(H2O) = 0.291 atm

p(N2) = 0.097 atm

p(NH3) = 0.612 atm