Question

A stream of air (21mole% O2, the rest N2) flowing at a rate of 30 kg/h is mixed with a stream of CO2. The CO2 enters the mixer at a rate of 23.6 m3/h at 200 oC and 1.5 bar. What is the mole % of CO2 in the product gas? a) Calculate the number of kg. moles of air flowing b) Calculate the number of kg. moles of CO2 c) Calculate the mole percent of CO2 in the mixture?

Answer 1

Given the mass of air = 30 Kg/hr = 30000 g / hr

Let the moles of air (both N2 and O2) in 30000 g be 'n' moles

Moles of O2 = 0.21 x n mol

Hence mass of O2 in air = (32 g/mol) x  0.21 x n mol = 6.72n g O2

Moles of N2 = 0.79 x n mol

Hence mass of N2 in air = (28 g/mol) x  0.79 x n mol = 22.12n g N2

Also mass of O2 + mass of N2 = 30000 g

=>  6.72n g O2 + 22.12n g N2 = 30000 g

=> nx(6.72 + 22.12) = 30000 g

=> n = 30000 / 28.84 = 1040 mol air(O2+N2)

For CO2: V = 23.6 m3/hr

T = 200 DegC = 200+273 = 473 K

P = 1.5 bar = 1.5 x 10⁵ Pa

Applying ideal gas equation

PV = nRT

=> n = PV / RT = 1.5 x 10⁵ Pa x 23.6 m3 / (8.314 JK-1mol-1 x 473K) = 900 mol CO2

Hence moles percent of CO2 in the product gas

= [(moles of CO2) / (moles of CO2 + moles of air)] x 100

= [(900 / (900 + 1040)] x 100 = 46.4 % (answer)

(a): As calculated above, moles of air flowing = 1040 mol air / hr = 1040 mol x ( 1 Kgmol / 1000 mol)

= 1.04 Kg.mol air / hr (answer)

(b): Moles of CO2 flowing = 900 mol / hr = 900 mol x (1 Kgmol / 1000 mol) = 0.900 Kg.mol CO2 / hr (answer)

(c): moles percent of CO2 in the mixture

= [(moles of CO2) / (moles of CO2 + moles of air)] x 100

= [(900 / (900 + 1040)] x 100 = 46.4 % (answer)