Question #181525

Dry products of combustion of a coal have the following volumetric composition 11 % CO2, 3.5 % O2 and 83.5 % N2, at atmospheric pressure 1 bar and temperature 27°C. Calculate : (i) The gravimetric analysis, (ii) The gas constant for the mixture, (iii) The equivalent molecular weight, (iv) Of the mixture specific heat.

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2021-04-16T07:47:14-0400

Mr(CO2) = 44 Kg/Kmol; Mr(O2) = 32 Kg/Kmol; Mr(N2) = 28 Kg/Kmol

m(CO2) = N(CO2) x Mr(CO2) = 11 Kmol x 44 Kg/Kmol = 484 Kg

m(O2) = N(O2) x Mr(O2) = 3.5 Kmol x 32 Kg/Kmol = 112 Kg

m(N2) = N(N2) x Mr(N2) = 83.5 Kmol x 28 Kg/Kmol = 2338 Kg

mtotal = 484 + 112 + 2338 = 2934 Kg

The mass fraction

mfCO2 = m(CO2) / mtotal = 484 Kg / 2934 Kg = 0.165

mfo2 = m(O2) / mtotal = 112 Kg / 2934 Kg = 0.038

mfN2 = m(N2) / mtotal = 2338 Kg / 2934 Kg = 0.797

Molar mass

Mrm = mtotal / Ntotal = 2934 Kg / 100 Kmol = 29.34 Kg/Kmol

The gas constant of the mixture

Rm = Ru / Mrm = 8.314 Kj/(KmolxK) / 29.34 Kg/Kmol = 0.2834 Kj/(KgxK)

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