Question

In: Chemistry

A 1.000-L vessel contains a mixture of hydrogen, carbon monoxide and methane gases in unknown proportions...

A 1.000-L vessel contains a mixture of hydrogen, carbon monoxide and methane gases in unknown proportions at 23.3-bar and 25 °C. The gaseous mixture is found to have a mass of 13.70-g and a higher heating value of 494.90 kJ/mol. What is the composition of the gas?

Solutions

Expert Solution

Volume V = 1 L = 0.001 m3

Pressure P = 23.3 bar = 2330 kpa

Temperature = 25 °C = 25+273.15 = 298.15 K

Mass M = 13.7 g = 0.0137 kg

Assume this as ideal mixture, for ideal gas mole%=vol%

PV=nRT

where n = M/Mw

n - no of moles

Mw - Moleculat weight of mixture

R - gas constant 8.314 kJ/kmol.K

Mw = MRT/PV

=0.0137*8.314*298.15/2330*0.001

Mw=14.575 g/gmol

Mw= xh2*Mwh2 + xco*Mwco + xch4*Mwch4

14.575 = xh2*2 + xco*28 + xch4*16   -------- (1)

xh2 - mole fraction of hydrogen

Mwh2 - Molecular weight of hydrogen, 2

xco - mole fraction of carbonmonoxide

Mwco - molecular weight of carbonmonoxide, 28

xch4 - mole faction of methane

Mwch4 - molecular weight of methane, 16

Higher heating value gas mixture HHVmix = xh2*HHVh2 + xco*HHVco + xch4*HHVch4

HHVh2 higher heating value of H2 = 284.162 kJ/mol

HHVco higher heating value of CO = 284.479 kJ/mol

HHVch4 higher heating value of CH4 = 886.15 kJ/mol

494.9 = xh2*284.162 + xco*284.479 + xch4*886.15 ---------- (2)

we know that sum of the fraction is equal to 1, so

1 = xh2 + xco+ xch4 --------(3)

Solving (1) , (2) & (3)

we get

xh2 = 0.3548

xco = 0.2952

xch4 = 0.35


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