19, A gas mixture consists of only three components-40 mol% A, 18.75 mass% B and

Question

19, A gas mixture consists of only three components-40 mol% A, 18.75 mass% B and 20 mol% C. The molar masses of A and C are 40 and 50 respectively. Find the molar mass of B and the molar mass of the mixture. 20. Unless told otherwise, assume that the composition of air is 79% nitrogen and 21% oxygen on a molar basis. (I expect you to remember this composition!) What is the composition of air on a mass basis? What would a composition on a volume% basis mean? And what would the volume% of 02 in air be? 21. Air is mixed with a flue gas (exhaust gases) from a furnace in order to cool the gas. The molar composition of the flue gas is 2% oxygen, 35% CO2, 10% water vapour and the rest is nitrogen. The nitrogen concentration in the gas after the air and flue gas been mixed is 71 mol%. The flow rate of the flue gas is 600 kg/hr. a) What is the molar flow rate of the air stream? b) What is the mass flow rate of oxygen in the mixed gas?

Explanation / Answer

Ans 20

Basis 100 mol of air

Moles of nitrogen = 79 mol

Mass of nitrogen = moles x molecular weight

= 79 mol x 28 g/mol = 2212 g

Moles of Oxygen = 21 mol

Mass of oxygen = moles x molecular weight

= 21 mol x 32 g/mol = 672 g

Total mass of air = 2212 + 672 = 2884 g

Mass% of Nitrogen = mass of nitrogen *100/total mass

= 2212*100/2884

= 76.7 %

Mass% of oxygen = mass of oxygen *100/total mass

= 672*100/2884

= 23.3 %

Volume % = mol %

Lets prove it

For a component, ideal gas equation at constant pressure and temperature

PV1 = n1RT

For a mixture

PV = nRT

V1/V = n1/n

Volume fraction = mol fraction

Volume % = mol %

Composition of air on volume % will remain the same.

Volume % of O2 = 21%

Volume % of N2 = 79%

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