A stream of air (21 mole% o2, the rest n2) flowing at a rate of 10.0 kg/h is mixed with a stream of co2. the co2 enters the mixer at a rate of 20:0 m3/h at 150°c and 1.5 bar. what is the mole percent of co2 in the product stream?

Question

Juliet Brock

Chemistry

18 March, 16:28

A stream of air (21 mole% o2, the rest n2) flowing at a rate of 10.0 kg/h is mixed with a stream of co2. the co2 enters the mixer at a rate of 20:0 m3/h at 150°c and 1.5 bar. what is the mole percent of co2 in the product stream?

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Eric · Answer 1

Molecular weight of air = 29 g/mole, moles of air entering = 5/29 kgmoles/hr = 0.1724 kg moles/hr

Moles of N2 = 0.21*0.1724=0.036 kg moles/hr, moles of N2 = 0.79*0.1724 = 0.1362 kgmoles/hr

Moles of CO2 can be calculated from gas law equation, n = PV/RT

V = 25 m3/hr = 25 * 1000 L/hr, P = 2 bar = 2*0.9869 atm,=1.9738 atm R = 0.0821 L. atm/mole. K T = 100 deg. c = 100+?273.15 = 373.15K

n = number of moles of CO2 = 1.9738*25*1000 / (0.0821*373.15) = 1611 moles/hr = 1.611 kg moles/hr

total moles of mixture = 0.1724 + 1.611 = 1.7834 kg moles/hr

Moles % CO2 in the mixed stram = 100*1.611/1.7834 = 90.33%

Answer is 90.33%