Question

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?

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%