Question

A single charcoal briquette has a mass of 0.1 kg, and is made up of almost entirely carbon. I want to burn that briquette in air in a sealed oven. When the combustion is complete, the gases in the oven will include CO2, O2, and N2, and the mole fraction of CO2 in this exhaust is 5%. How much air is needed (before the combustion)? Give your answer as kg, kg mole, and as a volume in m3 . Assume that p = 1 atm, and T = 400 ˚C

Answer 1

The combustion reaction can be represented as

C+O2----> CO2, mass of C= 0.1 kg ,moles of C= 0.1/12=0.00833 kg moles

moles of CO2 formed = 0.00833 moles This correspond to 5% of exhausy gases. So flow rat of exhaust gases = 0.00833/0.05= 0.167 kg moles of exhaust gases

for combusiing 0.00833 moles of C oxygen required= 0.00833 moles

moles of air =0.00833/0.21 ( since air contains 21% oxygen)= 0.0397 kg moles

siince outlet contains oxygen oxygen is suppled in excess, let exces percentage of air =x

Air supplied =(1+x/100)*0.0397

N2 = (1+x/100)*0.0397*0.79=(1+x/100)*0.0313 Oxygen =(1+x/100)*0.0397*0.21= 0.00833*(1+x/100)

Oxygen used = 0.00833 oxygen remaining = 0.00833*(1+x/100) -0.00833= 0.00833*x/100

Exhaust gases contains 0.0313*(1+x/100) N2 and oxygen =0.00833*x/100

Total of gases of oxygen and nitrogen in the exhaust= 0.0313*(1+x/100))+ 0.00833*x/100 = 0.167*0.95

0.0313 +0.0313*x/100 +0.00833x/100 =0.159

0.0313*x/100+0.00833*x/100 =0.159-0.0313=0.1277

x(0.0313+0.00833)= 0.1277*100 =12.77

x*0.03963= 12.77 x= 322%

so oxygen is supplied 322% excess

Oxygen remaining =0.00833*3.22 =0.0268 N2= 0.167-0.0268=0.1402 moles

air needed = 0.0397 kg moles air suppied =0.0397*4.22= 0.167 kg moles

air supplied in mass= 0.167*29 kg=4.843 kg and volume =0.167*1000*0.08206*(400+273.15)/1 ( V= nRT/P)=9225 L