Heptane, C7H16, is being burned with oxygen, O2, to produce carbon dioxide, CO2,

Question

Heptane, C7H16, is being burned with oxygen, O2, to produce carbon dioxide, CO2, and water by the reaction:

C7H16 + 11O2 7CO2 + 8H2O.

Suppose that 750 kg/hr of C7H16 are reacted with 3200 kg/hr O2 to produce 2310 kg/hr of CO2.

a. What is the limiting reactant? Does the choice of the limiting reactant make sense for and industrial process that employs this reaction? Why or why not?

b. What is the percentage excess of the non-limiting reactant?

c. What is the conversion of C7H16 to CO2 and H2O?

d. What is the conversion of O2 to CO2 and H2O?

e. How many kg/hr of water are produced during the reaction?

f. Suppose we use air (79 mole% N2 and 21 mole% O2, molecular weight of 29) in place of O2 in the process. Why would we choose to use air instead of O2? How many kg/hr of air are required to deliver the 3200 kg/hr of O2 to the process?

Explanation / Answer

C7H16 + 11O2 7CO2 + 8H2O

nO OF mole of c7h16 = 750*10^3/100.21 = 7484.28 mole/hr

No of mole of o2 = 3200*10^3/32 = 100000 mole/hr

limiting reagent is c7h16

so that,

No of moles CO2 produced   = 7484.28*7 = 52389.96 mole/hr

mass of Co2 produced = 52389.96*44 = 2305.1582 kg/hr

the mass of given CO2 in hr = 2310 kg it is qual to theoretical vaue.

so that given one correct.

b. Non limiting reactant = O2 = 100000-(7484.28*11) = 17672.92 mole

mass of O2 excess = 17672.92*32 = 2634.466 kg

c. 1 mole c7h16 = 7 mole co2 = 8 mole H2O

d. 11 mole O2 = 7 mole co2 = 8 mole H2O

e. mass of water produced = 7484.28*8*18 = 1077.73632 kg/hr

f. 3200 kg/hr of O2 = 100000 mole/hr

to deliver 100000 mole/hr of O2 mole of air required = 100000*100/21

         = 476190.48 mole/hr

massof air required = 476190.48 *29 = 13809.524 kg/hr

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