2. [Stoichiometry, combustion] The treatment of a waste stream, containing prima

Question

2. [Stoichiometry, combustion] The treatment of a waste stream, containing primarily of chlorobenzene (C6HsCI), consists in reusing the waste as fuel for a boiler in a plant. The balanced equation for complete combustion of chlorobenzene with a stoichiometric amount of air (using air composition of 20.9 vol% 02 and 79.1 vol% N2) is given by: CcH5Cl + 7(O2+ 3.78 N2)-> 6 CO2 + 2 H2O + 26.46 N2 + HCl In the real process, the feed flow rate of C6HsCl is 100 kg/hr and the waste is incinerated in 25% excess air (molar basis). The combustion products (flue gases) are released to the atmosphere via a chimney/stack. a) Write the balanced chemical equation describing the complete combustion of chlorobenzene in 25% excess air; b) For every kilogram of fuel burned, how many kilograms of air are fed to the boiler? Ans: 10.73 kg

Explanation / Answer

Part a

Molar flow rate of C6H5Cl = mass/molecular weight

= (100kg/hr) / (112.56 kg/kmol)

= 0.88842 kmol/hr

From the stoichiometry of the reaction

Moles of air required = 7 + 7*3.78

= 33.46 kmol air / kmol C6H5Cl

For burning of 0.88842 kmol/hr of C6H5Cl

Moles of air required = 33.46 x 0.88842 = 29.727 kmol/hr

With 25% excess air

Moles of air fed = 1.25 x 29.727 = 37.158 kmol/hr

Moles of O2 fed = 0.209 x 37.158 = 7.766 kmol/hr

Moles of N2 fed = 0.791 x 11.971 = 29.354 kmol/hr

In the outlet gas stream

Moles of CO2 = 6 x moles of C6H5Cl = 6 x 0.88842

= 5.331 kmol/hr

Moles of H2O = 2 x moles of C6H5Cl = 2 x 0.88842

= 1.777 kmol/hr

Moles of HCl = 0.88842 kmol/hr

Moles of N2 = 29.354 kmol/hr

Moles of O2 = moles of O2 fed - moles of O2 reacted

= 7.766 - 0.88842*7

= 1.547 kmol/hr

Balanced chemical reaction

0.88842 C6H5Cl + 7.766 (O2 + 3.779 N2) = 5.331 CO2 + 1.777 H2O + 1.547 O2 + 29.354 N2 + 0.88842 HCl

Part b

Moles of air fed to boiler = 37.158 kmol/hr

Kg of air fed = 37.158 kmol/hr x 29 kg/kmol = 1077.582 kg/hr

Kg of fuel burned = 100 kg/hr

Kg of air fed / kg of fuel burned = 1077.582/100

= 10.775 kg air fed / kg fuel

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.