Answer / kangchuentat

REACTION ENGINEERING - ANSWER 13.1 : (a) Theoretical oxygen gas = 100 kmol C (1/1) = 100 kmol of oxygen gas. (b) Theoretical air = (100 kmol) (1 / 0.21) = 476.2 kmol. (c) Amount of air required 50 % excess = 476.2 kmol (1.5) = 714.3 kmol. (d) Carbon consumption to produce carbon dioxide gas = 0.8 x 100 = 80 kmol = oxygen consumption to produce carbon dioxide gas. Carbon consumption to produce carbon monoxide gas = 0.2 x 100 = 20 kmol = 2 times oxygen consumption to produce carbon monoxide gas, then oxygen consumption to produce carbon monoxide gas = 20 / 2 = 10 kmol. Total oxygen gas required = 80 + 10 = 90 kmol. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Explain why does catalytic converter used in an automobile?

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