A chemical process is used to convet toulene (MW=92) and hydrogen to benzene (MW

Question

A chemical process is used to convet toulene (MW=92) and hydrogen to benzene (MW=78) and methane (CH4) by the reaction : Toulene+H2->Benzene+CH4

Two streams enter the process. The first input stream is pure liquid toulene, which enters at a rate of 40gmol/s. The second input stream is a mixture of H2 (95 mole%) and CH4 (5mole%). The flow rate of H2 in the second stream is equal to 200 gmol/s.

Two streams leave the process. The first output stream contains only liquid benzene (product) and toulene (uncoverted reactant). The second output stream gaseous H2 and CH4. If the concersion of toulene in the process is 75% (ie,. 25% id unreacted), what are the molar flow rate and mass fraction of benzene in the liquid output stream?

Hint: Like many situations in life, this problem may provide more information than you need to answer the given question.

Explanation / Answer

Molar rate of benzene = same as the amount of toluene reacting = 30 gmol/s

toluene unreacted = 10ggmol/s

Hence molar fraction = 30*78/(30*78+10*92)

= 0.71

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