A chemical process to convert toluene (MW = 92) and hydrogen to benzene (MW = 78

Question

A chemical process to convert toluene (MW = 92) and hydrogen to benzene (MW = 78) and methane by the reaction. C7H8 + H2 -> C6H6 + CH4 Two streams enter the process, the first stream is pure toluene (40 gmol/s) The second stream is a mixture of H2 (95 mol%) and CH4 (5 mol%) . The flow rate of the H2 in the second stream is 200 gmol/s. Two streams leave the process. The first output stream contains only liquid benzene (product) and toluene (unreacted reactant). The second output stream contains gaseous H2 and CH4. If the conversion of toluene in the process is 75% ( 25% is unreacted) what is the molar flow rate and mass fraction of benzene in the liquid output stream?

Explanation / Answer

if 25% of toulene is unreacted that means that 75% of benzene has been formed

If number of moles of toulene is x(initially)

moles of toluene now=0.25x :unreacted

moles of benzene=0.75x

mass fraction of benzene=0.25*molar mass of benzene/(0.25*molar mass of benzene+0.25*molar mass of toluene)=0.458

Molar flow rate of liquid benzene=40 gmols/s(same as that of toluene which acts as the limiting reagent in this case)

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