Toluene $(C_6{H}_{5}C{H}_3)$ is hydrogenated according to the two simultaneous

chemical reactions

$C_6{H}_{5}C{H}_3+H_2{C}_6{H}_6+C{H}_4$

$2C_6{H}_{5}C{H}_3+H_2$longrightarrow$(C_6{H}_6{)}_2+2C{H}_4$

Initially, the reactor contains $40\%$ toluene and $60\%H_2(\%$ mole$).$ At

the end of the operation, the reactor contains $10\%$ toluene and $30\%C{H}_4$

$(\%$ mole$).$ Find:

a$.$ The mole fraction of the diphenyl, $(C_6{H}_5{)}_2,$ at the end of the operation

b$.$ The conversion of $H_2$

c$.$ Solve $($a$)$ and $($b$)$ when the following final composition is specified: The

reactor contains $10\%$ toluene and mole$).$