Problem $6.$ Benzaldehyde is produced from toluene in the catalytic reaction. An unwanted side reaction produces $C{O}_2.$ The reactions are as follows:

$C_6{H}_{5}C{H}_3+O_2{C}_6{H}_5$CHO$+H_{2}O$

$C_6{H}_{5}C{H}_3+9O_{2}7C{O}_2+4H_{2}O$

Dry air and toluene enter the process as fresh feed. Air is also supplied in $100\%$ excess $($hint: two moles of $O_2$ fed per mole of toluene fed$)$ Of the toluene fed to the reactor$($not the same amount as is in the fresh feed to the process$)13\%$ reacts to form benzaldehyde and $0\mathrm{.}5\%$ reacts with $O_2$ to produce $C{O}_2$ and $H_{2}O.$ The product gas leaves the reactor and enters a separation unit that separates the unreacted toluene from the remaining products and condenses the water plus benzaldehyde and the gases are vented. Over a four hour period, $24kg$ of water is condensed from the system. All of the produced benzaldehyde, soluble in water, leaves the condenser with the $H_{2}O.$ Assume complete condensation of the water$($no water is recycled to the reactor$).$ Unreacted toluene is mixed with the fresh feed and enters the reactor. Product gases include all the $C{O}_2$ unreacted $O_2$ and $N_2($none of the gas is recycled$)$