$((((($explain the moles part specifically by details $)))))))$Question $2$ The combustion of ethane is given by: $C_2{H}_6(g)+3\mathrm{.}5O_2(g)2C{O}_2(g)+3H_{2}O$

$(g),$ where the oxygen is provided by air injection at the reactor inlet. The reaction is carried out

with $40\%$ excess of oxygen, and, in these conditions, the $C_2{H}_6$ conversion is $100\%.$ Both ethane

and air enter the combustion chamber at $25C$ and atmospheric pressure. The product stream

leaves the reactor at $700C,$ also at atmospheric pressure. Consider the air composition to be

Feed consists of $1mol{e}_2{H}_6,3\mathrm{.}5*1\mathrm{.}4=4\mathrm{.}9mol{e}_2$ and $\frac{79}{21}*4\mathrm{.}9=18\mathrm{.}43$ mole $N_2.$

$O_2$ and $18\mathrm{.}43$ mole $N_2.$

$79\%N_2$ and $21\%O_2.$

c$.$ How much heat can be withdrawn from the reactor per mole of ethane combusted?

${}_r{H}_{298}^0={\displaystyle \underset{i}{\overset{?}{}}}{v}_i{}_f{H}_{298}^0(N2$ does not react $)$

${}_r{H}_{298}^0=2*(-393509)+3*(-241818)-1*(-83820)-0=-1\mathrm{.}429*{10}^{6}J$

$H_P=({\displaystyle \underset{i}{\overset{?}{}}}{n}_i($:$C_{p,i}$:${)}_H)*(T-T_0)=({\displaystyle \underset{i}{\overset{?}{}}}{n}_i($:$C_{p,i}$:${)}_H)*(973-298)$

with: $\frac{(\text{:}{C}_P\text{:}{)}_H}{R}=A+\frac{B}{2}{T}_0(+1)+\frac{C}{3}{T}_0^2({}^2++1)+\frac{D}{T_0^2},=\frac{T}{T_0}=3\mathrm{.}26$

and $A=n_i{A}_i=86\mathrm{.}88,B={}_i{B}_i=18\mathrm{.}08*{10}^{-3},C=n_i{C}_i=0$ and

$A=0*1\mathrm{.}131+1\mathrm{.}4*3\mathrm{.}639+18\mathrm{.}43*3\mathrm{.}28+2*5\mathrm{.}457+3*3\mathrm{.}47=86\mathrm{.}8$

$B=0*1\mathrm{.}131+1\mathrm{.}4*0\mathrm{.}5*{10}^{-3}+18\mathrm{.}43*0\mathrm{.}593*{10}^{-3}+2*1\mathrm{.}045*{10}^{-3}+3*1\mathrm{.}45*{10}^{-3}=18\mathrm{.}06*{10}^{-3}$

${\displaystyle \underset{i}{\overset{?}{}}}{n}_i($:$C_{p,i}$:${)}_H=8\mathrm{.}314*(86\mathrm{.}88+\frac{0\mathrm{.}0188}{2}298*4\mathrm{.}26)=821\mathrm{.}5$

$H_P=({\displaystyle \underset{i}{\overset{?}{}}}{n}_i($:$C_{p,i}$:${)}_H)*(973-298)=821\mathrm{.}5*675=554,510J$

$Q=H_R+{}_r{H}_{298}^o+H_P=0-1\mathrm{.}429*{10}^6+554,510=-874\mathrm{.}5kJ5$