Answer to Question #100643 in General Chemistry for Amin

Question

The orsat analysis of fuel gas from a boiler house chimney gives CO2=11.4%, O2=4.2%, N2=84.4% (mole%). Assuming that complete combustion has taken place. calculate the excess air and find C:H ratio in the fuel.

Answer 1

Let us first suppose that nitrogen from the air would not be oxidize through the combustion and that the molar ratio of the air nitrogen to the air oxygen is close to 37 : 10. We also have to check the total percentage of the outgoing gases:

$Total\%=\phi_{O_2}+\phi_{N_2}+\phi_{CO_2}=4.2+84.4+11.4=100\%;$

This means that all water formed was somehow condensed through the analysis.

Now, let us suppose that there are some $n$ moles of the outgoing gas mixture. Thus, the amount in moles of all gases formed can be expressed as:

$n_{res.O_2}=0.042n;$

$n_{N_2}=0.844n;$

$n_{CO_2}=0.114n.$

The total amount of the oxygen consumed for the combustion can be found on the base of the amount of nitrogen in the mixture:

$n_{O_2}=\frac{n_{N_2}}{3.7}=\frac{0.844n}{3.7}=0.228n;$

Now, we are not able to found the amount of the excess oxygen and/or excess air, while we can calculate the percentage of the oxygen and/or air in excess:

$\eta_{ex.air}=\eta_{ex.O_2}=\frac{n_{res.O_2}}{n_{O_2}}*100\%=\frac{0.042n}{0.228n}*100\%=18.4\%.$

Now, we must have some data about the amount in moles of carbon and hydrogen in the fuel burned:

$n_{C}=0.114n;$

$n_{O/CO_2}=2*n_{C}=2*0.114n=0.228n;$

$n_{O/H_2O}=2*n_{O_2}-n_{O/CO_2}-2*n_{res.O_2}=2*0.228n-0.228n-2*0.042n=0.144n;$

$n_{H}=2*n_{O/H_2O}=0.144n*2=0.288n;$

Finally, we would calculate the molar ratio of C to H and also the mass ratio of C to H:

$\frac{n_{C}}{n_{H}}=\frac{0.114n}{0.288n}=0.4=\frac{2}{5};$

$\frac{m_{C}}{m_{H}}=\frac{n_{C}*M(C)}{n_{H}*M(H)}=\frac{0.114n*12g/mol}{0.288n*1g/mol}=\frac{19}{4};$

We can answer that the molar ratio of C to H in the fuel is 2 : 5 and the mass ratio - 19 : 4.