What is the mole fraction of $\ce{C2H6} / \ce{C3H8}$ in mixture, if both compounds reacted with $\ce{O2}$ in stoichiometric ratio? Mass of produced $\ce{CO2}$ is $1.819$ times larger than the mass of $\ce{H2O}$.

$$\ce{C2H6_{(g)} + O2_{(g)} -> CO2_{(g)} + H2O_{(g)}}$$

$$\ce{C3H8_{(g)} + O2_{(g)} -> CO2_{(g)} + H2O_{(g)}}$$

I balanced the equations and tried the following:


Which led me to the wrong result.

Could anyone kick me in the right direction?


It's always helpful to show as much work as you can so that we can see where you went off track, In any case it appears you made an effort, so let's start at the beginning.

First we need to write out the balanced equations \begin{aligned} \ce{&C2H6(g) + $3.5$~ O2(g) -> 2 CO2(g) + 3 H2O(g)} \\ \ce{&C3H8(g) + 5 O2(g) -> 3 CO2(g) + 4 H2O(g)} \end{aligned}

Second we're given that

mass ($\ce{CO2}$) = 1.819 mass ($\ce{H2O}$)

we can converting this to a molar ratio

$$\ce{\frac{[mass ~(CO2)]}{[$44.009$]}~=~ $1.819$ * \frac{[mass~ (H2O)]}{[$18.015$]}}$$

and we find that the molar ratio of $$\ce{\frac{[moles~(CO2)]}{[moles~(H2O)]}~=~ $0.744604$}$$

Now, just to get a feel for things, let's look at the ethane case and we see that we have a mole ratio of $\ce{CO2~/~ H2O}$ of 2/3, whereas in the propane case the ratio is 3/4. The observed molar ratio of 0.74 indicates that the propane reaction is the major contributor. We can write a general equation $$\ce{\frac{[2/3 + (r * 3/4)]}{[1+r]}~=~$0.744604$}$$ where "r" is the molar ratio of propane to ethane being burned. Solving for "r" yields $$\ce{r~=~$13.59$}$$

In other words $$\ce{moles~\frac{[C2H6]}{[C3H8]}~=~ \frac{[1]}{[$13.59$]}}$$

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  • $\begingroup$ Thank you for your effort. I completely forgot to look at the bigger picture, to see which reaction is the major contributor. The correct result is $1/9,19$. Could it be the fault of over simplification of $\ce{\frac[2/3 + (r * 3/4)][1+r]~=~$0.74$}$ ? $\endgroup$ – xan Aug 17 '14 at 18:29
  • $\begingroup$ I get a different value: $\ce{\frac{[C2H6]}{[C3H8]}} \approx \frac{1}{11.26}$. I let $A$ represent the moles of ethane, $B$ represent the moles of propane, and solved $(2A + 3B) \cdot 44.01 = (3A + 4B) \cdot 18.02 \cdot 1.819$ for either variable. Plugging in any valid pair of values for $(A, B)$ derived from the above relation into the $\ce{CO2}$/$\ce{H2O}$ mass ratio equation gives $1.819$, while using the ratio of $\frac{1}{7}$ yields $1.812$. $\endgroup$ – Greg E. Aug 17 '14 at 18:50
  • $\begingroup$ I did mine with greater precision and now I get 13.59. The answer is obtained by dividing 2 very small numbers, so the answer is very extremely sensitive to the atomic weights you use for the elements and the number of decimal places you carry along. $\endgroup$ – ron Aug 17 '14 at 19:27
  • $\begingroup$ It really comes down on how you round up the results.I appreciate both your help. $\endgroup$ – xan Aug 17 '14 at 19:29
  • $\begingroup$ Yes, the instructor or book made a poor choice in setting up a problem that is so sensitive to numeric input choice and rounding. $\endgroup$ – ron Aug 17 '14 at 19:30

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