What is the mole fraction in mixture before combustion?
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{&amp;C2H6(g) + $3.5$~ O2(g) -&gt; 2 CO2(g) + 3 H2O(g)} \\
\ce{&amp;C3H8(g) + 5 O2(g) -&gt; 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 $$\ce{moles~\frac{[C2H6]}{[C3H8]}~=~ \frac{[1]}{[$13.59$]}}$$