Answer
$y_{co_{2}}= 0.4444\\y_{co_{2}}= 0.5556$
Work Step by Step
The combustion equation in this case is $$
\mathrm{C}_4 \mathrm{H}_{10}+6.5 \mathrm{O}_2 \longrightarrow 4 \mathrm{CO}_2+5 \mathrm{H}_2 \mathrm{O}
$$ The total mole of the products are $4+5=9 \mathrm{kmol}$. Then the mole fractions are
$$
\begin{aligned}
& y_{\mathrm{CO} 2}=\frac{N_{\mathrm{CO} 2}}{N_{\text {total }}}=\frac{4 \mathrm{kmol}}{9 \mathrm{kmol}}=\mathbf{0 . 4 4 4 4} \\
& y_{\mathrm{CO} 2}=\frac{N_{\mathrm{H} 2 \mathrm{O}}}{N_{\text {total }}}=\frac{5 \mathrm{kmol}}{9 \mathrm{kmol}}=\mathbf{0 . 5 5 5 6}
\end{aligned}
$$ Also, $$
N_{\mathrm{CO} 2}=\mathbf{4} \mathbf{k m o l C O}=\mathbf{k m o l C}_{\mathbf{4}} \mathbf{H}_{10}
$$
