20ml mixture containing methane and ethane in the volume ratio of x:y when subjected to combustion, volume of ${\mathrm{CO}}_2$  liberated is 35ml. What will be the volume of ${\mathrm{CO}}_2$ liberated when the mole ratio of methane to ethane is y :x

Let the volume of ethane = ${\mathrm{V}}_1$
  Volume of methane = $(20-{\mathrm{V}}_1)$
$\begin{array}{l}\\ {\mathrm{CH}}_4+ {\mathrm{O}}_2\to {\mathrm{CO}}_2+ 2{\mathrm{H}}_2\mathrm{O}\\ \left(20-{\mathrm{V}}_1\right) \left(20-{\mathrm{V}}_1\right)\end{array}$
$\begin{array}{l}{\mathrm{C}}_2{\mathrm{H}}_6+\frac{7}{2}{\mathrm{O}}_2\to 2{\mathrm{CO}}_2+3{\mathrm{H}}_2\mathrm{O}\\ {\mathrm{V}}_{1}2{\mathrm{V}}_1\end{array}$
$\mathrm{Given} 20-{\mathrm{V}}_1+2{\mathrm{V}}_1=35\mathrm{ml}$
        ${\mathrm{V}}_1=15\mathrm{ml}$

Volume of methane = (20-15) = 5 ml
               Now if the mole ratio is y : x
$\mathrm{Then} {\mathrm{V}}_{{\mathrm{CH}}_4}=15\mathrm{ml}$
${\mathrm{V}}_{{\mathrm{C}}_2{\mathrm{H}}_6}=5\mathrm{ml}$
$\begin{array}{l}{\mathrm{CH}}_4+2{\mathrm{O}}_2\to {\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}\\ 15\mathrm{ml} \text{\hspace{0.17em}15\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ml}\end{array}$
$$\begin{gathered} {\mathrm{C}}_2{\mathrm{H}}_6+\frac{7}{2}{\mathrm{O}}_2\to 2{\mathrm{CO}}_2+3{\mathrm{H}}_2\mathrm{O} \\ 5\mathrm{ml} 10\mathrm{ml} \end{gathered}$$
${\mathrm{V}}_{{\mathrm{CO}}_2}=25\mathrm{ml}$