Chemistry - Eudiometry Question with Solution | TestHub

Let $\mathrm{x}\mathrm{mL}{\mathrm{C}}_4{\mathrm{H}}_{10}$ in mixture So, $(3-x)mL{\mathrm{C}}_3{\mathrm{H}}_8$ in mixture $\therefore {\mathrm{C}}_3{\mathrm{H}}_{8(\mathrm{g})}+5{\mathrm{O}}_{2(\mathrm{g})}\to 3{\mathrm{CO}}_{2(\mathrm{g})}+4{\mathrm{H}}_2{\mathrm{O}}_{(\ell )}$

$(3-x)\mathrm{L}$ of ${\mathrm{C}}_3{\mathrm{H}}_8$ produce $3(3-x)\mathrm{L}$ of ${\mathrm{CO}}_2$

$\therefore {\mathrm{C}}_4{\mathrm{H}}_{10(\mathrm{g})}+\frac{13}{2}{\mathrm{O}}_{2(\mathrm{g})}\to 4{\mathrm{CO}}_{2(\mathrm{g})}+5{\mathrm{H}}_2{\mathrm{O}}_{(\ell )}$

' X ' L of ${\mathrm{C}}_4{\mathrm{H}}_{10}$ produce 4 xL of ${\mathrm{CO}}_2$

Total ${\mathrm{CO}}_2$ produced $=3(3-\mathrm{x})+4\mathrm{x}=11\mathrm{L}$

$\mathrm{X}=2\mathrm{L}$ So, mixture contains 2 L butane $\&1\mathrm{L}$ propane Thus, ratio of volume of butane to propane $=2:1$ Correct answer is ' B '