A 36 mL mixture of an alkene and propane required 171 " mL of " O_2 for complete combustion and yielded 109 " mL of " CO_2 (all volume measured at same temperature and presasure). Calculate the molecular formula of olefin and composition of the mixture by volume.

A 36 mL mixture of an alkene and propane required 171 " mL of " O_2 fo

1.	A 36 mL mixture of an alkene and propane required 171 " mL of " O_2 for complete combustion and yielded 109 " mL of " CO_2 (all volume measured at same temperature and presasure). Calculate the molecular formula of olefin and composition of the mixture by volume.
Answer» <html><body><p></p>Solution :`x " mL of " C_(n)H_(<a href="https://interviewquestions.tuteehub.com/tag/2n-300431" style="font-weight:bold;" target="_blank" title="Click to know more about 2N">2N</a>)` <br/> `(36-x)" mL of " C_3H_8` <br/> `C_(n)H_(2n)+(3n)/(2)O_2tonCO_2+nH_2O` <br/> `C_(3)H_(8)+5O_2to3CO_2+4H_2O` <br/> Volume of `CO_2=nx+3(36-x)=108` <br/> `impliesn=3 or x=0` (impossible) <br/> Volume of `O_2` used`=<a href="https://interviewquestions.tuteehub.com/tag/171-1796636" style="font-weight:bold;" target="_blank" title="Click to know more about 171">171</a>` mL <br/> `thereforex((3n)/(2))+5(36-x)=171` <br/> Substituting `n=3,ximplies18mL` <br/> The <a href="https://interviewquestions.tuteehub.com/tag/hydrocarbon-1034072" style="font-weight:bold;" target="_blank" title="Click to know more about HYDROCARBON">HYDROCARBON</a> is `C_3H_6` and the mixture is `50%` composition by volume because `x=18mL`</body></html>