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`

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» `x " mL of " C_(n)H_(2n)` `(36-x)" mL of " C_3H_8` `C_(n)H_(2n)+(3n)/(2)O_2tonCO_2+nH_2O` `C_(3)H_(8)+5O_2to3CO_2+4H_2O` Volume of `CO_2=nx+3(36-x)=108` `impliesn=3 or x=0` (impossible) Volume of `O_2` used`=171` mL `thereforex((3n)/(2))+5(36-x)=171` Substituting `n=3,ximplies18mL` The hydrocarbon is `C_3H_6` and the mixture is `50%` composition by volume because `x=18mL`

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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.

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