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90 mL of oxygen is required for complete combustion of unsaturated 20mL gaseous hydrocarbon, hydrocarbon is ?

Answer»

Solution :`("Volume of Hydrocarbon")/("Volume of" O_(2))=(2)/(3n)` (for Alkene)
`("Volume of Hydrocarbon")/("Volume of" O_(2))=(2)/(3n-1)` (for alkyne)
By PUTTING the values in above fomulae we can find the hydrocarbon for which `n` is natural number. `
(20)/(90)=(2)/(3n) n=3` So hydrocarbon is Propene `[C_(3)H_(6)]`.


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