What mass of propene `(CH_(3)-CH=CH_(2))` is obtained from `34*0g` of 1-iodopropane by treating with ethanolic KOH if the yield of propene is 36 percent?

Explanation

The elimination reaction taking place is
`underset((3xx12+7xx1+127=170u.=170g))underset("1-Iodopropane")(CH_(3)-CH_(2)-CH_(2)-I+KOH(alc.))tounderset((3xx12+6xx1=42u.=42g))underset("Propane")(CH_(3)-CH=CH_(2)+KI+H_(2)O`
170g of 1-iodopropane give propene =42g
34g of 1-iodopropane give propene`=(42)/(170)xx34=8*4g`
As the yield of propene is only 36% therefore, the actual mass of propene obtained.
`=(8*4xx36)/(100)=3*0.24g`