Question

What mass of propene is obtained from 34.0 g of 1-iodo-propane on treating with ethanolic koh, if the yield is 36%?

Answer 1

CH3CH2CH2I + alc.KOH -----> CH3CH=CH2 + KI + H2O

MOLECULAR MASS OF IODOPROPANE = 3*12 + 7*1 + 127= 170
MOLECULAR MASS OF PROPENE =  3*12 + 6*1 = 42
170g OF IODOPROPANE GIVES 42g OF PROPENE
SO 34g OF IODOPROPANE GIVES (42/170)*34 = 8.4g
ACTUAL YIELD IS 36% 
SO (36/100)*8.4 = 3.024g

Answer 2

$\fbox \textcolor{green}{Answer = 3.02\:g}$

$\textcolor{red}{Explanation :-}$

$170 \: of \: iodopropane\:gives\:propane\:=\:42\:g$

$34\:g\:of\:iodopropane\:gives\:propane\:=$

$\frac{42}{170}\times{34}\:=\: 8.4\:g$

$The\:percentage \:yield \:=\:36\%$

$Therefore, \textcolor{violet}{mass\:of\:propane}\:obtained\:=\:$

$\frac{8.4\times{36}}{100}\:=\:3.02\:g$