Q. When 32.25 g of ethyl chloride is subjected to dehydrohalogenation reaction the yield of the alkene formed is 50%. The mass of the product formed is (atomic mass of chlorine is 35.5)
A) 14 g
B) 28 g
C) 64.5 g
D) 7 g
Explanation
(d)
$C_{2}H_{5}Cl\overset{Dehydrohalogenation\ \ }{\rightarrow}C_{2}H_{4}$
(24+5+35.5) (24+4)
64.5 g of $C_{2}H_{5}Cl$ forms=28 g$C_{2}H_{4}$
$\therefore 32.25\ g\ of\ C_{2}H_{5}Cl\$will form$= \frac{28}{64.5} \times 32.25$
$= 14\ g\ C_{2}H_{5}$
yield of alkene $= 50\%\ of\ 14\ g$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{50}{100} \times 14 = 7g$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.