Hydrocarbons Question 413
Question: How many gm of bromine will react with 21 gm $ C_3H_6 $ [MP PET 1985]
Options:
A) 80
B) 160
C) 240
D) 320
Show Answer
Answer:
Correct Answer: A
Solution:
$ \underset{\text{42 }gms}{\mathop{\underset{\text{1 mole}}{\mathop{\underset{Propane}{\mathop{CH_3-CH}}}}}}=CH_2+\underset{\underset{42gms}{\mathop{1mole}}}{\mathop{Br_2}}\to CH_3-\underset{\text{1, 2-dibromo propane}}{\mathop{\underset{Br}{\mathop{\underset{|}{\mathop{CH_2}}}}-\underset{Br}{\mathop{\underset{|}{\mathop{CH_2}}}}}} $
$ \because $ 42 gms of propene reacts with 160 gms of bromine.
$ \therefore $ 21gms of propene $ \frac{160}{42}\times 21=80gms $ .
BETA
