Question

How many gram of bromine will react with $21\mathrm{gm}$ of Propene (${\mathrm{C}}_3{\mathrm{H}}_6$)?

• A. 80
• B. 160
• C. 240
• D. 320

Answer 1

The correct option is A

80

The explanation for the correct option:-

Option (A) 80

Step 1: Chemical Reaction involved

$\underset{\mathrm{Propene}}{{\mathrm{CH}}_3-\mathrm{CH}={\mathrm{CH}}_2}+\underset{\mathrm{Bromine}\mathrm{gas}}{{\mathrm{Br}}_2}\to \underset{1,1\mathrm{Dibromopropane}}{{\mathrm{CH}}_3-{\mathrm{CH}}_2\mathrm{Br}-{\mathrm{CH}}_2\mathrm{Br}}$

So, $42\mathrm{gm}$ of Propene reacts with $160\mathrm{gms}$ of Bromine.

Step 2: Calculation of the amount of Bromine

From the mole concept, $42\mathrm{gm}$ of propene reacts with $160\mathrm{gms}$ of bromine.

$21\mathrm{gm}$ of Propene will react:

$\frac{160}{42}\times 21$

= $80\mathrm{gms}$

Therefore, $80\mathrm{gms}$ of Bromine will react with $21\mathrm{gms}$ of Propene. Option (A) is correct.