How many gm of bromine will react with 21 , gm C_3 H_6

MCQ

How many gm of bromine will react with $21\, gm$ ${C_3}{H_6}$

✓
$80$
B
$160$
C
$240$
D
$320$

✓

Answer

Correct option: A.

$80$

a
(a) $\mathop {\mathop {\mathop {C{H_3} - CH}\limits_{{\rm{Propane}}} }\limits_{{\rm{1 \,mole}}} }\limits_{{\rm{42 }}\,gms} = \,\,C{H_2}\,\, + \mathop {B{r_2}}\limits_{\mathop {1\,{\rm{mole}}}\limits_{42\,{\rm{ }}gms} } \to C{H_3} - \mathop {\mathop {\mathop {C{H_2}}\limits_{|\,\,\,\,\,\,\,\,} }\limits_{Br\,\,\,\,} - \mathop {\mathop {C{H_2}}\limits_{|\,\,\,\,\,\,\,\,} }\limits_{Br\,\,\,\,} }\limits_{{\rm{1, 2 - dibromo propane}}} $

$42\, gm$ of propene reacts with $160\, gms$ of bromine.

$\therefore $ $21\,gms$ of propene $\frac{{160}}{{42}} \times \,21 = 80\,gms$.

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