MCQ
On reaction of $38.1\, g$ iodine and $28.4\, g$ chlorine, mixture of $ICl$ and $ICl_3$ is obtained then total number of moles after reaction will be
- A$0.45$
- ✓$0.40$
- C$0.30$
- D$0.25$
✓
Answer
Correct option: B.
$0.40$
b
$\mathrm{I}_{2}+2 \mathrm{Cl}_{2} \rightarrow \mathrm{ICl}+\mathrm{ICl}_{3}$
$\mathrm{I}_{2}+2 \mathrm{Cl}_{2} \rightarrow \mathrm{ICl}+\mathrm{ICl}_{3}$
Initially $\left(\frac{38.1}{254}=0.15\, \mathrm{mol}\right)\left(\frac{28.4}{71}=0.4\, \mathrm{mol}\right) 0\,\,\,\,0$
$\downarrow$
$L.R.$
Remaining $0.15-0.25$ $0.4-0.3$ $0+0.15$
Moles $=0$ $=0.15\,mol$ $=0.15\,mol$
$\text { Total moles after reaction } =0.1+0.15+0.15$
$=0.40 \,\mathrm{moles}$
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