A) 0.1, 0.1
B) 0.2, 0.2
C) 0.1, 0.2
D) 0.2, 0.1
Correct Answer: A
Solution :
[a] \[{{I}_{2}}+2C{{l}_{2}}\xrightarrow{{}}ICl+IC{{l}_{3}}\]No. of moles | \[\frac{25.4}{254}\] | \[\frac{14.2}{71}\] | 0 | 0 |
initially | 0.1 | 0.2 | 0 | 0.1 |
No. of moles after reaction | 0 | 0 | 0.1 | 0.1 |
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