A) 1.2, 1.6
B) 1.8, 1.0
C) 0.4, 2.4
D) 0.8, 2.0
Correct Answer: A
Solution :
\[{{H}_{2}}\] | + | \[{{I}_{2}}\] | \[\rightleftarrows \] | \[2HI\] | |
Initially | 1 mol | 2 mol | 2 mol | ||
at eq. | 0.2 mol | 2-0.8 | \[2\times 0.8\] | ||
= 1.2 mol | =1.6 mol |
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