A) \[1.2,\,\,1.6\]
B) \[1.8,\,\,1.0\]
C) \[0.4,\,\,2.4\]
D) \[0.8,\,\,2.0\]
Correct Answer: A
Solution :
\[\begin{matrix} \begin{align} & Initial \\ & at\,\,eq. \\ \end{align} & \underset{\begin{smallmatrix} 1\,\,mol \\ 1.08 \\ 0.2\,\,mol \end{smallmatrix}}{\mathop{{{H}_{2}}}}\,+\underset{\begin{smallmatrix} 1\,\,mol \\ 2-0.8 \\ =1.2\,\,mol \end{smallmatrix}}{\mathop{{{I}_{2}}}}\,\underset{\begin{smallmatrix} 2\,\,moles \\ 2\times 0.8 \\ =1.6\,\,moles \end{smallmatrix}}{\mathop{2HI}}\, \\ \end{matrix}\] Amount of \[{{H}_{2}}\] (and hence\[{{I}_{2}}\]) consumed \[=1-0.2=0.8\]
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