A) 2.0
B) 1.0
C) 0.334
D) None of these
Correct Answer: B
Solution :
\[\underset{\text{co}{{\text{n}}^{\text{n}}}\,\text{at}\,\text{e}{{\text{q}}^{\text{m}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2-x}{\mathop{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{N}_{2}}{{O}_{5}}(g)}}\,\underset{x-y}{\mathop{{{N}_{2}}{{O}_{3}}(g)}}\,+\underset{x+y}{\mathop{{{O}_{2}}(g)}}\,\] \[\,\,\,\underset{co{{n}^{n}}at\,e{{q}^{m}}}{\mathop{{{N}_{2}}{{O}_{3}}(g)}}\,\underset{x-y}{\mathop{}}\,\underset{y}{\mathop{{{N}_{2}}O(g)}}\,+\underset{y+x}{\mathop{{{O}_{2}}(g)}}\,\] \[2.5=\frac{2.5\times (x-y)}{(2-x)}\] \[x-y=2-x\] or \[2x-y=2\] & as per given\[[{{O}_{2}}(g)]=x+y=2.5\] \[x=1.5\] and \[[{{N}_{2}}O(g)]=y=1.0\]
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