A) \[7.14\]
B) \[14.28\]
C) \[28.57\]
D) \[33.33\]
Correct Answer: B
Solution :
At \[\begin{matrix} {} & {{H}_{2}}(g)+C{{O}_{2}}(g)\rightleftharpoons CO(g)+{{H}_{2}}O(g) \\ e{{q}^{m}} & 0.25-x\,\,0.25-x\,\,x\,\,x \\ \end{matrix}\] \[{{K}_{p}}=0.16=\frac{{{x}^{2}}}{{{(0.25-x)}^{2}}}\Rightarrow 0.4=\frac{x}{0.25-x}\Rightarrow 0.1-0.4=x\]\[x=0.0714\] Mole% of \[CO(g)=\frac{0.0714}{0.50}\times 100=14.28\]You need to login to perform this action.
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