A) 7.14
B) 14.28
C) 28.57
D) 33.33
Correct Answer: B
Solution :
\[\underset{0.25-x}{\mathop{{{H}_{2}}(g)+}}\,\underset{0.25-x}{\mathop{C{{O}_{2}}(g)}}\,CO\underset{x}{\mathop{(g)}}\,+{{H}_{2}}O\underset{x}{\mathop{(g)}}\,\] At \[\text{e}{{\text{q}}^{\text{m}}}\] \[{{K}_{p}}=0.16=\frac{{{x}^{2}}}{{{(0.25-x)}^{2}}}\Rightarrow 0.4=\frac{x}{0.25-x}\] \[\Rightarrow 0.1-0.4x=x\] \[x=0.0714\] 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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