Answer: Moles of CO present in the flask are 0.175
Explanation:
Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.
[tex]2CO(g)+O_2(g)\rightleftharpoons 2CO_2(g)[/tex]
Equilibrium concentration of [tex]CO_2[/tex] = [tex]\frac{moles}{volume}=\frac{0.400}{2.50L}=0.16M[/tex]
Equilibrium concentration of [tex]O_2[/tex] = [tex]\frac{moles}{volume}=\frac{0.100}{2.50L}=0.04M[/tex]
The expression for equilibrium constant for this reaction will be,
[tex]K_c=\frac{[CO_2]^2}{[CO]^2\times [O_2]}[/tex]
[tex]1.4\times 10^2=\frac{(0.16)^2}{[CO]^2\times 0.04}[/tex]
[tex][CO]=0.07M[/tex]
Equilibrium concentration of [tex]CO[/tex] = [tex]\frac{moles}{volume}=\frac{moles}{2.50L}[/tex]
[tex]0.07=\frac{moles}{2.50L}[/tex]
[tex]moles=0.175[/tex]
Thus moles of CO present in the flask are 0.175