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superman1987
When the balanced equation for this reaction is:
2CO(g) + O2(g)→ 2CO2(g)
and according to Kc formula = concentration of the products/concentration of the reactant
Kc = [CO2]^2 / [CO]^2 [O2]
by getting the concentration of CO2 = 0.4 (no.of moles) / 3 (volume per L ) =0.13
and the concentration of O2 = 0.1 (no.of moles of O2) / 3 (volume per L)=0.03
and we have Kc value (given) = 1.4 x 10^2 
So by substitution in the Kc formula:
1.4x10^2 = 0.13^2 / [Co]^2 * 0.03
 
[CO]^2 = 0.13^2 / (1.4x10^2 * 0.03)= 0.004
[CO]= √0.004=  0.63 
that means moles of CO = 0.63 x 3 (volume L) = 1.89 moles

The equilibrium constant is the proportion of the equilibrium concentration of the product to the reactants. The moles of CO present in a flask is 1.89 moles.

What are moles?

Moles are the product of the molar concentration and the volume of the solution. It is given in mol.

The balanced chemical reaction can be shown as:

[tex]\rm 2CO(g) + O_{2}(g) \rightarrow 2CO_{2}(g)[/tex]

The equilibrium constant is given as,

[tex]\rm K_{c} = \dfrac{[CO_{2}]^{2}}{[CO_{2}]^{2} [O_{2}]}[/tex]

The concentration of carbon dioxide is calculated as:

[tex]\dfrac{0.4 \;\rm mol}{3\;\rm L} = 0.13[/tex]

The concentration of oxygen is calculated as:

[tex]\dfrac{0.1 \;\rm mol}{3\;\rm L} = 0.03[/tex]

Substituting values in the formula of the equilibrium constant:

[tex]\begin{aligned}{1.4 \times 10^{2} &= \rm \dfrac{0.13^{2}}{ [CO]^{2} \times 0.03}\\\\\rm [CO] &= \sqrt{0.004} \\\\&= 0.63 \end{aligned}[/tex]

The moles of carbon monoxide will be  [tex]0.63 \times 3 = 1.89\;\rm moles.[/tex]

Therefore, 1.89 moles of carbon monoxide are present in the flask.

Learn more about moles here:

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