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For the equilibrium,

If the value of Kc is 3, the percentage by mass of iso-butanein the equilibriummixture would be:-

a) 75%

b) 90%

c) 30%

d) 60%

Kindly answer sir/mam.

_{c}for the above reaction is represented as :

${K}_{c}=\frac{\left[isobu\mathrm{tan}e\right]}{\left[bu\mathrm{tan}e\right]}\phantom{\rule{0ex}{0ex}}3=\frac{\left[isobu\mathrm{tan}e\right]}{\left[bu\mathrm{tan}e\right]}\phantom{\rule{0ex}{0ex}}\left[isobu\mathrm{tan}e\right]=3\left[bu\mathrm{tan}e\right]\phantom{\rule{0ex}{0ex}}$

2. So concentration of isobutane is thrice of butane in a 100% solution of the two.

Thus isobutane = 75%

butane = 25%

Correct answer is (a).

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