Question

FOR THE EQUILIBRIUM BUTANE TO ISOBUTANE IF THE VALUE OF KC IS 3.0 THE PERCENTAGE BY MASS OF ISOBUTANE IN THE EQUILIBRIUM MIXTURE WOULD BE

Answer 1

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Answer 2

Answer: 75% of isobutane will be present in the equilibrium mixture.

Explanation: These two compounds are the structural isomers and are always present in equilibrium.

$Butane\rightleftharpoons Isobutane$

$k_c$ (equilibrium constant) for this reaction is written as:

$k_c=\frac{[Isobutane]}{[Butane]}$

It is given that, $k_c=3.0$

We can write,

$3=\frac{[Isobutane]}{[Butane]}$

$[Isobutane]=3\times [Butane]$

For a given reaction,

[Butane] + [Isobutane] = 1

[Butane] + 3[Butane] = 1

$[Butane]=\frac{1}{4}$

[Butane] = 0.25

[Isobutane] = 0.75

Mass percentage can be calculated as:

$\text{Mass }\%=\frac{\text{Mass of the component}}{\text{Total mass of the solution}}\times 100$

$\text{Mass }\%\text{ of isobutane}=\frac{0.75}{1}\times 100$

Mass percentage of Isobutane = 75%