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Explanation:

equilibrium constant

Kc = [ C ]² / [ A ] [ B ]

= .5² / .2  x 3

= .4167

Let moles of A to be added be n

concentration of A unreacted becomes .2 + n M

increase of product C by .2 M will require use of A  and B be .1 M

So unreacted A = .2 + n - .1 = n + .1

Kc = [ C ]² / [ A ] [ B ]

.4167 = .7² / ( n + .1 ) ( 3 - .1 )

n + .1 = .4

n = . 3 moles .

So .3 moles of A to be added .

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