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Initially, 0.600 mol of A is present in a 4.00-L solution.

2A[aq]->2B[aq]+C[aq]

At equilibrium, 0.200 mol of C is present. Calculate K.

1 answer

K= [B]^2*[C}/[A]^2

A=(.6-x)/4
B= (x)
C=(x/2)=.2

use C to solve for x, then solve for the rest.

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