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At a certain temperature, the equilibrium constant, Kc, for this reaction is 53.3 At this temp, 0.3000 mol of H2 and 0.300 mol of I2 were placed in a 1L container to react. What concentration of HI is present at equilibrium? H2+I2<-> 2HI

Timothy Norman

in Chemistry

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Daniel King on February 14, 2018

Your answer is: Kc = [HI]2 / {[H2] [I2]} 53.3 = (2x)2 / {(0.400 M - x)(0.400 M - x)} sqrt(53.3) = 2x / (0.400 M - x) 2.92 - 7.30 x = 2x x = 0.314 M and [HI] = 2x = 0.628 M:) hope this helped. (:


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