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(B) The reaction is $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$.Initial moles: $H_2 = 0.3$,$I_2 = 0.3$,$HI = 0$.Let the amount of $H_2$ reacted be $x

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$0.06$

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(B) The reaction is $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$.Initial moles: $H_2 = 0.3$,$I_2 = 0.3$,$HI = 0$.Let the amount of $H_2$ reacted be $x \ mol$. At equilibrium,moles are: $H_2 = (0.3 - x)$,$I_2 = (0.3 - x)$,$HI = 2x$.Since the volume is $10 \ L$,the concentrations are $[H_2] = \frac{0.3-x}{10}$,$[I_2] = \frac{0.3-x}{10}$,$[HI] = \frac{2x}{10}$.$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(\frac{2x}{10})^2}{(\frac{0.3-x}{10})(\frac{0.3-x}{10})} = \frac{4x^2}{(0.3-x)^2} = 64$.Taking the square root on both sides: $\frac{2x}{0.3-x} = 8$.$2x = 8(0.3 - x) \implies 2x = 2.4 - 8x \implies 10x = 2.4 \implies x = 0.24 \ mol$.Amount of unreacted $I_2$ at equilibrium $= 0.3 - x = 0.3 - 0.24 = 0.06 \ mol$.

$A$ mixture of $0.3 \ mol$ of $H_2$ and $0.3 \ mol$ of $I_2$ is allowed to react in a $10 \ L$ evacuated flask at $500 \ ^oC$. The reaction is $H_2 + I_2 \rightleftharpoons 2HI$,and the equilibrium constant $K_c$ is found to be $64$. The amount of unreacted $I_2$ at equilibrium is $...... \ mol$.

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