A mixture of `0.3` mole of `H_(2)` and `0.3` mole of `I_(2)` is allowed to react in a 10 litre evacuated flask at `500^(@)`C. The reaction is `H_(2) + I_(2) hArr 2HI`, the K is found to be 64. The amount of unreacted `I_(2)` at equilibrium is

Options

A. 0.15 mole
B. 0.06 mole
C. 0.03 mole
D. 0.2 mole

Answer

B

Explanation

`H_(2)+I_(2) Leftrightarrow 2HI`
`{:(,"Initial", 0.3, 0.3, 0),(,R & F,x,x,2x),(,"at eq",(0.3-x),(0.3-x),2x):}`
`64=((2x)^(2))/((0.3-x)^(2)), x=0.24`
unreacted `I_(2)=0.3 -0.24=0.06`