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(g) + I_(2)(g) hArr 2HI(g)`
`{:(0.3//10,0.3//10,0,"at start"),((0.3)/10-x/10,(0.3)/10-x/10,(2x)/10,"at equilibrium"):}`
`K_C = ((2x //10)^2)/(((0.3-x)/(10))^2)`
`64=(4x^2)/((0.3 -x)^2)`
`x=0.24`
Concentration of unreacted `I_2 = (0.3)/10 -x/10`
= 0.006 mole/L
Amount of unreacted `I_2 = 0.006 xx10 = 0.06 ` moles.