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

`K_(c) = ([HI]^(2))/([H_(2)][I_(2)]), 64 = (x^(2))/(0.03xx0.03)`
`x^(2) = 64xx9xx10^(-4)`
`x = 8xx3xx10^(-2) = 0.24`
x is the amount of HI at equilibrium amount of `I_(2)` at equilibrium will be `0.30 - 0.24 = 0.06`