If 33%HI decomposes at a particular temperature,Kc for the reaction 2HI gives H2 + I2?

we have
2HI ------H2+I2
degree of dissociation=.33
let 2 mole of HI taken initially
mole remain after time when 33 percent decomposes
2-2*.33-mole of HI=2-.66 =1.34 mole
mole of H2=2*.33/2 =.33
mole of I2=2*.33/2=.33
Kc at a particular temperature=[H2][I2]/[HI]^2
=.33*.33/(1.34)^2
=.1089/1.7956
=.060648