The question : Contain SO

_{2} 1 mole, O

_{2} 2 moles, SO

_{3} 0.5 mol in a close system size 5.0 dm

^{3}Which occurs this reaction 2SO

_{2} + O

_{2} 2SO

_{3} when the system is in equilibrium state, there are SO

_{3} 40% by moles. Find out equilibrium constant (k)

I try to solve it :

[SO

_{2}] = 1 ÷ 5 = 0.2 M

[O

_{2}] = 2 ÷ 5 = 0.4 M

[SO

_{3}] = 0.5 ÷ 5 = 0.1 M

SO

_{3} 40% by mole = (1 + 2 + 0.5) × 40% = 1.4 mole

[SO

_{3}] = 1.4 ÷ 5 = 0.28 M

so..

[SO

_{3}]

0.1

+ 0.18 = 0.28 M

[SO

_{2}]

0.2

- 0.18 = 0.02 M

[O

_{2}]

0.4

- 0.09 = 0.31 M

k = [SO

_{3}]

^{2} ÷ [So

_{2}]

^{2}[O

_{2}]

= (0.28)

^{2} ÷ (0.02)

^{2}(0.31) = 652.26

but my key book said that the answer is 76.9, what is wrong with my way to solve this question?