What is the value for δs°reaction for the following reaction, given the standard entropy values c6h12o6(s) + 6o2(g) —-> 6co2(g) + 6h2o(l)

Standard entropy of the compounds of interest are as follows, 
= 212.1 J/K.mol
= 205.0 J/K.mol
= 213.6 J/K.mol
= 69.9 J/K.mol

Now for the reaction: 
C6H12O6(s) + 6O2(g) —->6CO2(g) + 6H2O(l)

ΔS°reaction = ∑ - ∑
∴ ΔS°reaction =( 6 + 6  ) - ( + 6 = 205.0 J/K.mol)
∴ ΔS°reaction = [6 (213.6) + 6 (69.9 )] - [(212.1) + 6(205.0)]
∴ ΔS°reaction = 258.9

Well to my expert knowledge and my exquisite calculations i think the the negative ion # 412 is the key to sucess