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chemistry (Pls Bob check for me)

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0.5g of fuming h2s04 oleum is diluted with water. This solution is completely neutralized by 26.7ml of 0.4N naoh. Find the percentage of free so3 in the sample

My calculation
h2s04 = 98g/mol
s03 = 80g/mol
98/2 = 49
80/2 = 40
26.7ml x 0.4n = 10.68ml - 0.01067L
x/49 + x/40 (0.5-x) = 0.01068L Please help me The answer 20.6%

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