chemistry (Pls Bob check for me)

A 0.5g of fuming h2so4 oleum is diluted with water. This solution is completedly neutralized by 26.7ml of 0.4N naoh. Find the percentage of free so3 in the sample

The asnwer 20.6%

My calculation
h2so4= 98/2 = 49
so3= 80/2 = 40
26.7ml x 0.4N = 10.68ml - 0.01068L

x/49 + x/40 (0.5-X) Bob helps me

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  1. Eq. H2SO4 = Eq of NaOH
    = (26.7 x 0.4)/1000
    = 10.68 mEq.

    H2SO4 + SO3 + H2O -----> 2(H2SO4)
    >---------< is oleum.
    ==> SO3 + H2SO4 -----> H2SO4
    ∴Eq of SO3 = 1/2 of mEq. total H2SO4
    ==>mEq. of Oleum = Eq of H2SO4 + Eq of SO3
    = Eq of total H2SO4
    = 10.68/1000 (bec. 1 mEq. = 1/1000 Eq.)
    Let x be mass of SO2 in Oleum. ==> mass of H2SO4 = (0.5-x)
    Eq. of Oleum = (0.5 - x)/49 + x/40 = 10.68/1000
    Solving.... x = 0.1836g

    Percentage = 0.1836/0.5 x 100 = 20.73%

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