Chemistry help plz!!

41 mL of an aqueous solution of iodine
is titrated with 0.046 M Na2S2O3(aq), with
starch as the indicator:
I2(aq) + 2 S2O2−3(aq) −→2 I−(aq) + S4O2−6(aq)
The blue color of the starch-iodine complex
disappears when 43.65 mL of the thiosulfate
solution had been added. What is the molar
concentration of I2 in the original solution?
Answer in units of mol/L

You've got the ionic equation wrong. It is actually as follows:
I₂(aq)+2S₂O₃²⁻(aq)→2I⁻(aq)+S₄O₆²⁻(aq)

Number of moles of S₂O₃²⁻ added
= ([S₂O₃²⁻])(volume of S₂O₃²⁻ solution added in L)
= (0.046)(43.65x10⁻³)
= 2.0079x10⁻³

When the blue colour of the starch-iodine complex disappears, it implies that iodine has been reduced and all the iodine has reacted. According to the equation, the ratio of the number of moles of I₂ required to that of S₂O₃²⁻ required is 1:2. Therefore, the number of moles of I₂ that were originally present
= 1/2(number of moles of S₂O₃²⁻)
= 1/2(2.0079x10⁻³)
= 1.00395x10⁻³

Thus, [I₂] in the original solution
= number of moles of I₂/volume of the solution in L
= 1.00395x10⁻³/41x10⁻³
= 0.024486585 mol/L
= 0.024mol/L correct to 2sf

I hope this helps and feel free to send me an e-mail if you have any doubts!