1.5 mmol of I2 react completely with sodium thiosulphate (I2 + 2S2O3^2- -> 2I- + S4O6^2-). The amount of Na2S2O3 consumed is:

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1.5 mmol of $I X_{2}$ react completely with sodium thiosulphate ( $I X_{2} + 2 S X_{2} O X_{3} X^{2 -} 2 I X^{-} + S X_{4} O X_{6} X^{2 -}$ ). The amount of $Na X_{2} S X_{2} O X_{3}$ consumed is:

The stoichiometry is 1 mol

I X_{2}

: 2 mol thiosulphate, so the amount of thiosulphate =

2 \times 1.5 = 3.0

mmol.