$$\(3 \quad \mathrm{H}_{2}(\mathrm{~g})\)$$ and $$\(\mathrm{I}_{2}(\mathrm{~g})\)$$ react to form $$\(\mathrm{HI}(\mathrm{g})\)$$ in a reversible reaction, as shown in equation 1. In three separate experiments, a student combines different amounts of two or more of the gases from equation 1. In each experiment, the gases are left to reach equilibrium at a given temperature. In experiment 1, $$\(\mathrm{H}_{2}(\mathrm{~g})\)$$ and $$\(\mathrm{I}_{2}(\mathrm{~g})\)$$ are combined. Water is then added to the equilibrium mixture to produce $$\(1.00 \mathrm{dm}^{3}\)$$ of solution $$\(\mathbf{A}\)$$. The amount of $$\(\mathrm{I}_{2}(\mathrm{aq})\)$$ present in $$\(1.00 \mathrm{dm}^{3}\)$$ of solution $$\(\mathbf{A}\)$$ is found by titration with $$\(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(\mathrm{aq})\)$$. Exactly $$\(32.90 \mathrm{~cm}^{3}\)$$ of $$\(0.200 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\)$$ reacts with all of the $$\(\mathrm{I}_{2}\)$$ in a $$\(25.0 \mathrm{~cm}^{3}\)$$ sample of solution A. $$\[ \mathrm{I}_{2}(\mathrm{aq})+2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq}) \rightarrow 2 \mathrm{I}^{-}(\mathrm{aq})+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(\mathrm{aq}) \]$$ (i) Calculate the amount, in mol , of $$\(\mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}\)$$ that reacts in the titration. $$\[ \text { amount of } \mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}=\ldots \ldots \ldots \ldots \ldots \ldots . . . . . . . . . . . . . . . . \mathrm{mol} \text { [1] } \]$$ (ii) Use your answer to (a)(i) to calculate the amount, in mol, of $$\(I_{2}\)$$ present in $$\(1.00 \mathrm{dm}^{3}\)$$ of solution A. (If you were unable to obtain an answer in (a)(i), then use $$\(9.42 \times 10^{-3} \mathrm{~mol}\)$$. This is not the correct answer.) amount of $$\(I_{2}\)$$ in $$\(1.00 \mathrm{dm}^{3}\)$$ of solution $$\(\mathbf{A}=\)$$ mol

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