Answer:
[tex]\large \boxed{\math{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}[/tex]
Explanation:
The I₂ is the common substance in the two equations.
(1) IO₃⁻ + 5I⁻ + 6H⁺ ⟶ 3I₂ + 3H₂O
{2) I₂ + 2S₂O₃²⁻ ⟶ 2I⁻ + S₄O₆²⁻
From Equation (1), the molar ratio of iodate to iodine is
[tex]\dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{3}{1}[/tex]
From Equation (2), the molar ratio of iodine to thiosulfate is
[tex]\dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} = \dfrac{2}{1}[/tex]
Combining the two ratios, we get
[tex]\text{Stoichiometric factor} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{IO}_{3}^{-}} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} \times \dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{2}{1} \times \dfrac{3}{1} = \mathbf{\dfrac{6}{1}}\\\\\text{The stoichiometric factor is $\large \boxed{\mathbf{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}$}[/tex]
