Question

Part III.The two reactions involved in quantitatively determining the amount of iodate in solution are: IO3-(aq) 5 I-(aq) 6 H (aq) --> 3 I2(aq) 3 H2O(l) followed by reaction of the I2: I2(aq) 2 S2O32- --> 2 I-(aq) S4O62-(aq). What is the stoichiometric factor, that is the number of moles of Na2S2O3 reacting with one mole of KIO3

Answer 1

`\large \boxed{\math{\frac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}`

Step-by-step 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

`\frac{\text{I}_(2)}{\text{IO}_(3)^(-)} = (3)/(1)`

From Equation (2), the molar ratio of iodine to thiosulfate is

`\frac{\text{S$_(2)$O}_(3)^(2-)}{\text{I}_(2)} = (2)/(1)`

Combining the two ratios, we get

`\text{Stoichiometric factor} = \frac{\text{S$_(2)$O}_(3)^(2-)}{\text{IO}_(3)^(-)} = \frac{\text{S$_(2)$O}_(3)^(2-)}{\text{I}_(2)} * \frac{\text{I}_(2)}{\text{IO}_(3)^(-)} = (2)/(1) * (3)/(1) = \mathbf{(6)/(1)}\\\\\text{The stoichiometric factor is $\large \boxed{\mathbf{\frac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}$}`