Question

the redox titration reached the endpoint when 0.01924 mol triiodide solution were added.
i₃⁻(aq) + sn²⁺(aq) → sn⁴⁺(aq) + 3i⁻(aq)
how many moles of tin(ii) ions were used in the titration?
? moles sn²⁺

Explanation:

Step1: Identify the stoichiometry

From the balanced equation $\ce{I_3^- (aq) + Sn^{2+} (aq) -> Sn^{4+} (aq) + 3I^- (aq)}$, the mole ratio of $\ce{I_3^-}$ to $\ce{Sn^{2+}}$ is 1:1.

Step2: Calculate moles of $\ce{Sn^{2+}}$

Given moles of $\ce{I_3^-}$ is 0.01924 mol. Since the mole ratio of $\ce{Sn^{2+}}$ to $\ce{I_3^-}$ is 1:1, moles of $\ce{Sn^{2+}}$ = moles of $\ce{I_3^-}$ = 0.01924 mol.

Answer:

0.01924