Reaction and Mole Calculation
Step 1: Reaction Equation
The balanced chemical reaction is:
\[ 2 \text{KI} + 2 \text{K}_3[\text{Fe(CN)}_6] \xrightarrow{\text{H}^+} \text{I}_2 + 2 \text{K}_4[\text{Fe(CN)}_6] \]
Step 2: Mole Calculation
From the reaction, 2 moles of KI react with 2 moles of \( \text{K}_3[\text{Fe(CN)}_6] \) to produce 1 mole of \( \text{I}_2 \) and 2 moles of \( \text{K}_4[\text{Fe(CN)}_6] \).
The stoichiometric ratio between KI and \( \text{K}_3[\text{Fe(CN)}_6] \) is 1:1.
If 2 moles of \( \text{K}_3[\text{Fe(CN)}_6] \) are used, 2 moles of KI are required.
To solve the problem, we need to determine the number of moles of potassium iodide required to produce 2 moles of complex \( P \) when potassium iodide reacts with potassium ferricyanide.
1. Reaction and complex formation:
- Potassium ferricyanide: \( \mathrm{K_3[Fe(CN)_6]} \)
- Potassium iodide: \( \mathrm{KI} \)
- Reaction forms complex \( P \) reversibly.
- In strong acidic medium, equilibrium shifts completely towards \( P \).
2. Stoichiometry:
- The complex \( P \) is formed by the reaction of potassium ferricyanide with potassium iodide.
- Each mole of potassium ferricyanide requires 1 mole of potassium iodide to form 1 mole of complex \( P \) (typical for such redox/complex formation reactions).
- To form 2 moles of \( P \), 2 moles of potassium iodide are required.
Final Answer:
The number of moles of potassium iodide required to produce 2 moles of \( P \) is \(\boxed{2}\).
Let $ S $ denote the locus of the point of intersection of the pair of lines $$ 4x - 3y = 12\alpha,\quad 4\alpha x + 3\alpha y = 12, $$ where $ \alpha $ varies over the set of non-zero real numbers. Let $ T $ be the tangent to $ S $ passing through the points $ (p, 0) $ and $ (0, q) $, $ q > 0 $, and parallel to the line $ 4x - \frac{3}{\sqrt{2}} y = 0 $.
Then the value of $ pq $ is