For the reaction,
`3Zn^(2+)+2K_(4)[Fe(CN)_(6)]rarr K_(2)Zn_(3)[Fe(CN)_(6)]_(2)+6K^(+)`, what will be the equivalent weight of `K_(4)[Fe(CN)_(6)]`, if the molecular weight of `K_(4)[Fe(CN)_(6)]` is M ?

Options

A. M
B. `(M)/(2)`
C. `(M)/(3)`
D. `(M)/(4)`

Answer

To find the equivalent weight of \( K_4[Fe(CN)_6] \) in the given reaction, we can follow these steps:

### Step 1: Understand the Reaction
The reaction is:
\[ 3Zn^{2+} + 2K_4[Fe(CN)_6] \rightarrow K_2Zn_3[Fe(CN)_6]_2 + 6K^+ \]

### Step 2: Identify the Change in Valence
In this reaction, each \( K_4[Fe(CN)_6] \) provides a total of 6 \( K^+ \) ions when 2 moles of \( K_4[Fe(CN)_6] \) are used. This indicates that for every 2 moles of \( K_4[Fe(CN)_6] \), 6 moles of \( K^+ \) are produced.

### Step 3: Determine the n-factor
The n-factor is defined as the number of moles of reactive species produced or consumed per mole of the compound. Here, since 2 moles of \( K_4[Fe(CN)_6] \) yield 6 moles of \( K^+ \), we can calculate the n-factor:
\[ \text{n-factor} = \frac{\text{moles of } K^+}{\text{moles of } K_4[Fe(CN)_6]} = \frac{6}{2} = 3 \]

### Step 4: Calculate Equivalent Weight
The equivalent weight is calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{n-factor}} \]
Given that the molecular weight of \( K_4[Fe(CN)_6] \) is \( M \), we substitute into the formula:
\[ \text{Equivalent Weight} = \frac{M}{3} \]

### Final Answer
Thus, the equivalent weight of \( K_4[Fe(CN)_6] \) is:
\[ \frac{M}{3} \]