How much amount of `CuSO_(4).5H_(2)O` required for liberation of `2.55g I_(2)` when titrated with KI

Options

A. 2.5 gm
B. 4.99 gm
C. 2.4 gm
D. 1.2 gm

Answer

To solve the problem of how much amount of `CuSO4·5H2O` is required to liberate `2.55 g` of `I2` when titrated with `KI`, we will follow these steps:

### Step 1: Calculate the number of moles of `I2`
To find the number of moles of iodine (`I2`), we use the formula:

\[ \text{Number of moles} (n) = \frac{\text{mass}}{\text{molar mass}} \]

The molar mass of `I2` is approximately `253.8 g/mol`.

Given:
- Mass of `I2` = `2.55 g`

Calculating the moles of `I2`:

\[ n_{I2} = \frac{2.55 \, \text{g}}{253.8 \, \text{g/mol}} \approx 0.01 \, \text{mol} \]

### Step 2: Write the balanced chemical equation
The balanced reaction between `CuSO4·5H2O` and `KI` can be represented as:

\[ 2 \, \text{CuSO4·5H2O} + 4 \, \text{KI} \rightarrow \text{Cu2I2} + 2 \, \text{K2SO4} + \text{I2} \]

From this equation, we see that `2 moles` of `CuSO4·5H2O` produce `1 mole` of `I2`.

### Step 3: Calculate the moles of `CuSO4·5H2O` needed
Since `2 moles` of `CuSO4·5H2O` produce `1 mole` of `I2`, we can find the moles of `CuSO4·5H2O` required to produce `0.01 moles` of `I2`:

\[ \text{Moles of } CuSO4·5H2O = 2 \times n_{I2} = 2 \times 0.01 \, \text{mol} = 0.02 \, \text{mol} \]

### Step 4: Calculate the mass of `CuSO4·5H2O`
To find the mass of `CuSO4·5H2O`, we use the formula:

\[ \text{Mass} (W) = \text{Number of moles} (n) \times \text{Molar mass} (M) \]

The molar mass of `CuSO4·5H2O` is approximately `249.6 g/mol`.

Calculating the mass:

\[ W = n_{CuSO4·5H2O} \times M_{CuSO4·5H2O} = 0.02 \, \text{mol} \times 249.6 \, \text{g/mol} \approx 4.99 \, \text{g} \]

### Conclusion
Thus, the amount of `CuSO4·5H2O` required to liberate `2.55 g` of `I2` is approximately `4.99 g`.

### Final Answer
**4.99 g of `CuSO4·5H2O` is required.**

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