25.4 gm of iodine and 14.2 gm of chlorine are made to react
completely to yield mixture of ICI and `ICl_(3)` Ratio of moles of ICI &
`ICl_(3)` formed is (Atomic mass I: 127, Cl=35.5)

Explanation

To solve the problem of determining the ratio of moles of ICl and ICl3 formed from the reaction of iodine and chlorine, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between iodine (I2) and chlorine (Cl2) can be represented as:
\[ \text{I}_2 + 2\text{Cl}_2 \rightarrow \text{ICl} + \text{ICl}_3 \]

### Step 2: Calculate the molar masses
- Molar mass of I2:
\[ \text{Molar mass of I}_2 = 2 \times 127 = 254 \text{ g/mol} \]

- Molar mass of Cl2:
\[ \text{Molar mass of Cl}_2 = 2 \times 35.5 = 71 \text{ g/mol} \]

### Step 3: Calculate the number of moles of I2 and Cl2
- Moles of I2:
\[ \text{Moles of I}_2 = \frac{25.4 \text{ g}}{254 \text{ g/mol}} = 0.1 \text{ moles} \]

- Moles of Cl2:
\[ \text{Moles of Cl}_2 = \frac{14.2 \text{ g}}{71 \text{ g/mol}} = 0.2 \text{ moles} \]

### Step 4: Determine the limiting reactant
From the balanced equation, we see that 1 mole of I2 reacts with 2 moles of Cl2. Therefore, for 0.1 moles of I2, we would need:
\[ 0.1 \text{ moles of I}_2 \times 2 = 0.2 \text{ moles of Cl}_2 \]
Since we have exactly 0.2 moles of Cl2 available, neither reactant is in excess, and both will be consumed completely.

### Step 5: Calculate the moles of ICl and ICl3 produced
From the balanced equation:
- For every 1 mole of I2, 1 mole of ICl and 1 mole of ICl3 is produced.
Thus, from 0.1 moles of I2, we will produce:
- Moles of ICl = 0.1
- Moles of ICl3 = 0.1

### Step 6: Calculate the ratio of moles of ICl to ICl3
The ratio of moles of ICl to ICl3 is:
\[ \text{Ratio} = \frac{\text{Moles of ICl}}{\text{Moles of ICl}_3} = \frac{0.1}{0.1} = 1:1 \]

### Final Answer
The ratio of moles of ICl to ICl3 formed is **1:1**.

---