When tetracarbonylnickel(0) is heated , it dissociates into its components. If 5 moles of this compound is heated and the resulting gaseous component is absorbed by sufficient amount of `I_(2)O_(5)`, liberating `I_(2)`. What volume of 4M Hypo solution will be required to react with this `I_(2) ` : `Ni(CO)_(4)overset(A)rarr Ni + 4 CO`

Answer

To solve the question step by step, we will follow the decomposition of tetracarbonylnickel(0) and the subsequent reactions involving iodine and hypo solution.

### Step 1: Write the decomposition reaction of tetracarbonylnickel(0)
The decomposition of tetracarbonylnickel(0) can be represented as:
\[ \text{Ni(CO)}_4 \xrightarrow{\text{heat}} \text{Ni} + 4 \text{CO} \]

### Step 2: Calculate the moles of carbon monoxide produced
Given that 5 moles of Ni(CO)4 are heated, we can calculate the moles of CO produced:
- From the balanced equation, 1 mole of Ni(CO)4 produces 4 moles of CO.
- Therefore, 5 moles of Ni(CO)4 will produce:
\[ 5 \text{ moles Ni(CO)}_4 \times 4 \text{ moles CO/mole Ni(CO)}_4 = 20 \text{ moles CO} \]

### Step 3: Write the reaction of I2O5 with carbon monoxide
The reaction between iodine pentoxide (I2O5) and carbon monoxide (CO) can be represented as:
\[ \text{I}_2\text{O}_5 + 5 \text{CO} \rightarrow \text{I}_2 + 5 \text{CO}_2 \]

### Step 4: Calculate the moles of I2 produced
From the balanced equation, 5 moles of CO produce 1 mole of I2. Therefore, the moles of I2 produced from 20 moles of CO will be:
\[ \frac{20 \text{ moles CO}}{5} = 4 \text{ moles I}_2 \]

### Step 5: Write the reaction of I2 with hypo solution
The reaction of iodine (I2) with sodium thiosulfate (hypo solution) is given by:
\[ \text{I}_2 + 2 \text{Na}_2\text{S}_2\text{O}_3 \rightarrow 2 \text{NaI} + \text{Na}_2\text{S}_4\text{O}_6 \]

### Step 6: Calculate the moles of hypo solution required
From the balanced equation, 1 mole of I2 reacts with 2 moles of sodium thiosulfate. Therefore, for 4 moles of I2, the moles of hypo solution required will be:
\[ 4 \text{ moles I}_2 \times 2 \text{ moles Na}_2\text{S}_2\text{O}_3/\text{mole I}_2 = 8 \text{ moles Na}_2\text{S}_2\text{O}_3 \]

### Step 7: Calculate the volume of hypo solution required
Given that the concentration of the hypo solution is 4 M (molar), we can use the formula:
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
Rearranging gives:
\[ \text{Volume (L)} = \frac{\text{moles of solute}}{\text{Molarity}} \]
Substituting the values:
\[ \text{Volume} = \frac{8 \text{ moles}}{4 \text{ M}} = 2 \text{ L} \]

### Final Answer
The volume of 4 M hypo solution required to react with the liberated I2 is **2 liters**.

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