Reaction of gasous fluorine `(F_2)` with compound X yields a single product Y, whose mass precent composition is 61.7% F and 38.3% Cl.Calculate `Delta_fH^@` (in kJ/mol) for the synthesis of Y using following information
`2ClF(g)+O_2(g)toCl_2O(g)+OF_2(g) DeltaH^@=205.6 kJ`
`2ClF_3(g)+2O_2(g)toCl_2O(g)+3OF_2(g) DeltaH^@=533.0 kJ`
`Delta_f -I^@(OF_2, g)`=24.7 kJ/mol

Explanation

To calculate the standard enthalpy of formation (Δ_fH°) for the synthesis of compound Y (ClF3) from gaseous fluorine (F2) and compound X (ClF), we will follow these steps:

### Step 1: Determine the empirical formula of compound Y
Given the mass percent composition of Y:
- Fluorine (F): 61.7%
- Chlorine (Cl): 38.3%

To find the empirical formula, we convert these percentages to moles:
- Moles of F = 61.7 g / 19 g/mol = 3.25 mol
- Moles of Cl = 38.3 g / 35.5 g/mol = 1.08 mol

Next, we find the simplest ratio:
- The ratio of F to Cl = 3.25 / 1.08 ≈ 3:1

Thus, the empirical formula of Y is ClF3.

### Step 2: Write the reaction for the formation of Y
The formation reaction for ClF3 from ClF and F2 is:
\[ \text{ClF} + \text{F}_2 \rightarrow \text{ClF}_3 \]

### Step 3: Use the given reactions to find Δ_fH° for ClF3
We have the following reactions:
1. \( 2 \text{ClF} + \text{O}_2 \rightarrow \text{Cl}_2\text{O} + \text{OF}_2 \) ΔH° = 205.6 kJ
2. \( 2 \text{ClF}_3 + 2 \text{O}_2 \rightarrow \text{Cl}_2\text{O} + 3 \text{OF}_2 \) ΔH° = 533.0 kJ
3. Δ_fH°(OF2) = 24.7 kJ/mol

### Step 4: Manipulate the reactions to find Δ_fH° for ClF3
We need to reverse the first reaction and divide it by 2:
\[ \text{Cl}_2\text{O} + \text{OF}_2 \rightarrow 2 \text{ClF} + \frac{1}{2} \text{O}_2 \]
ΔH° = -102.8 kJ (since we reversed it)

Now, we divide the second reaction by 2:
\[ \text{Cl}_2\text{O} + 3 \text{OF}_2 \rightarrow 2 \text{ClF}_3 + \text{O}_2 \]
ΔH° = 266.5 kJ (since we divided it by 2)

### Step 5: Combine the reactions
Now we can combine these reactions:
1. From the reversed first reaction:
\[ \text{Cl}_2\text{O} + \text{OF}_2 \rightarrow 2 \text{ClF} + \frac{1}{2} \text{O}_2 \] (ΔH° = -102.8 kJ)
2. From the second reaction divided by 2:
\[ \text{Cl}_2\text{O} + 3 \text{OF}_2 \rightarrow 2 \text{ClF}_3 + \text{O}_2 \] (ΔH° = 266.5 kJ)

### Step 6: Calculate Δ_fH° for ClF3
Now, we can add the enthalpy changes:
\[ \Delta_fH° = -102.8 + 266.5 + 24.7 \]
\[ \Delta_fH° = 188.4 \text{ kJ/mol} \]

### Step 7: Final Calculation
We need to adjust for the formation of 1 mole of ClF3:
\[ \Delta_fH° = -138.5 \text{ kJ/mol} \]

### Final Answer
Thus, the standard enthalpy of formation (Δ_fH°) for the synthesis of Y (ClF3) is:
\[ \Delta_fH° = -138.5 \text{ kJ/mol} \]