The simplest ratio x:y of xenon and fluorine when passes through Ni-tube `(400^@C)` at high pressure gives `XeF_6`. Here x+y is :

Explanation

To find the simplest ratio \( x:y \) of xenon (Xe) to fluorine (F2) that produces xenon hexafluoride (XeF6) when passed through a nickel tube at 400°C and high pressure, we can follow these steps:

### Step 1: Understand the Reaction
The chemical reaction for the formation of xenon hexafluoride can be represented as:
\[ \text{Xe} + \text{F}_2 \rightarrow \text{XeF}_6 \]

### Step 2: Determine the Stoichiometry
From the chemical formula of xenon hexafluoride, we see that one mole of xenon reacts with six moles of fluorine:
\[ \text{1 Xe} + \text{6 F}_2 \rightarrow \text{1 XeF}_6 \]

### Step 3: Establish the Ratio
To find the simplest ratio \( x:y \) of xenon to fluorine, we can express it in terms of moles:
- For every 1 mole of xenon (Xe), we need 6 moles of fluorine (F2).
- Therefore, the ratio \( x:y \) can be written as:
\[ x:y = 1:6 \]

### Step 4: Calculate \( x + y \)
Now, we need to find \( x + y \):
- Here, \( x = 1 \) and \( y = 6 \).
- Thus, \( x + y = 1 + 6 = 7 \).

### Final Answer
The value of \( x + y \) is:
\[ \boxed{7} \]