In Mond's process for refining of nickel, the following reactions occur. The value of 'x' is --------.
`Ni+xCOoverset("330 K - 350 K")rarr No(CO)_(x)overset("450 K - 407 K")rarr Ni_((s))+xCO_((g))`

Answer

To find the value of 'x' in the given reactions of Mond's process for refining nickel, we can follow these steps:

### Step 1: Write down the first reaction
The first reaction in Mond's process is:
\[ \text{Ni} + x \text{CO} \overset{330 \, \text{K} - 350 \, \text{K}}{\rightarrow} \text{Ni(CO)}_x \]

### Step 2: Identify the product
In this reaction, nickel (Ni) reacts with carbon monoxide (CO) to form nickel tetracarbonyl. The chemical formula for nickel tetracarbonyl is:
\[ \text{Ni(CO)}_4 \]

### Step 3: Determine the value of 'x'
Since the product is nickel tetracarbonyl, we can see that it contains 4 carbon monoxide (CO) molecules. Therefore, we can conclude that:
\[ x = 4 \]

### Step 4: Write down the second reaction
The second reaction involves the decomposition of nickel tetracarbonyl at a higher temperature:
\[ \text{Ni(CO)}_4 \overset{450 \, \text{K} - 470 \, \text{K}}{\rightarrow} \text{Ni}_{(s)} + x \text{CO}_{(g)} \]

### Step 5: Balance the second reaction
From the decomposition of nickel tetracarbonyl, we know that one mole of nickel tetracarbonyl produces one mole of solid nickel and releases 4 moles of carbon monoxide. Thus, we can write:
\[ \text{Ni(CO)}_4 \rightarrow \text{Ni}_{(s)} + 4 \text{CO}_{(g)} \]

### Conclusion
From the above reactions, we can see that the value of 'x' in both reactions is 4. Therefore, the final answer is:
\[ x = 4 \]

### Summary
The value of 'x' is 4.

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