Fluoro carbon polymers can be made by fluorinationg polythene.
(i) `(CH_(2))_(n) + 4n Co F_(3) rarr (CF_(2))_(n) + 2 nHF + 4 nCoF_(2)`
Where `n` is large integer. The `CoF_(3)` can be regenarted by the above reaction.
(ii) `2CoF_(2) + F_(2) rarr 2CoF_(3)`
If `HF` can be recovered and electrolyzed to `H_(2)` and if `F_(2)`, is used for regenerating `CoF_(3)`, what is the net consuption of `F_(2)` for `1.0 g` of `(CF_(2))_(n)`.

Options

A. `1.0 g`
B. `1.26 g`
C. `0.76 g`
D. `1.5 g`

Explanation

To solve the problem, we need to determine the net consumption of \( F_2 \) for the production of \( 1.0 \, g \) of \( (CF_2)_n \). Let's break down the solution step by step.

### Step 1: Understand the Reactions
We have two reactions:
1. \( (CH_2)_n + 4n \, CoF_3 \rightarrow (CF_2)_n + 2n \, HF + 4n \, CoF_2 \)
2. \( 2 \, CoF_2 + F_2 \rightarrow 2 \, CoF_3 \)

From the first reaction, we see that 1 mole of \( (CF_2)_n \) is produced from 4 moles of \( CoF_3 \).

### Step 2: Relate \( CoF_2 \) and \( F_2 \)
From the second reaction, we can see that:
- 2 moles of \( CoF_2 \) produce 1 mole of \( F_2 \).
- Therefore, 1 mole of \( CoF_2 \) will require \( \frac{1}{2} \) mole of \( F_2 \).

### Step 3: Calculate the Moles of \( F_2 \) Needed for 1 mole of \( (CF_2)_n \)
From the first reaction, we know that:
- 1 mole of \( (CF_2)_n \) is produced from 4 moles of \( CoF_2 \).
- Thus, the moles of \( F_2 \) required for 4 moles of \( CoF_2 \) can be calculated as follows:
\[ \text{Moles of } F_2 = 4 \, \text{moles of } CoF_2 \times \frac{1}{2} \, \text{mole of } F_2/\text{mole of } CoF_2 = 2 \, \text{moles of } F_2 \]

### Step 4: Calculate the Molar Mass of \( (CF_2)_n \)
The molar mass of \( (CF_2)_n \) can be calculated as follows:
- Molar mass of \( C \) = 12 g/mol
- Molar mass of \( F \) = 19 g/mol
- Therefore, the molar mass of \( (CF_2)_n \) is:
\[ \text{Molar mass} = n \times 12 + 2n \times 19 = 12n + 38n = 50n \, \text{g/mol} \]

### Step 5: Calculate the Moles of \( (CF_2)_n \) in 1 g
Using the molar mass calculated:
- Moles of \( (CF_2)_n \) in 1 g:
\[ \text{Moles} = \frac{1 \, \text{g}}{50n \, \text{g/mol}} = \frac{1}{50n} \, \text{moles} \]

### Step 6: Calculate the Moles of \( F_2 \) Required for 1 g of \( (CF_2)_n \)
Now, using the relationship established earlier:
- Moles of \( F_2 \) needed:
\[ \text{Moles of } F_2 = \left(\frac{1}{50n} \, \text{moles of } (CF_2)_n\right) \times 2n = \frac{2}{50} = \frac{1}{25} \, \text{moles of } F_2 \]

### Step 7: Calculate the Mass of \( F_2 \)
Now, we can find the mass of \( F_2 \):
- Molar mass of \( F_2 \) = 38 g/mol
- Therefore, the mass of \( F_2 \):
\[ \text{Mass} = \text{Moles} \times \text{Molar mass} = \frac{1}{25} \times 38 = \frac{38}{25} = 1.52 \, \text{g} \]

### Final Answer
The net consumption of \( F_2 \) for the production of \( 1.0 \, g \) of \( (CF_2)_n \) is **1.52 g**.