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 the `HF` formed in reactionn (i) cannot be reused, calculate the weight of `F_(2)` consumed by `1.0 g` of `(CF_(2))_(n)` produced.

Options

A. `2.0 g`
B. `2.52 g`
C. `1.52 g`
D. `3.0 g`

Explanation

To solve the problem, we need to calculate the weight of \( F_2 \) consumed by \( 1.0 \, g \) of \( (CF_2)_n \) produced in the reaction. Let's break it down step by step.

### Step 1: Understand the Reaction
The first reaction is:
\[ (CH_2)_n + 4n \, CoF_3 \rightarrow (CF_2)_n + 2n \, HF + 4n \, CoF_2 \]
From this reaction, we can see that for every mole of \( (CF_2)_n \) produced, \( 4n \) moles of \( CoF_2 \) are generated.

### Step 2: Relate \( CoF_2 \) to \( F_2 \)
The second reaction is:
\[ 2 \, CoF_2 + F_2 \rightarrow 2 \, CoF_3 \]
From this reaction, we can deduce that \( 1 \, mole \) of \( F_2 \) is needed to produce \( 2 \, moles \) of \( CoF_3 \) from \( 2 \, moles \) of \( CoF_2 \). Therefore, \( 1 \, mole \) of \( CoF_2 \) requires \( \frac{1}{2} \, mole \) of \( F_2 \).

### Step 3: Calculate Moles of \( F_2 \) Required
Since \( 4n \) moles of \( CoF_2 \) are produced from \( 1 \, mole \) of \( (CF_2)_n \), we can calculate the moles of \( F_2 \) required:
\[ \text{Moles of } F_2 = \frac{1}{2} \times 4n = 2n \, moles \]

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

### Step 5: Calculate Moles of \( (CF_2)_n \) from 1 g
Now, we can find the number of moles of \( (CF_2)_n \) in \( 1.0 \, g \):
\[ \text{Number of moles of } (CF_2)_n = \frac{1.0 \, g}{50n \, g/mol} = \frac{1}{50n} \, moles \]

### Step 6: Calculate Moles of \( F_2 \) from Moles of \( (CF_2)_n \)
Using the moles of \( (CF_2)_n \) calculated:
\[ \text{Moles of } F_2 = 2n \times \frac{1}{50n} = \frac{2}{50} = 0.04 \, moles \]

### Step 7: Calculate Mass of \( F_2 \)
Finally, we can calculate the mass of \( F_2 \):
\[ \text{Molar mass of } F_2 = 2 \times 19 = 38 \, g/mol \]
\[ \text{Mass of } F_2 = \text{moles} \times \text{molar mass} = 0.04 \, moles \times 38 \, g/mol = 1.52 \, g \]

### Final Answer
The weight of \( F_2 \) consumed by \( 1.0 \, g \) of \( (CF_2)_n \) produced is \( 1.52 \, g \).