A 10g mixture of iso-butane and iso-butene requires 20g of `Br_2`(in `CCl_4`) for complete addition. If 10g of the mixture is catalytically hydrogenated and the entire alkane is monobrominated in the presence of light at `127^(@)`C, how much of it would be formed?
(Atom weight of bromine=80)

Options

A. 24.21g
B. 20.0g
C. 30.0g
D. 12g

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Determine the Composition of the Mixture
We have a 10 g mixture of isobutene (C4H8) and isobutane (C4H10). Let:
- A = mass of isobutane (g)
- B = mass of isobutene (g)

From the problem, we know:
\[ A + B = 10 \, \text{g} \]

### Step 2: Calculate the Amount of Isobutene
It is given that 20 g of Br2 is required for the complete addition reaction. The molar mass of Br2 is 160 g/mol, and the molar mass of isobutene (C4H8) is 56 g/mol.

Using stoichiometry, we can find out how much isobutene is present in the mixture. The reaction is:
\[ \text{C}_4\text{H}_8 + \text{Br}_2 \rightarrow \text{C}_4\text{H}_8\text{Br}_2 \]

From the stoichiometry, 160 g of Br2 reacts with 56 g of isobutene:
\[ 20 \, \text{g of Br}_2 \text{ will react with } \frac{56 \, \text{g}}{160 \, \text{g}} \times 20 \, \text{g} = 7 \, \text{g of isobutene} \]

Thus, we can conclude:
\[ B = 7 \, \text{g} \]

### Step 3: Calculate the Amount of Isobutane
Now, substituting B into the equation from Step 1:
\[ A + 7 \, \text{g} = 10 \, \text{g} \]
\[ A = 10 \, \text{g} - 7 \, \text{g} = 3 \, \text{g} \]

### Step 4: Hydrogenation of Isobutene
Next, we need to hydrogenate the isobutene to form isobutane. The reaction is:
\[ \text{C}_4\text{H}_8 + \text{H}_2 \rightarrow \text{C}_4\text{H}_{10} \]

From stoichiometry, 56 g of isobutene produces 58 g of isobutane:
\[ 7 \, \text{g of isobutene} \text{ will produce } \frac{58 \, \text{g}}{56 \, \text{g}} \times 7 \, \text{g} = 7.25 \, \text{g of isobutane} \]

### Step 5: Total Isobutane
Now, adding the isobutane from the mixture and the hydrogenated isobutene:
\[ \text{Total isobutane} = 3 \, \text{g} + 7.25 \, \text{g} = 10.25 \, \text{g} \]

### Step 6: Monobromination of Isobutane
Now, we need to find out how much of this isobutane will undergo monobromination. The reaction is:
\[ \text{C}_4\text{H}_{10} + \text{Br}_2 \rightarrow \text{C}_4\text{H}_{9}\text{Br} \]

Using stoichiometry, the molar mass of isobutane is 58 g/mol and the molar mass of monobrominated isobutane is 137 g/mol:
\[ 58 \, \text{g of isobutane will produce } 137 \, \text{g of brominated product} \]

Thus, for 10.25 g of isobutane:
\[ \text{Amount of brominated product} = \frac{137 \, \text{g}}{58 \, \text{g}} \times 10.25 \, \text{g} = 24.21 \, \text{g} \]

### Final Answer
The total amount of monobrominated isobutane formed is approximately **24.21 g**.

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