One of the reactions used in the petroleum industry for improving octance number of fuels is
`C_(7) H_(14) rarr C_(7)H_(8) + 3H_(2)`
The two hydrocarbons `C_(7) H_(14)` and `C_(7)H_(8)` are liquid, `H_(2)` formed is gas. Whatis the percentage reduction in liquid weight accompanying the completion of the above reaction?

Options

A. `~~ 1%`
B. `~~ 3%`
C. `~~ 5%`
D. `~~ 6%`

Explanation

To solve the problem, we need to determine the percentage reduction in liquid weight that occurs during the reaction:

\[ C_{7}H_{14} \rightarrow C_{7}H_{8} + 3H_{2} \]

### Step 1: Calculate the molar masses of the compounds involved.

- **Molar mass of \( C_{7}H_{14} \)**:
- Carbon (C) has a molar mass of 12 g/mol.
- Hydrogen (H) has a molar mass of 1 g/mol.

\[ \text{Molar mass of } C_{7}H_{14} = (7 \times 12) + (14 \times 1) = 84 + 14 = 98 \text{ g/mol} \]

- **Molar mass of \( C_{7}H_{8} \)**:

\[ \text{Molar mass of } C_{7}H_{8} = (7 \times 12) + (8 \times 1) = 84 + 8 = 92 \text{ g/mol} \]

### Step 2: Determine the reduction in weight.

- **Initial weight of liquid**:
- This is the weight of \( C_{7}H_{14} \), which is 98 g.

- **Final weight of liquid**:
- This is the weight of \( C_{7}H_{8} \), which is 92 g.

- **Reduction in weight**:

\[ \text{Reduction in weight} = \text{Initial weight} - \text{Final weight} = 98 \text{ g} - 92 \text{ g} = 6 \text{ g} \]

### Step 3: Calculate the percentage reduction in weight.

- **Percentage reduction**:

\[ \text{Percentage reduction} = \left( \frac{\text{Reduction in weight}}{\text{Initial weight}} \right) \times 100 \]

\[ \text{Percentage reduction} = \left( \frac{6 \text{ g}}{98 \text{ g}} \right) \times 100 \approx 6.12\% \]

### Step 4: Final answer.

Since we are looking for the percentage reduction in liquid weight, we can round this to:

\[ \text{Percentage reduction} \approx 6\% \]

### Conclusion:
The percentage reduction in liquid weight accompanying the completion of the reaction is approximately **6%**.

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