`n-`Butance `(C_(4)H_(10))` is produced by monobromation of `C_(2)H_(6)` followed by the Wurtz reaction. Calculate the volume of ethane at `STP` requried to produce `55 g` of n-butane. The bromination takes place with 90% yield and the Wurtz reaction with 85% yield.

Options

A. `27.75 L`
B. `55.5 L`
C. `111 L`
D. `5.55 L`

Explanation

To solve the problem, we need to calculate the volume of ethane required to produce 55 g of n-butane, taking into account the yields of the reactions involved. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the molar mass of n-butane (C₄H₁₀)
The molar mass of n-butane can be calculated as follows:
- Carbon (C) has an atomic mass of approximately 12 g/mol.
- Hydrogen (H) has an atomic mass of approximately 1 g/mol.

\[ \text{Molar mass of } C_4H_{10} = (4 \times 12) + (10 \times 1) = 48 + 10 = 58 \text{ g/mol} \]

### Step 2: Calculate the number of moles of n-butane produced
Using the mass of n-butane produced (55 g) and its molar mass (58 g/mol), we can find the number of moles:

\[ \text{Number of moles of } C_4H_{10} = \frac{\text{mass}}{\text{molar mass}} = \frac{55 \text{ g}}{58 \text{ g/mol}} \approx 0.948 \text{ moles} \]

### Step 3: Determine the moles of ethane required
The production of n-butane from ethane involves a Wurtz reaction, where 2 moles of ethane produce 1 mole of n-butane. Therefore, the moles of ethane required can be calculated as:

\[ \text{Moles of } C_2H_6 = 2 \times \text{moles of } C_4H_{10} = 2 \times 0.948 \approx 1.896 \text{ moles} \]

### Step 4: Calculate the volume of ethane at STP
At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 liters. Thus, the volume of ethane required can be calculated as:

\[ \text{Volume of } C_2H_6 = \text{moles} \times \text{volume at STP} = 1.896 \text{ moles} \times 22.4 \text{ L/mol} \approx 42.5 \text{ liters} \]

### Step 5: Adjust for yields of the reactions
Since the bromination has a yield of 90% and the Wurtz reaction has a yield of 85%, we need to adjust the volume calculated for these yields:

\[ \text{Adjusted volume} = \frac{\text{Volume at 100% yield}}{\text{Yield of bromination} \times \text{Yield of Wurtz}} \]

Converting the percentages to fractions:

\[ \text{Adjusted volume} = \frac{42.5 \text{ L}}{0.90 \times 0.85} = \frac{42.5 \text{ L}}{0.765} \approx 55.6 \text{ liters} \]

### Final Answer
The volume of ethane required at STP to produce 55 g of n-butane is approximately **55.6 liters**.

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