What volume of `H_(2)` at `273 K` and 1 atm will be consumed in obtaining `21.6 g` of elemental boron (atomic mass of `B = 10.8)` from the reduction of `BCl_(3)` with `H_(2)`.

Options

A. `89.6 L`
B. `67.2 L`
C. `44.8 L`
D. `22.4 L`

Explanation

To solve the problem, we need to determine the volume of hydrogen gas (H₂) consumed in the reduction of boron trichloride (BCl₃) to obtain 21.6 g of elemental boron (B).

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The reduction of BCl₃ with H₂ can be represented as:
\[ \text{BCl}_3 + 1.5 \text{H}_2 \rightarrow \text{B} + 3 \text{HCl} \]
This shows that 1 mole of B requires 1.5 moles of H₂.

2. **Calculate Moles of Boron (B):**
To find the moles of boron produced from 21.6 g, we use the formula:
\[ \text{Moles of B} = \frac{\text{mass}}{\text{molar mass}} \]
The molar mass of boron (B) is 10.8 g/mol. Therefore:
\[ \text{Moles of B} = \frac{21.6 \, \text{g}}{10.8 \, \text{g/mol}} = 2 \, \text{moles} \]

3. **Determine Moles of H₂ Required:**
From the balanced equation, we know that 1 mole of B requires 1.5 moles of H₂. Thus, for 2 moles of B:
\[ \text{Moles of H}_2 = 2 \, \text{moles of B} \times 1.5 \, \text{moles of H}_2/\text{mole of B} = 3 \, \text{moles of H}_2 \]

4. **Calculate Volume of H₂ at STP:**
At Standard Temperature and Pressure (STP: 273 K and 1 atm), 1 mole of an ideal gas occupies 22.4 liters. Therefore, the volume of H₂ consumed is:
\[ \text{Volume of H}_2 = \text{moles of H}_2 \times 22.4 \, \text{L/mol} = 3 \, \text{moles} \times 22.4 \, \text{L/mol} = 67.2 \, \text{L} \]

5. **Final Answer:**
The volume of H₂ consumed in obtaining 21.6 g of elemental boron is **67.2 liters**.