What volume of hydrogen gas at 273K and 1 atm pressure will be consumed in obtaining 21.6g of elemential boron (Atomic mass = 10.8) from the reduction of boron trichloride by hydrogen?

Options

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

Answer

To solve the problem, we need to determine the volume of hydrogen gas consumed in the reduction of boron trichloride (BCl₃) to obtain elemental boron (B). The balanced chemical equation for this reaction is:

\[ 2BCl_3 + 3H_2 \rightarrow 2B + 6HCl \]

### Step 1: Calculate the number of moles of boron produced.
Given that the atomic mass of boron (B) is 10.8 g/mol, we can calculate the number of moles of boron in 21.6 g.

\[ \text{Number of moles of B} = \frac{\text{mass of B}}{\text{molar mass of B}} = \frac{21.6 \, \text{g}}{10.8 \, \text{g/mol}} = 2 \, \text{mol} \]

### Step 2: Determine the moles of hydrogen gas required.
From the balanced equation, we see that 2 moles of B are produced from 3 moles of H₂. Therefore, for 2 moles of B, we need 3 moles of H₂.

\[ \text{Moles of H}_2 = \frac{3}{2} \times \text{moles of B} = \frac{3}{2} \times 2 \, \text{mol} = 3 \, \text{mol} \]

### Step 3: Calculate the volume of hydrogen gas at STP.
At standard temperature and pressure (STP: 273 K and 1 atm), 1 mole of any ideal gas occupies 22.4 liters. Therefore, the volume of hydrogen gas required can be calculated as follows:

\[ \text{Volume of H}_2 = \text{moles of H}_2 \times \text{volume of 1 mole at STP} = 3 \, \text{mol} \times 22.4 \, \text{L/mol} = 67.2 \, \text{L} \]

### Final Answer:
The volume of hydrogen gas consumed is **67.2 liters**.

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