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Step-by-Step Solution
Step 1: Identify the Balanced Chemical Equation
The reaction given is:
$$2\,\text{BCl}_{3} + 3\,\text{H}_{2} \to 2\,\text{B} + 6\,\text{HCl}.$$
This tells us that for every 2 moles of boron trichloride, 3 moles of hydrogen gas are required, producing 2 moles of boron and 6 moles of hydrogen chloride.
Step 2: Relate Moles of Hydrogen to Moles of Boron
From the balanced equation:
3 moles of $ \text{H}_{2} $ produce 2 moles of $ \text{B} $.
Thus, if we know how many moles of boron we need, we can find how many moles (and thus volume) of hydrogen gas are required.
Step 3: Calculate the Number of Moles of Boron
We want to prepare 21.6 g of boron. The atomic mass of boron is given as 10.8. Therefore, the number of moles of boron is:
$$\text{Moles of } \text{B} = \frac{\text{Mass of B}}{\text{Atomic mass of B}} = \frac{21.6}{10.8} = 2 \text{ moles}.$$
Step 4: Determine the Required Moles of Hydrogen
According to the reaction stoichiometry:
2 moles of $ \text{B} $ correspond to 3 moles of $ \text{H}_{2} $.
Because we are producing exactly 2 moles of boron, the required moles of hydrogen are 3.
Step 5: Convert the Moles of Hydrogen to Volume at NTP
At NTP (273 K and 1 atm), 1 mole of any ideal gas occupies 22.4 L. We have 3 moles of $ \text{H}_{2} $, so the volume of hydrogen needed is:
$$\text{Volume of } \text{H}_{2} = 3 \times 22.4 \text{ L} = 67.2 \text{ L}.$$
Step 6: State the Final Answer
Therefore, the volume of hydrogen gas required at 273 K and 1 atm to obtain 21.6 g of boron is 67.2 L.
