Question

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

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

Answer 1

Answer: (A)

67.2 L

Given: Mass of elemental boron $=$ 21.6 g
Temperature $=$ 273 K
Pressure $=$ 1 atm
Solution: The balanced chemical equation for the reduction of boron dichloride is

$BC{l}_3+1.5H_2\to B+3HCl$
From the equation, it is clear that 1 mole of boron formation requires 1.5 moles of hydrogen gas.
Let's calculate the moles of boron from the given mass of boron.

No. of moles $=$ $\frac{Mass}{Molarmass}$

$=$ $\frac{21.6g}{10.8gmo{l}^{-1}}$

$=$ $2.00mol$

So, to obtain 2.00 mol of boron, 3.00 mol of hydrogen gas is required.
One mole of hydrogen gas at 273 K and 1 atm occupies a volume of $22.4L$.

Therefore, volume of hydrogen gas $=$ $3.00mol\times 22.4Lmo{l}^{-1}$
$=$ $67.2L$