Question

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 $BC{l}_3$ with $H_2$?

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

Answer 1

Answer: (B)

$67.2$ L

$BC{l}_3+1.5H_2\to B+3HC{l}_2$

From the above reaction, 1 mole of B require 1.5 mole of $H_2$.

Moles of boron obtained = $\frac{weight}{molecularweight}=\frac{21.6}{10.8}=2$.

2 mole of B is formed, which means $2\times 1.5=3$ mole of $H_2$ is consumed.

At 273 K and 1 atm (STP), 1 mole of ideal gas occupies 22.4 L.

$\therefore$ The volume of $H_2$ consumed$=3\times 22.4=67.2L$.