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

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Solution

The underlying reaction is:
$2 B C l_{3} + 3 H_{2} ⟶ 2 B + 6 H C l$
Mass of boron formed = 21.6 g,
Moles of boron: $\frac{mass}{molar mass} = \frac{21.6 g}{10.8 g/mol} = 2 mol$
Two moles of boron are formed, which means that 3 mol of hydrogen gas is consumed from the stoichiometry above.
Since, 1 mol of gas at STP occupies 22.4 L, 3 moles would occupy:
$3 \times 22.4 = 67.2 L$

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