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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 the reduction of boron trichloride by hydrogen?

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\text{The balanced chemical equation for the reduction of boron dichloride is}

                                      B C l_{3}+1.5 H_{2} \rightarrow B+3 H C l

\\\text{From the equation, it is clear that 1 mole of boron formation requires 1.5 moles of hydrogen gas.} \\ \text{Let's calculate the moles of boron from the given mass of boron.}

\begin{aligned} \text { No. of moles } &=\frac{\text {Mass}}{\text {Molar mass}} \\\\ &=\frac{21.6 \mathrm{g}}{10.8 \mathrm{gmol}^{-1}} \\\\ &=2.00 \mathrm{mol} \end{aligned}

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

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

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