The reduction of boron trichloride (BCl_3) by | vedclass

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(D) The balanced chemical equation for the reduction of boron trichloride by hydrogen is: $2BCl_3(g) + 3H_2(g) \rightarrow 2B(s) + 6HCl(g)$ From the s

Vedclass · Accepted Answer

$67.2$

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(D) The balanced chemical equation for the reduction of boron trichloride by hydrogen is: $2BCl_3(g) + 3H_2(g) ightarrow 2B(s) + 6HCl(g)$ From the stoichiometry,$2 \ mol$ of $B$ are produced by $3 \ mol$ of $H_2$. Moles of $B$ produced = $\frac{	ext{mass}}{	ext{atomic mass}} = \frac{21.6 \ g}{10.8 \ g/mol} = 2 \ mol$. Since $2 \ mol$ of $B$ require $3 \ mol$ of $H_2$,the moles of $H_2$ required is $3 \ mol$. At $STP$ ($273 \ K$ and $1 \ atm$),$1 \ mol$ of any ideal gas occupies $22.4 \ L$. Therefore,volume of $H_2$ = $3 \ mol 	imes 22.4 \ L/mol = 67.2 \ L$.

The reduction of boron trichloride $(BCl_3)$ by hydrogen yields elemental boron (atomic mass $10.8 \ g/mol$). If $21.6 \ g$ of boron is produced,what is the volume of hydrogen gas required at $273 \ K$ and $1 \ atm$ pressure (in $L$)?

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