What volume of H_2 gas in liters at 1 , text{ | vedclass

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(C) The balanced chemical equation for the reduction of boron trichloride is: $BCl_3 + \frac{3}{2}H_2 \rightarrow B + 3HCl$ Calculate the moles of bor

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$67.2$

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(C) The balanced chemical equation for the reduction of boron trichloride is: $BCl_3 + \frac{3}{2}H_2 ightarrow B + 3HCl$ Calculate the moles of boron $(B)$ produced: $n(B) = \frac{21.6 \, g}{10.8 \, g/mol} = 2 \, mol$ From the stoichiometry of the reaction,$1 \, mol$ of $B$ requires $\frac{3}{2} \, mol$ of $H_2$. Therefore,$2 \, mol$ of $B$ requires $2 	imes \frac{3}{2} = 3 \, mol$ of $H_2$. At $STP$ ($1 \, 	ext{atm}$,$273 \, K$),the volume of $1 \, mol$ of gas is $22.4 \, L$. Volume of $H_2 = 3 \, mol 	imes 22.4 \, L/mol = 67.2 \, L$.

List-$I$	List-$II$ (At $STP$)
$(A)$ $10 \ g \ CaCO_3 \xrightarrow{\Delta} \text{decomposition}$	$(i)$ $0.224 \ L \ CO_2$
$(B)$ $1.06 \ g \ Na_2CO_3 \xrightarrow{\text{Excess } HCl}$	$(ii)$ $4.48 \ L \ CO_2$
$(C)$ $2.4 \ g \ C \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iii)$ $0.448 \ L \ CO_2$
$(D)$ $0.56 \ g \ CO \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iv)$ $2.24 \ L \ CO_2$
	$(v)$ $22.4 \ L \ CO_2$

What volume of $H_2$ gas in liters at $1 \, \text{atm}$ pressure and $273 \, K$ temperature is required to reduce $21.6 \, g$ of boron trichloride $(BCl_3)$ (atomic weight of $B = 10.8$) to obtain boron?

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