How many liters of H_2 gas will be evolved at | vedclass

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(C) For $HCl$,normality $(N)$ is equal to molarity $(M)$ because the n-factor is $1$.So,$0.5 \, N \, HCl = 0.5 \, M \, HCl$.Number of moles of $HCl =

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$2.8$

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(C) For $HCl$,normality $(N)$ is equal to molarity $(M)$ because the n-factor is $1$.So,$0.5 \, N \, HCl = 0.5 \, M \, HCl$.Number of moles of $HCl = 	ext{Molarity} 	imes 	ext{Volume in Liters} = 0.5 	imes 0.5 = 0.25 \, 	ext{mol}$.The balanced chemical equation is: $Zn + 2HCl ightarrow ZnCl_2 + H_2$.According to the stoichiometry,$2 \, 	ext{moles}$ of $HCl$ produce $1 \, 	ext{mole}$ of $H_2$ gas.Therefore,$0.25 \, 	ext{moles}$ of $HCl$ will produce $\frac{0.25}{2} = 0.125 \, 	ext{moles}$ of $H_2$.Volume of $H_2$ at $STP = 	ext{moles} 	imes 22.4 \, L = 0.125 	imes 22.4 = 2.8 \, L$.

How many liters of $H_2$ gas will be evolved at $STP$ by the reaction of zinc metal with $500 \, mL$ of $0.5 \, N \, HCl$?

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