According to the reaction,Mg_{(s)} + 2HCl_{(a | vedclass

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(B) The number of moles of a gas $(n)$ is given by the formula: $n = \frac{	ext{Volume of gas at STP}}{22.4 \ dm^3 \ mol^{-1}}$.From the given volume

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$4.8$

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(B) The number of moles of a gas $(n)$ is given by the formula: $n = \frac{	ext{Volume of gas at STP}}{22.4 \ dm^3 \ mol^{-1}}$.From the given volume,$n(H_2) = \frac{4.48 \ dm^3}{22.4 \ dm^3 \ mol^{-1}} = 0.2 \ mol$.According to the balanced chemical equation: $Mg_{(s)} + 2HCl_{(aq)} \longrightarrow MgCl_{2_{(aq)}} + H_{2_{(g)}} \uparrow$.$1 \ mol$ of $Mg$ produces $1 \ mol$ of $H_2$ gas.Therefore,the moles of $Mg$ required to produce $0.2 \ mol$ of $H_2$ gas is $0.2 \ mol$.Mass of $Mg = 	ext{moles} 	imes 	ext{molar mass} = 0.2 \ mol 	imes 24 \ g \ mol^{-1} = 4.8 \ g$.

According to the reaction,$Mg_{(s)} + 2HCl_{(aq)} \longrightarrow MgCl_{2_{(aq)}} + H_{2_{(g)}} \uparrow$,calculate the mass of $Mg$ required to liberate $4.48 \ dm^3$ of $H_2$ gas at $STP$. (Molar mass of $Mg = 24 \ g \ mol^{-1}$) (in $g$)

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