The volume of hydrogen liberated at STP by tr | vedclass

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(A) The balanced chemical equation is: $Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g) \uparrow$Calculate the moles of magnesium $(Mg)$:$n(Mg) = \fr

Vedclass · Accepted Answer

$224$

Vedclass · Answer

(A) The balanced chemical equation is: $Mg(s) + 2HCl(aq) ightarrow MgCl_2(aq) + H_2(g) \uparrow$Calculate the moles of magnesium $(Mg)$:$n(Mg) = \frac{	ext{mass}}{	ext{molar mass}} = \frac{2.4 \ g}{24 \ g \ mol^{-1}} = 0.1 \ mol$From the stoichiometry of the reaction,$1 \ mol$ of $Mg$ produces $1 \ mol$ of $H_2$ gas.Therefore,$0.1 \ mol$ of $Mg$ produces $0.1 \ mol$ of $H_2$ gas.Calculate the volume of $H_2$ at $STP$:$V = n 	imes 	ext{molar volume} = 0.1 \ mol 	imes 22.4 \ L \ mol^{-1} = 2.24 \ L$Convert $2.24 \ L$ to the form $x 	imes 10^{-2} \ L$:$2.24 \ L = 224 	imes 10^{-2} \ L$Thus,the value of $x$ is $224$.

The volume of hydrogen liberated at $STP$ by treating $2.4 \ g$ of magnesium with excess of hydrochloric acid is $x \times 10^{-2} \ L$. Find the value of $x$. Given: Molar volume of gas is $22.4 \ L$ at $STP$. Molar mass of magnesium is $24 \ g \ mol^{-1}$.

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