Mass of magnesium required to produce 220 mL | vedclass

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(C) The balanced chemical equation is: $Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)$According to the stoichiometry,$1 \ mol$ of $Mg$ produces $1

Vedclass · Accepted Answer

$235.7 \ mg$

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(C) The balanced chemical equation is: $Mg(s) + 2HCl(aq) ightarrow MgCl_2(aq) + H_2(g)$According to the stoichiometry,$1 \ mol$ of $Mg$ produces $1 \ mol$ of $H_2$ gas.At $STP$,$1 \ mol$ of any gas occupies $22.4 \ L$ or $22400 \ mL$.Number of moles of $H_2$ produced = $\frac{220 \ mL}{22400 \ mL \ mol^{-1}} = 0.00982 \ mol$.Since $1 \ mol$ of $H_2$ requires $1 \ mol$ of $Mg$,moles of $Mg$ required = $0.00982 \ mol$.Mass of $Mg$ = $	ext{moles} 	imes 	ext{molar mass} = 0.00982 \ mol 	imes 24 \ g \ mol^{-1} = 0.2357 \ g$.Converting to milligrams: $0.2357 \ g 	imes 1000 = 235.7 \ mg$.

Mass of magnesium required to produce $220 \ mL$ of hydrogen gas at $STP$ on reaction with excess of dil. $HCl$ is (Given: Molar mass of $Mg$ is $24 \ g \ mol^{-1}$)

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