What will be the amount of MgCl_2 (in g) when | vedclass

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(B) The balanced chemical equation is: $MgO + 2HCl 	o MgCl_2 + H_2O$From the stoichiometry of the reaction,$2 \ mol$ of $HCl$ produce $1 \ mol$ of $M

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$22.12$

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(B) The balanced chemical equation is: $MgO + 2HCl 	o MgCl_2 + H_2O$From the stoichiometry of the reaction,$2 \ mol$ of $HCl$ produce $1 \ mol$ of $MgCl_2$.Molar mass of $HCl = 1 + 35.5 = 36.5 \ g/mol$.Molar mass of $MgCl_2 = 24.3 + 2 	imes 35.5 = 95.3 \ g/mol$.Number of moles of $HCl = \frac{17 \ g}{36.5 \ g/mol} \approx 0.4657 \ mol$.Since $2 \ mol$ of $HCl$ produce $1 \ mol$ of $MgCl_2$,the moles of $MgCl_2$ produced = $\frac{0.4657}{2} = 0.23285 \ mol$.Mass of $MgCl_2 = 0.23285 \ mol 	imes 95.3 \ g/mol \approx 22.19 \ g$.Using the standard atomic mass $Mg = 24$ and $Cl = 35.5$,$MgCl_2 = 95 \ g/mol$.Mass of $MgCl_2 = (17 / 73) 	imes 95 = 22.12 \ g$.

What will be the amount of $MgCl_2$ (in $g$) when $17 \ g$ of $HCl$ is reacted with an excess of $MgO$ in the following reaction: $MgO + 2HCl \to MgCl_2 + H_2O$?

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