In the laboratory,chlorine is prepared by the | vedclass

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(A) The balanced chemical equation is:$4HCl_{(aq)} + MnO_{2(s)} 	o 2H_2O_{(l)} + MnCl_{2(aq)} + Cl_{2(g)}$Molar mass of $MnO_2 = 54.94 + 2 	imes 16

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$8.39$

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(A) The balanced chemical equation is:$4HCl_{(aq)} + MnO_{2(s)} 	o 2H_2O_{(l)} + MnCl_{2(aq)} + Cl_{2(g)}$Molar mass of $MnO_2 = 54.94 + 2 	imes 16 = 86.94 \approx 87 \ g/mol$.Molar mass of $HCl = 1 + 35.5 = 36.5 \ g/mol$.From the stoichiometry of the reaction,$1 \ mol$ of $MnO_2$ $(87 \ g)$ reacts with $4 \ mol$ of $HCl$ $(4 	imes 36.5 = 146 \ g)$.Therefore,for $5.0 \ g$ of $MnO_2$,the amount of $HCl$ required is:$	ext{Mass of } HCl = \frac{146 \ g \ HCl}{87 \ g \ MnO_2} 	imes 5.0 \ g \ MnO_2 = 8.39 \ g$ of $HCl$.

In the laboratory,chlorine is prepared by the reaction of manganese dioxide $(MnO_2)$ with aqueous hydrochloric acid $(HCl)$ as follows:
$4HCl_{(aq)} + MnO_{2(s)} \to 2H_2O_{(l)} + MnCl_{2(aq)} + Cl_{2(g)}$
How many grams of $HCl$ will react with $5.0 \ g$ of manganese dioxide (in $g$)?

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In the laboratory,chlorine is prepared by the reaction of manganese dioxide $(MnO_2)$ with aqueous hydrochloric acid $(HCl)$ as follows:$4HCl_{(aq)} + MnO_{2(s)} \to 2H_2O_{(l)} + MnCl_{2(aq)} + Cl_{2(g)}$How many grams of $HCl$ will react with $5.0 \ g$ of manganese dioxide (in $g$)?