Chlorine is prepared in the laboratory by treating manganese dioxide $(MnO_{2})$ with aqueous hydrochloric acid according to the reaction
$4HCl_{(aq)} + MnO_{2(s)} \rightarrow 2H_{2}O_{(l)} + MnCl_{2(aq)} + Cl_{2(g)}$
How many grams of $HCl$ react with $5.0\ g$ of manganese dioxide?
$4HCl_{(aq)} + MnO_{2(s)} \rightarrow 2H_{2}O_{(l)} + MnCl_{2(aq)} + Cl_{2(g)}$
How many grams of $HCl$ react with $5.0\ g$ of manganese dioxide?
(8.4 G) The balanced chemical equation is: $4HCl_{(aq)} + MnO_{2(s)} \rightarrow 2H_{2}O_{(l)} + MnCl_{2(aq)} + Cl_{2(g)}$
Molar mass of $MnO_{2} = 54.94 + 2 \times 16.00 = 70.94 \approx 87\ g/mol$.
Molar mass of $HCl = 1.008 + 35.45 = 36.458 \approx 36.5\ g/mol$.
From the stoichiometry,$1\ mol$ of $MnO_{2}$ $(87\ g)$ reacts with $4\ mol$ of $HCl$ $(4 \times 36.5 = 146\ g)$.
Therefore,$5.0\ g$ of $MnO_{2}$ will react with:
$= \frac{146\ g}{87\ g} \times 5.0\ g = 8.39\ g \approx 8.4\ g$ of $HCl$.
Molar mass of $MnO_{2} = 54.94 + 2 \times 16.00 = 70.94 \approx 87\ g/mol$.
Molar mass of $HCl = 1.008 + 35.45 = 36.458 \approx 36.5\ g/mol$.
From the stoichiometry,$1\ mol$ of $MnO_{2}$ $(87\ g)$ reacts with $4\ mol$ of $HCl$ $(4 \times 36.5 = 146\ g)$.
Therefore,$5.0\ g$ of $MnO_{2}$ will react with:
$= \frac{146\ g}{87\ g} \times 5.0\ g = 8.39\ g \approx 8.4\ g$ of $HCl$.
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