Question.
Chlorine is prepared in the laboratory by treating manganese dioxide $\left(\mathrm{MnO}_{2}\right)$ with aqueous hydrochloric acid according to the reaction
$4 \mathrm{HCl}_{(a q)}+\mathrm{MnO}_{2(s)} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}_{(l)}+\mathrm{MnCl}_{2(a q)}+\mathrm{Cl}_{2(g)}$
How many grams of HCl react with 5.0 g of manganese dioxide?
Chlorine is prepared in the laboratory by treating manganese dioxide $\left(\mathrm{MnO}_{2}\right)$ with aqueous hydrochloric acid according to the reaction
$4 \mathrm{HCl}_{(a q)}+\mathrm{MnO}_{2(s)} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}_{(l)}+\mathrm{MnCl}_{2(a q)}+\mathrm{Cl}_{2(g)}$
How many grams of HCl react with 5.0 g of manganese dioxide?
Solution:
$1 \mathrm{~mol}[55+2 \times 16=87 \mathrm{~g}] \mathrm{MnO}_{2}$ reacts completely with $4 \mathrm{~mol}[4 \times 36.5=146 \mathrm{~g}]$ of $\mathrm{HCl}$.
$5.0 \mathrm{~g}$ of $\mathrm{MnO}_{2}$ will react with
$=\frac{146 \mathrm{~g}}{87 \mathrm{~g}} \times 5.0 \mathrm{~g}$ of $\mathrm{HCl}$
$=8.4 \mathrm{~g}$ of $\mathrm{HCl}$
Hence, $8.4 \mathrm{~g}$ of $\mathrm{HCl}$ will react completely with $5.0 \mathrm{~g}$ of manganese dioxide.
$1 \mathrm{~mol}[55+2 \times 16=87 \mathrm{~g}] \mathrm{MnO}_{2}$ reacts completely with $4 \mathrm{~mol}[4 \times 36.5=146 \mathrm{~g}]$ of $\mathrm{HCl}$.
$5.0 \mathrm{~g}$ of $\mathrm{MnO}_{2}$ will react with
$=\frac{146 \mathrm{~g}}{87 \mathrm{~g}} \times 5.0 \mathrm{~g}$ of $\mathrm{HCl}$
$=8.4 \mathrm{~g}$ of $\mathrm{HCl}$
Hence, $8.4 \mathrm{~g}$ of $\mathrm{HCl}$ will react completely with $5.0 \mathrm{~g}$ of manganese dioxide.