Question

Chlorine is prepared in the laboratory by treating manganese dioxide($Mn{O}_2$) with aqueous hydrochloric acid according to the reaction:

$4HCl\left(aq\right)+Mn{O}_2\left(s\right)\to 2H_{2}O\left(l\right)+MnC{l}_2\left(aq\right)+C{l}_2\left(g\right)$

The amount of $HCl$ that react with 5.0 g of manganese dioxide is:

• A. $8.39g$
• B. $83.9g$
• C. $167.8g$
• D. $16.78g$

Answer 1

The correct option is A $8.39g$

The molecular weights of $HCl$ and $Mn{O}_2$ are $36.5$ g/mol and $86.9$ g/mol respectively.

$4HCl\left(aq\right)+Mn{O}_2\left(s\right)\to 2H_{2}O\left(l\right)+MnC{l}_2\left(aq\right)+C{l}_2\left(g\right)$

$4$ moles $(4\times 36.5=146g)$ of $HCl$ reacts with one mole ($86.9$ g) of $Mn{O}_2$.

Hence, $5.0$ g of $Mn{O}_2$ will react with $\frac{146}{86.9}\times 5=8.39g$ of $HCl$.