Question

Chlorine is prepared in the laboratory by treating manganese dioxide $\left( {Mn{O_2}} \right)$ with aqueous hydrochloric acid according to the reaction.
$4HCl\left( {aq} \right) + Mn{O_2}\left( s \right) \to 2{H_2}O\left( l \right) + MnC{l_2}\left( {aq} \right) + C{l_2}$
How many grams of $HCl$ react with $5.0g$ of manganese dioxide?

Answer 1

When manganese oxide reacts with hydrochloric acid, it gives manganese chloride, water and chlorine gas. Here, in this question the reaction equation given is well balanced, thus we can calculate how many moles of hydrochloric acid react with how many moles of manganese dioxide, and subsequently it can be converted to grams.

Complete step by step answer:
According to the reaction given in question:
$4HCl\left( {aq} \right) + Mn{O_2}\left( s \right) \to 2{H_2}O\left( l \right) + MnC{l_2}\left( {aq} \right) + C{l_2}$, this equation depicts that $1$ mole of manganese oxide $\left( {Mn{O_2}} \right)$ reacts with $4$ moles of hydrochloric acid $\left( {HCl} \right)$ to form $1$mole of manganese chloride$\left( {MnC{l_2}} \right)$,$1$ mole of chlorine gas and $2$ moles of water.
We know that, molecular weight of $Mn{O_2}$ $= 87g/{\text{mol}}$:
$\Rightarrow 87g$ of $Mn{O_2}$ has $1$ mole
$1$ $g$ of $Mn{O_2}$ has $\dfrac{1}{87}$ mole
Therefore, $5g$ of $Mn{O_2}$ has $\dfrac{5}{{87}}$ moles $= 0.0574$ moles
From the reaction, we can say that
$1$ mole of $Mn{O_2}$ reacts with $4$ moles of $HCl$
$0.0574$ moles of $Mn{O_2}$ reacts with $\left( {4 \times 0.0574} \right)$ moles of $HCl$
$0.0574$ moles of $Mn{O_2}$ can be written as $5gm$ of $Mn{O_2}$.
So, $0.23$ moles of $HCl$ will react with $5gm$ of $Mn{O_2}$.
The molecular weight of $HCl$ $= 36.5g/{\text{mol}}$
Amount of $HCl$ in $0.23$ mole $= 0.23 \times 36.5 = 8.4g$ of $HCl$
Thus, it is clear that $8.4gm$ of $HCl$ will react with $5gm$ of $Mn{O_2}$.

Note:
Chlorine to be evolved on a large scale can also be produced by electrolysis of concentrated solution of sodium chloride in water. By electrolysis of fused sodium chloride, which also produces metallic sodium, chlorine is again evolved at the anode.