Question

Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq), as described by the chemical equation MnO2(s)+4HCl(aq)⟶MnCl(aq)+2H2O(l)+Cl2(g)

How much MnO2(s) should be added to excess HCl(aq)to obtain 185 mL Cl2(g)at 25 °C and 795 Torr?

Answer 1

We first find the number of moles of $Cl_2$ to be obtained.

$PV=nRT$

$P=$ $795Tore\times 1atm\over760 Torr$ $=1.046Torr$

$V=0.185L$

$n=x$ $Mol$

$R=0.0821Latm/Kmol(Constant)$

$T=298K$

$n=$ $moles=$ $PV\over RT$ $=$ $(1.046)(0.185)\over (0.0821)(298)$ $=0.00791moles Cl_2$

Moles and grams of $MnO_2$ needed from the stoichometry equation are;

$0.00791mol Cl_2$ $\times$ $1mol MnO_2\over 1mol Cl_2$ $=0.687gMnO_2$