Answer to Question #167070 in General Chemistry for Nada

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;

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