Answer to Question #267261 in General Chemistry for jmtz

Question

Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq)

HCl(aq), as described by the chemical equation

MnO

2

(s)+4HCl(aq)⟶MnCl

2

(aq)+2H

2

O(l)+Cl

2

(g)

MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g)

How much MnO

2

(s)

MnO2(s) should be added to excess HCl(aq)

HCl(aq) to obtain 175 mL Cl

2

(g)

175 mL Cl2(g) at 25 °C and 765 Torr

Answer 1

From the equation and the statement "excess HCl" means MnO2 is the limiting reagent.

But we have to get the mass of Cl2 produced since we were given in volume.

But to do that we have to get the no of mole in the formula: PV=nRT

n=PV/RT

n=(765×0.175)/(62.36×298)

n=0.007

Mass=mole × molar mass

Mass=0.007×71

Mass=0.5g

If 87g of MnO2 produces 71g of Cl2

Then Xg of ........................... 0.5 of Cl2

X=0.61g

Therefore 0.61g of MnO2 will be very added to to excess HCl to give 175mL of Cl2