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)⟶MnCl2(aq)+2H2O(l)+Cl2(g)

How much MnO2(s) should be added to excess HCl(aq) to obtain 105 mL Cl2(g) at 25 °C and 795 Torr ?

Answer 1

According to the question Volume of Cl₂ = V = 105 mL = 0.105 L

Temperature = T = 25 C = 273 + 25 = 298 K

Pressure P = 795 torr = 795 / 760 = 1.04 atm

Now we Calculate the number of Moles of Cl₂

We know that

PV = n RT

n = PV / RT

=> n = ( 1.04 * 0.105 ) / ( 0.082 * 298 )

=> n = 4.4*10^-3

MnO₂ + 4HCl ⟶ MnCl2 + 2H₂O + Cl₂

Strochiometrically ,

1 moles Cl₂ produce form 1 mole MnO₂

So 4.4*10^-3  moles Cl₂ produce form 4.4*10^-3  mole MnO₂

Molecular Mass of MnO₂ = 87 g /mol

So Mass of MnO₂ = Moles * Molecular Mass

=> Mass of MnO₂ = 4.4*10^-3 * 87

=> Mass of MnO₂ = 0.3828 g

So The 0.3828 g of MnO₂ should be added

Thank you

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