Question

Chlorine gas (Cl₂) can be produced by reaction of hydrochloride acid with manganese oxide. What is the volume of chlorine gas is produced by adding 1 g of MnO₂ to one litre of 1 M HCl?

MnO₂ + 4HCl -> MnCl₂ + Cl₂ + 2H₂O

How much chlorine gas is produced when instead of 1 g MnO₂ this time 50 g MnO₂ is added?

Answer 1

MnO₂ + 4HCl -> MnCl₂ + Cl₂ + 2H₂O

no of moles of MnO2 = W/G.M.Wt

                                    = 1/87   = 0.0115 moles

no of moles of HCl   = molarity * volume in L

                               = 1*1 = 1moles

from balanced equation

4 moles of HCl react with 1 mole of MnO2

1 mole of HCl react with = 1*1/4 = 0.25 moles of MnO2

    mnO2 is limiting reactant

MnO₂ + 4HCl -> MnCl₂ + Cl₂ + 2H₂O

1 moles of MnO2 react with HCl to gives 1 mole of Cl2

1mole of MnO2 react with HCL to gives 22.4L of Cl2

0.0115moles of MnO2 react with HCl to gives = 22.4*0.0115/1   = 0.2576L of Cl2

no of moles of MnO2 = 50/87   = 0.57 moles

this time HCl is limiting reactant

4 moles of HCl react with MnO2 to gives 1 mole of Cl2

4 mole of HCl react with MnO2 to gives 71g of Cl2

1 moles of HCl react with MnO2 to gives = 71*1/4   = 17.75g of Cl2