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 --> MnCl2 (aq) + 2H2O (l) + Cl2 (g) How much MnO2(s) should be added to excess HCl(aq) to obtain 405 mL of Cl2(g) at 25 �C and 805 Torr?

Answer 1

Calculation of number of moles of Cl₂ obtained :

We know that ideal gas equation is PV = nRT

Where

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

P = pressure = 805 torr = 805x0.00131 atm = 1.054 atm

n = No . of moles = ?

R = gas constant = 0.0821 L atm / mol - K

V= Volume of the gas = 405 mL = 0.405 L

Plug the values we get n = (PV) / (RT)

                                      = 0.01746 moles

The balanced reaction is MnO₂(s) + 4HCl --> MnCl₂ (aq) + 2H₂O (l) + Cl₂ (g)

From the balanced reaction ,

1 mole of Cl₂ obtained by the reaction of 1 mole of MnO₂

0.01746 moles of Cl2 obtained by the reaction of 0.01746 moles of MnO₂

MOlar mass of MnO₂ = 54.9+(2x16) = 86.9 g/mol

So mass of MnO₂ is , m = number of moles x molar mass

                                    = 0.01746 mol x 86.9(g/mol)

                                   = 1.517 g

Therefore the mass of MnO₂ added is 1.517 g