MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g) How much MnO2(s) should be added to excess HCl(aq) to obtain 195 mL Cl2(g) at 25 °C and 765 Torr ?

Question

I am confused on how to get the mass of MnO2

Answer 1

Kaitlyn,

Before you can do any calculations, you must have a correctly balanced equation. This is a redox reaction which can be seen by the fact that Mn goes from 4+ on the left side to 2+ on the right side, and Cl goes from 1- on the left to zero on the right. So, first, let's balance this redox reaction.

MnO₂ --> Mn²⁺ ... reduction reaction

MnO₂ --> Mn²⁺ + 2H₂O ... balanced for oxygen

MnO₂ + 4H⁺ --> Mn²⁺ + 2H₂O ... balanced for hydrogen using acid (HCl)

MnO₂ + 4H⁺+ 2e- --> Mn²⁺ + 2H₂O ... balanced for charge and is the final balanced reduction equation

Cl^- --> Cl₂ ... oxidation reaction

2Cl^- --> Cl₂ ... balanced for Cl

2Cl^- --> Cl₂ + 2e- ... balanced for charge and is the final balanced oxidation equation

Overall balanced redox equation:

MnO₂ + 4H⁺+ 2e- + 2Cl^- --> Mn²⁺ + 2H₂O + Cl₂ + 2e-

Now, we use the conditions given and the ideal gas law to find the moles of Cl₂ desired, and then convert that to moles and grams of MnO₂ needed.

PV = nRT (use atm for P and liters for V and Kelvin for T)

n = PV/RT = (765 torr/760 torr/atm)(0.195 L) / (0.0821 Latm/Kmol)(298K)

n = 0.0080 mols Cl₂

From the mol ratio of the balanced equation, we now find mols of MnO₂ needed:

0.0080 mols Cl₂ x 1 mol MnO₂ / mol Cl₂ = 0.0080 mols MnO₂

mass MnO₂ needed = 0.0080 mols x 86.9 g/mol = 0.697 g MnO₂