Engineering
Chemistry
Kinetic Theory of Gases KTG
Law of Chemical Equivalence
Question

In an experiment of analysis of oxalate salt, a solution of oxalate gives white precipitate with CaCl2 solution. This precipitate is dissolved in dilute H2SO4 and decolourises KMnO4 (acidified) solution as the following reactions.
Na2C2O4 + CaCl2 → CaC2O4↓ + 2NaCl
                                     white

CaC2O4 + H2SO4 → CaSO4 + H2C2O4

2KMnO4 + 3H2SO4 + 5H2C2O4 → 2MnSO4 + K2SO4 + 8H2O + 10CO2

If one mol of Na2C2O4 is taken and reacted with sufficient amount of other reactants such as to give 100% reaction yield at each step and the liberated CO2 is kept in a container of  volume 'V' at pressure 'P' , temperature 'T'. Then Urms of CO2 gas in container will be
 

3PV2MCO2

3RTMCO2

3PVMCO2

3RT2MCO2

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Solution

1 mol Na2C2O4 → 1 mol CaC2O4 [from reaction (i)]
1 mol CaC2O4 → 1 mol H2C2O4 [from reaction (ii)]
1 mol H2C2O4 → 2 mol CO2 [from reaction (iii)]
   ⇒ nCO2 = 2
    Urms  = 3RTM=3PVnM
    For CO2 gas 
    Urms = 3RTMCO2=3PV2MCO2