ChemistryEquivalent weight of Pb(NO 3 ) 2 in the equation 2 Pb ( NO 3 ) 2 → ∆ 2 PbO + 4 NO 2 + O 2 is

Equivalent weight of  Pb(NO3)2 in the equation

2Pb(NO3)2→∆ 2PbO+4NO2+O2 is

  1. A
    M2
  2. B
    M3
  3. C
    M4
  4. D
    2M3

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    Solution:

    2Pb(N+5O-23)2 â†’∆ 2PbO+4N+4O2 +O02

    'N' atom Undergoes reduction

    'O' atom Undergoes Oxidation

    Total increase in Ox. state of 'N' per 2 moles of lead nitrate = 20-16=4

    EW of Pb(NO3)2 = 2M/4=M/2

    Alternate method:

                     Total decrease in Ox. state of Oxygen per 2 moles of lead nitrate = [-20-(-24)]=4

    EW of Pb(NO3)2 = 2M/4=M/2

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