The equation below shows the decomposition of lead nitrate. how many grams of oxygen are produced when 11.5 g no 2 is formed? 2pb(no 3 ) 2 ( s ) → 2pbo( s ) + 4no 2 ( g ) + o 2 ( g )

4.0 g of oxygen are produced. 

Explanation: 

The reaction is 
2Pb(NO₃)₂(s) → 2PbO(s) + 4NO₂(g) + O₂(g)

Moles (mol) = mass (g) / molar mass (g/mol)

Mass of NO₂ = 11.5 g 
Molar mass of NO₂ = 46.00 g/mol
Moles of NO₂ = 11.5 g / 46.00 g/mol
                      = 0.25 mol
                        

The stoichiometric ratio between NO₂ and O₂ is 2 : 1. 
Hence,
          moles of O₂ formed = moles of formed NO₂ / 2
                                            = 0.25 / 2
                                            = 0.125 mol

Molar mass of O₂ = 32 g/mol
Hence, mass of O₂ = moles x molar mass
                                 = 0.125 mol x 32 g/mol
                                 = 4.0 g

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