The equation below shows the decomposition of lead nitrate. How many grams of oxygen are produced when 11.5 g NO2 is formed? 2Pb (NO3) 2 (s) - - > 2PbO (s) + 4NO2 (g) + O2 (g

We can calculate for the number of moles of NO2 using its molar mass 46.0055 g/mol:

11.5 g NO2 (1 mol NO2 / 46.0055 g NO2) = 0.250 mol NO2

We use the mole ratio of O2 and NO2 from their coefficients in the balanced chemical equation

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

which is one mole O2 is to react with four moles of NO2, to compute for the number of moles of oxygen:

0.250 mol NO2 (1 mol O2 / 4 mol NO2) = 0.0625 mol O2

We can now calculate for the mass of oxygen O2 using its molar mass:

0.0625 mol O2 (31.998 g O2 / 1 mol O2) = 2.0 grams O2

Therefore, 2.0 grams of oxygen is produced.