Question

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

Answer 1

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.