Question

38. When nitrogen dioxide (NO2) from car exhaust combines with water in the air it forms nitric acid (HNO3), which causes acid rain, and nitrogen oxide. 3NO2(g) + H2O (l)  2HNO3 (aq) + NO(g) (a) How many molecules of NO2 are needed to react with 0.250 mol H2O? (b) How many grams of HNO3 are produced when 60.0 g NO2 completely react? (c) How many grams of HNO3 can be produced if 225 g NO2 is mixed with 55.2 g H2O?

Answer 1

(a) How many molecules of NO2 are needed to react with 0.250 mol H2O?

3NO2:1H2O then 0.25*3 = 0.75 mol are needed

(b) How many grams of HNO3 are produced when 60.0 g NO2 completely react?

mol NO2= mass/MW = 60/46.0055 = 1.30419189

then since ratio is 3NO2: 2HNO3, then 1.30419189*2/3 = 0.86946126 mol of HNO3

mass = mol*MW = 0.86946126*63.01 = 54.7847 g of HNO3

(c) How many grams of HNO3 can be produced if 225 g NO2 is mixed with 55.2 g H2O?

mol NO2 = mass/MW = 225/46.0055 = 4.890719 mol of NO2

mol of H2O = masS/MW = 55.2/18 = 3.0666

limiting reactant is water so

3.066 mol of H2O cna produce 1:2 2*3.06 = 6.12 mol of HNO3

mass = mol*MW = 6.12*63.01 = 385.6212 g of HNO3