The combustion of ammonia in the presence of excess oxygen yields no2 and h2o: 4 nh3 (g) + 7 o2 (g) → 4 no2 (g) + 6 h2o (g) the combustion of 43.9 g of ammonia produces ________ g of no2.
The combustion of ammonia in presence of excess oxygen yields NO2 and H2O. The molar mass of ammonia is 17.02 g/mol Therefore, moles of ammonia in 43.9 g = 43.9 /17.02 = 2.579 moles From the equation the mole ratio of ammonia to nitrogen iv oxide is 4:4 The molar mass of NO2 is 46 g/mol The number of moles of NO2 is the same as that of ammonia since they have equal ratio, = 2.579 moles Therefore, mass of NO2 = 2.579 moles ×46 = 118.634 g ≈ 119 g
