Question

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.

Answer 1

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