# 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

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
The answer would be 118.68 g.
Explanation for this is:4 moles of NH3 give 4 moles of NO2
so 1mole of NH3 will give 1 mole of NO2
43.9 grams of NH3 contains 2.58 moles
so 2.58 moles will be produced of NO2
which is 118.7 grams this the amount of oxygen that is used.

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