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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.
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The combustion of ammonia in presence of excess oxygen yields NO2 and H2O.The molar mass of ammonia is 17.02 g/molTherefore, moles of ammonia in 43.9 g          = 43.9 /17.02          = 2.579 molesFrom the equation the mole ratio of ammonia to nitrogen iv oxide is 4:4The molar mass of NO2 is 46 g/molThe number of moles of NO2 is the same as that of ammonia since they have equal ratio, = 2.579 molesTherefore, mass of NO2   = 2.579 moles ×46   = 118.634 g   ≈ 119 g... Show More
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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.... Show More
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