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Anonymous

Feb 01, 2025

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

How much NO2 is produced when 43.9 g of ammonia combusts in the presence of excess oxygen, according to the reaction:

4 NH3 (g) + 7 O2 (g) → 4 NO2 (g) + 6 H2O (g)?

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2 Answers

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Dr. Angelina Mohr

Feb 21, 2025

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
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A
Anonymous

Feb 18, 2025

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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