How many grams of O2 are needed to react with 84.5 g of NH3?
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Answer:
187.76 grams.
Explanation:
We first need to write the equation for the reaction as follows:
4NH3 + 5O2 — 4NO + 6H2O
We then calculate the moles of Ammonia in the reaction.
Molar Mass of ammonia = 14 + 4 × 1 = 18
Moles of ammonia = 84.5/18 = 4.694 moles.
The mole ratio of Ammonia to Oxygen is :
Ammonia : Oxygen = 4 : 5
The moles of Oxygen in the reaction is therefore :
5/4 × 4.694 = 5.8675 moles
Molar mass of oxygen = 16 × 2 = 32
Mass of oxygen = 32 × 5.8675 = 187.76 grams
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