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Dmitrij [34]
3 years ago
13

Ammonia will react with fluorine to produce dinitrogen tetrafluoride and hydrogen fluoride (used in production of aluminum, in u

ranium processing, and in frosting of light bulbs). 2nh3(g) + 5f2(g) ? n2f4(g) + 6hf(g). how many moles of nh3 are needed to react completely with 13.6 mol of f2?
Chemistry
2 answers:
Digiron [165]3 years ago
5 0
2NH₃ + 5F₂ ---> N₂F₄ + 6HF
2 mol...5 mol

2 mol NH₃ --- 5 mol F₂
X mol NH₃ ---13,6 mol F₂
X = (2×13,6)/5
X = 5,44 mol NH₃
Svet_ta [14]3 years ago
5 0

Answer:

n_{NH_3}=5.44molNH_3

Explanation:

Hello,

At first, let the undergoing chemical reaction to be rewritten:

2NH_3(g)+5F_2(g)-->N_2F_4(g)+6HF(g)

Based on it, we apply the stoichiometric relationship between ammonia and fluorine to determine the completely reacting moles of ammonia, taking into account the 2 to 5 mole ratio between ammonia and fluorine respectively as shown below:

n_{NH_3}=13.6molF_2*\frac{2molNH_3}{5molF_2}=5.44molNH_3}}

Best regards.

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