Question

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

Answer 1

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

Answer 2

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.