Respuesta :
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
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:
[tex]n_{NH_3}=5.44molNH_3[/tex]
Explanation:
Hello,
At first, let the undergoing chemical reaction to be rewritten:
[tex]2NH_3(g)+5F_2(g)-->N_2F_4(g)+6HF(g)[/tex]
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:
[tex]n_{NH_3}=13.6molF_2*\frac{2molNH_3}{5molF_2}=5.44molNH_3[/tex]}}
Best regards.