Respuesta :
The amount of dinitrogen tetrafluoride produced from 45.0 g of fluorine, and 22.0 g of ammonia is
The balanced reaction is given as:
2NH₃ + 5F₂ → N₂F₄ + 6HF
In a chemical reaction involving more than one reactants, the reactant which is consumed more is the limiting reactant.
So, clearly here the limiting reactant is F₂.
Number of moles of F₂ = 45 / 38 = 1.18 moles
So, now lets find out the number of moles of N₂F₄.
Number of moles of N₂F₄ = (1.18 moles of F₂) × (1 mol N₂F₄ / 5 mol F₂)
= 0.236 moles of N₂F₄
The mass of the N₂F₄ is,
Mass of N₂F₄ = Moles of N₂F₄ × Molar mass of N₂F₄
= 0.236 × 104 g
= 24.544 g
Hence, the mass of dinitrogen tetrafluoride is 24.544 g.
Learn more about dinitrogen tetrafluoride from the link given below.
https://brainly.com/question/2562772
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