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How many grams N2F4 can be produced from 225 g F,?​

1 Answer

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Answer:

308 g

Step-by-step explanation:

Data given:

mass of Fluorine (F₂) = 225 g

amount of N₂F₄ = ?

Solution:

First we look to the reaction in which Fluorine react with Nitrogen and make N₂F₄

Reaction:

2F₂ + N₂ -----------> N₂F₄

Now look at the reaction for mole ratio

2F₂ + N₂ -----------> N₂F₄

2 mole 1 mole

So it is 2:1 mole ratio of Fluorine to N₂F₄

As we Know

molar mass of F₂ = 2(19) = 38 g/mol

molar mass of N₂F₄ = 2(14) + 4(19) =

molar mass of N₂F₄ = 28 + 76 =104 g/mol

Now convert moles to gram

2F₂ + N₂ -----------> N₂F₄

2 mole (38 g/mol) 1 mole (104 g/mol)

76 g 104 g

So,

we come to know that 76 g of fluorine gives 104 g of N₂F₄ then how many grams of N₂F₄ will be produce by 225 grams of fluorine.

Apply unity formula

76 g of F₂ ≅ 104 g of N₂F₄

225 g of F₂ ≅ X of N₂F₄

Do cross multiplication

X of N₂F₄ = 104 g x 225 g / 76 g

X of N₂F₄ = 308 g

So,

308 g N₂F₄ can be produced from 225 g F₂

User Chris Mack
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