Aluminum fluoride reacts with oxygen gas according to the following BALANCED equation. 4 AlF3 + 3 O2 →2 Al2O3 + 6 F2 If 219 g of aluminum fluoride react with 48 g of oxygen gas, how many grams of fluorine gas will be produced?

Question

contestada

Respuesta :

anfabba15 anfabba15 · Answer 1 · 2019-12-11T00:14:44+01:00

Answer:

114 grams of fluorine gas will be produced

Explanation:

4 AlF₃ + 3O₂ → 2 Al₂O₃ + 6 F₂

First of all, to know the mass of any product, we have to find out the limiting reactant unless the statement says, that one is been in excess.

Moles = Mass / Molar mass

Moles AlF₃ = Mass AlF₃ / Molar Mass AlF₃

Moles AlF₃ = 219 g / 83.97 g/m = 2.60 moles

Moles O₂ = Mass O₂ / Molar mass O₂

Moles O₂ = 48g / 32g/m = 1.5 moles

By stoichiometry, 4 moles of fluoride reacts with 3 moles of O₂.

If I have 2.60 moles, how many moles of oxygen, do i need.

4 moles AlF₃ _____ 3 O₂

2.60 m AlF₃ _____ (2.6 .3) / 4 = 1.95 moles

I need 1.95 moles of O₂, but I only have 1.5 moles, so the O₂ is the limiting.

3 moles of O₂ ___ produces ___ 6 moles of F₂ (the double)

My 1.5 moles O₂ ___ will produce (1.5 .6) / 3 = 3 moles

Molar mass F₂ = 38 g/m

3 moles of F₂ are 3 m. 38g/m = 114 g