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
