Question

1. Balance the following reaction:

Al(s) + F₂ ---> AlF₃

How many grams of aluminum fluoride would be produced from 40.84 g of aluminum (use the balanced equation above)? The molar masses are: Al, 26.981; F, 18.998, g per mole, respectively. Leave the answer as a fraction but state how many significant figures it has.

(PS: Please show all the works! Thank you!)

Answer 1

The balanced chemical equation of the recation taking place between Aluminum anf Fluorine is

    2 Al (s) + 3 F₂ (g) = 2 AlF₃ (s)

Molar Mass of Aluminum is 26.981 g/mol

Molar Mass of AlF3 is [26.981 + (3*18.998)] g/mol = 83.975 g/mol

From the balanced equation 26.981 g of Aluminum gives 83.975 g of Aluminum Fluoride.

So, 40.84 g of Aluminum will give

      [40.84 * (83.975/26.981)] g = 127.11 g of Aluminum Fluoride