How many grams of Br2 are needed to form 66.1 g of AlBr3

Question

66.1 g AlBr3 write in conversion to calculate the grams of Br2

Answer 1

2Al + 3Br₂ ==> 2AlBr₃ ... balanced equation

66.1 g AlBr₃ x 1 mol AlBr₃/267 g x 3 mol Br₂/2 mol AlBr₃ x 160 g/mol Br₂ = 59.4 g Br₂ (3 sig. figs.)

Answer 2

The first step for stoichiometry reactions is to write the balanced chemical equation.

We know that we want AlBr₃ --> Br₂

2AlBr₃ --> 2Al + 3Br₂ (*note that it doesn't really matter the phase of the Al here since the problem does not ask for it)

Now, we can set up the stoichiometry problem. For these types of problems, I like to write out all the unit conversions and then add in the numbers. See work and answers here: https://drive.google.com/file/d/1V9hMa0rF89m-0u4XRxgRZltKHGZ1wcPk/view?usp=sharing