if an excess of Al and 7.9 mol of Br2 are reacted in equation 2Al + 3Br2= 2AlBr2, how many moles of AlBr2 will be formed assuming a 100% yield

Answer 1

Tiffany, Assuming the reaction runs to completion and there is an excess of Al you know that Bromine is your limiting reagent.

Stoichiometry will give you the amount of AlBr2 produced.

You begin with 7.9 mol Br2 to cancel the moles of Bromine and find moles of Aluminum Bromide you must multiply by the mole ratio found in the balanced chemical equation:

7.9mol Br2 x (2molAlBr2/3molBr2) = ___ mol AlBr2

This yields 5.3 moles Aluminum Bromide.

Answer 2

I can't help but wonder if that AlBr2 is a typo meant to be AlBr3.