How to balance this equation and find the mass...

100.0 grams of Hydrobromic acid reacts with 50.0 grams of Chlorine gas, in a single replacements, exothermic reaction.

2 HBr + Cl2 → 2 HCl + Br2

(100.0 g HBr) / (80.9120 g HBr/mol) = 1.23591 mol HBr
( 50.0 g Cl2) / (70.9064 g Cl2/mol) = 0.70515 mol Cl2

0.70515 mole of Cl2 would react completely with 0.70515 x (2/1) = 1.4103 mol HBr but there is not that much HBr present, so HBr is the limiting reactant.

(1.23591 mol HBr) x (2/2) x (36.4611 g HCl/ mol) = 45.1 g HCl
(1.23591 mol HBr) x (1/2) x (159.8082 g Br2/mol) = 98.8 g Br2

Choose your answer.

2 HBr(aq) + Cl2(g) --> 2 HCl(aq) + Br2(g)
Moles of HBr = 100 / 81 = 1.235 moles.
Moles of Br2 = 50 /71 = 0.704 moles
One mole of Cl2 requires 2 moles of HBr = 2 x 1.235 = 1.408 , but we have only 1.235 moles, so that makes the Cl2 too much or in excess. Therefore, the remaining calculations will depend on the moles of HBr.
You did not say the mass of what. If it is HCl then the moles of HCl produced would be 0.704 moles.
Grams of HCl = 0.704 x 36.5 = 25.70

2 HBr(g) + Cl2(g) -----------> Br2(l) + 2 HCl(g)
? g Br2 = 100.0 g HBr x 1 mol HBr/80.91 g x 1 mol Br2 / 2 mol HBr x 159.8 g Br2/mole Br2 = 98.75 g Br2