Balanced equation:

[ 2Al + 3Br_2 \rightarrow 2AlBr_3 ]

Given:

[ \text{Al}=15.8,g,\qquad Br_2=55.6,g ]

Molar masses:

[ Al=27,g/mol,\quad Br_2=160,g/mol,\quad AlBr_3=267,g/mol ]

[ \text{Moles of Al}=\frac{15.8}{27}=0.585,mol ]

[ \text{Moles of }Br_2=\frac{55.6}{160}=0.3475,mol ]

From the equation, (2) mol Al require (3) mol (Br_2).

For (0.585) mol Al, bromine needed:

[ 0.585\times\frac{3}{2}=0.878,mol\ Br_2 ]

But only (0.3475) mol (Br_2) is available.

[ \boxed{Br_2\text{ (bromine) is the limiting reactant}} ]

[ 3Br_2 \rightarrow 2AlBr_3 ]

[ \text{Moles of }AlBr_3=0.3475\times\frac{2}{3}=0.2317,mol ]

[ \text{Mass of }AlBr_3=0.2317\times267 ]

[ \boxed{\text{Mass of }AlBr_3=61.9,g} ]

Answer:
(a) (\boxed{Br_2}) is the limiting reactant.
(b) (\boxed{61.9,g}) of (AlBr_3) is formed.