How many grams of bromine are required to react completely with 11.95 grams aluminum chloride? alcl3+br2--> albr3+cl2

Molar mass:
 
( Br₂ ) = 159.80 g/mol

AlCl₃ = 133.34 g/mol

2 AlCl₃ + 3 Br₂ = 2 AlBr₃ + 3 Cl₂

2 x ( 133.34) g 3 x ( 159.80) g
11.95 ( mass Bromine )

mass Bromine = 11.95 x 3 x 159.80 / 2 x 133.34

mass Bromine = 5728.83 / 266.68

mass Bromine = 21.48 g

hope this helps!

First have to balance the equation of  AlCl3 + Br2 ---> Br3 + Cl2:

                                  2AlCl3 + 3Br3> 2AlBr3 + 2Cl2

                           1 mol AlCl3           3mol Br2         159.8 g Br2                         11.95g AlCl3   x        x        x   =  21.48g Br2
                          133.33 g AlCl3      2 mol AlCl3       1 mol Br2

(133.33 g = molar mass of AlCl3       3:2 ratio   
159.8 g = molar mass of Br2)

21.48 g Br2 required 

North atlantic current or north atlantic drift is the answer