Answer to Question #114243 in General Chemistry for Daniel

Question

Using the reaction 2 AlBr3 + 3 Cl2 = 2 AlCl3 + 3 Br2, how many grams of Br2 will be produced if 6.785 g of aluminum bromide were reacted with 3.955 g of chlorine? What is the limiting reagent
in this reaction?

Answer 1

n(AlBr3)=m(AlBr3)/(2*Mr(AlBr3))=6.785/(2*267)=0.0127 mol;

n(Br2)=m(Br2)/(3*Mr(Br2))=3.955/(3*71)=0.0186 mol;

AlBr3 - limiting reactant;

m(Br2)=m(AlBr3)*3*Mr(Br2)/(2*Mr(AlBr3))= 6,785*3*160/(2*267)=6.1 g;

Answer: the limiting reagent in this reaction is aluminum bromide ;

6.1 grams of Br2 are produced