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Based on the equation, how many grams of Br2 are required to react completely with 42.3 grams of AlCl3?

2AlCl3 + 3Br2 → 2AlBr3 + 3Cl2 (5 points)

66.5 grams

71.2 grams

76.1 grams

80.2 grams

The answer to your question is 76.1 g of Br₂

Explanation:

Data

mass of Br₂ = ?

mass of AlCl₃ = 42.3 g

Balanced chemical reaction

               2AlCl₃  +  3Br₂   ⇒   2AlBr₃  + 3Cl₂

Process

1.- Calculate the molar mass of   2AlCl₃  and  3Br₂

2AlCl₃  = 2[ 27 + (35.5 x 3)] = 267 g

3Br₂ = 80 x 6 = 480 g

2.- Use proportions to find the mass of Br₂ needed

                  267 g of AlCl₃ 480 g Br₂

                   42.3 g of AlCl₃  x

                   x = (42.3 x 480)/267

                 x = 76.04 g of Br₂ ≈ 76.1 g

i believe it is 40–40–20%

explanation: