How many grams of bromine, br2, are required to react completely with 37.4 grams of aluminum chloride, alcl3?

The reaction of bromine with aluminum chloride will bring about the formation of aluminum bromide and chlorine through a single-replacement reaction. This can be expressed in a balanced chemical reaction,

      3Br₂ + 2AlCl₃ --> 2AlBr₃ + 3Cl₂

This means that 3 moles of Br₂ which is equal to 159.1 g is required to react with 2 moles of AlCl₃ or equal to 266.68 g. Using this and the given above,

      mass of bromine = (37.4 grams AlCl3)(159.1 g of Bromine/266.68 g of AlCl3)
      Mass of bromine = 22.31 grams of Br₂

22.31 grams

127.57 moles

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

We have the final answer as

127.57 moles

Hope this helps you