how many grams of bromine, br2, are required to react completely with 37.4 grams of aluminum chloride, AlCl3?

1 answer:

8 0

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

<em>Answer: 22.31 grams</em>

You might be interested in

How many moles of aspartame are present in 1.00 mg of aspartame?

Diano4ka-milaya [45]

Answer:- There are $3.40*10^-^6$ moles.

Solution:- It is a unit conversion problem where we are asked to convert mg of aspartame to moles. Aspartame is $C_1_4H_1_8N_2O_5$ and it's molar mass is 294.31 grams per mole.