How many moles of Al are necessary to form 23.6 g of AlBr₃ from this reaction: 2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s) ?

0.085 moles of Al are required to form 23.6 g of AlBr₃.

Let's consider the following balanced equation for the synthesis reaction of AlBr₃.

2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s)

First, we will convert 23.6 g to moles using the molar mass of AlBr₃ (266.69 g/mol).

The molar ratio of Al to AlBr₃ is 2:2. The moles of Al required to form 0.0885  moles of AlBr₃ are:

0.085 moles of Al are required to form 23.6 g of AlBr₃.

You can learn more about stoichiometry here: link

0.088 mole of Al.

Explanation:

First, we shall determine the number of mole in 23.6 g of AlBr₃.

This is illustrated below:

Mass of AlBr₃ = 23.6 g

Molar Mass of AlBr₃ = 27 + 3(80) = 267 g/mol

Mole of AlBr₃ =.?

Mole = mass/Molar mass

Mole of AlBr₃ = 23.6 / 267

Mole of AlBr₃ = 0.088 mol

Next, we shall writing the balanced equation for the reaction.

This is given below:

2Al(s) + 3Br₂(l) → 2AlBr₃(s)

From the balanced equation above,

2 moles of Al reacted with 3 mole of Br₂ to 2 moles AlBr₃.

Finally, we shall determine the number of mole of Al needed for the reaction as follow:

From the balanced equation above,

2 moles of Al reacted to 2 moles AlBr₃.

Therefore, 0.088 mole of Al will also react to produce 0.088 mole of AlBr₃.

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