How many moles of Al are necessary to form 28.6 g of AlBr₃ from this reaction: 2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s) ?

The number of moles of Al that are necessary to form 28.6 g of AlBr₃ from

this reaction: 2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s) is 0.107 moles.

The balanced equation is as follows;

2Al(s) + 3Br₂(l) → 2AlBr₃(s)

molar mass of aluminium = 27 g

molar mass of Aluminium bromide = 27 + 80(3) = 27 + 240 = 267 g

2(267)g of aluminium bromide needs 2(27)g of aluminium

28.6 g of aluminium bromide will need ? of aluminium

mass of aluminium = 54 × 28.6 / 534

mass of aluminium = 1544.4  / 534

mass of aluminium = 2.89213483146

mass of aluminium ≈ 2.89 g

moles of aluminium required = mass /molar mass

moles of aluminium required = 2.89 / 27

moles of aluminium required = 0.10711610486

moles of aluminium required ≈ 0.107 moles

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0.107 mole of Al.

Explanation:

We'll begin by calculating the number of mole in 28.6 g of AlBr₃. This can be obtained as illustrated below:

Mass of AlBr₃ = 28.6 g

Molar mass of AlBr₃ = 27 + (3×80)

= 27 + 240

= 267 g/mol

Mole of AlBr₃ =?

Mole = mass /Molar mass

Mole of AlBr₃ = 28.6 / 267

Mole of AlBr₃ = 0.107 mole

Next, we shall write the balanced equation for the reaction. This is shown below:

2Al(s) + 3Br₂(l) —> 2AlBr₃(s)

From the balanced equation above,

2 mole of Al reacted to produce 2 moles of AlBr₃.

Finally, we shall determine the number of mole of Al needed to produce 28.6 g (i.e 0.107 mole) of AlBr₃.

This can be obtained as illustrated below:

From the balanced equation above,

2 mole of Al reacted to produce 2 moles of AlBr₃.

Therefore, 0.107 mole of Al will also react to produce 0.107 mole of AlBr₃.

Thus, 0.107 mole of Al is needed for the reaction.

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