Answer:
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.
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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