Answer:
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.
