Answer:
1. 91.56 g of Br2.
2. 101.86 g of AlBr3
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2Al + 3Br2 —> 2AlBr3
Next, we shall determine the masses of Al and Br2 that reacted and the mass of AlBr3 that is produced from the balanced equation. This is illustrated below:
Molar mass of Al = 27 g/mol
Mass of Al from the balanced equation = 2 × 27 = 54 g
Molar mass of Br2 = 2 × 80 = 160 g/mol
Mass of Br2 from the balanced equation = 3 × 160 = 480 g
Molar mass of AlBr3 = 27 + (3 × 80)
= 27 + 240
= 267 g/mol
Mass of AlBr3 from the balanced equation = 2 × 267 = 534 g.
From the balanced equation above,
54 g of Al reacted with 480 g of Br2 to produce 534 g of AlBr3.
1. Determination of the mass of bromine, Br2 that reacted.
This can be obtained as follow:
From the balanced equation above,
54 g of Al reacted with 480 g of Br2.
Therefore, 10.3 g of Al will react with = (10.3 × 480)/54 = 91.56 g of Br2.
Therefore, 91.56 g of Br2 took part in the reaction.
2. Determination of the mass of the compound, AlBr3 produced from the reaction.
The can be obtained as follow:
Since no amount of Al is remaining, Al is the limiting reactant. Thus, we can obtain the mass of AlBr3 produced from the reaction as follow:
From the balanced equation above,
54 g of Al reacted to produce 534 g of AlBr3.
Therefore, 10.3 g of Al will react to produce = (10.3 × 534)/54 = 101.86 g of AlBr3.
Thus, 101.86 g of AlBr3 were produced from the reaction.