In a flask, 10.3 g of aluminum reacted with 100.0 g of liquid bromine to form aluminum bromide. After the reaction, no aluminum
remained, and 8.5 grams of bromine remained unreacted. How many
grams of bromine reacted? How many grams of compound were
formed?

91.56grams of bromine reacted and 101.86 grams of Aluminium bromide reacted.

Balanced chemical equation for the reaction between aluminum and liquid bromide is given below:

2Al + 3Br2 —> 2AlBr3

Determining masses of Al and Br2

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.

Mass of bromine, Br2 that reacted.

54 g of Al reacted with 480 g of Br2, 10.3 g of Al will react with = (10.3 × 480)/54 = 91.56 g of Br2.

Therefore, 91.56 g of Br2 participated in the reaction.

Mass of the compound, AlBr3 produced from the reaction. 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:

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 is produced in the reaction.

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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.

more info ? ? i don't understand what the question is

explanation:

- aluminum and copper chloride.

- the chloride ion to separate the aluminum from the oxygen so that the aluminum can react with the copper ions (and the water molecules).