How many liters of bromine are needed to produce 12 miles of aluminum bromide?

0.929L of Br2

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2Al + 3Br2 —> 2AlBr3

Next, we shall determine the number of mole of Br2 needed to produce 12 moles of AlBr3. This is illustrated below:

From the balanced equation above,

3 moles of Br2 reacted to produce 2 moles of AlBr3.

Therefore, Xmol of Br2 will produce 12 moles of AlBr3 i.e

Xmol of Br2 = (3 x 12)/2

Xmol of Br2 = 18 moles

Therefore, 18 moles of Br2 reacted to produce 12 moles of AlBr3.

Next, we shall convert 18 moles of Br2 to grams. This is illustrated below:

Number of mole Br2 = 18 moles

Molar mass of Br2 = 2 x 80 = 160g/mol

Mass of Br2 =.?

Mass = mole x molar mass

Mass of Br2 = 18 x 160 = 2880g

Next, we determine the volume of Br2 used in the reaction as follow:

Mass of Br2 = 2880g

Density of Br2 = 3.1g/mL

Volume of Br2 =..?

Density = Mass /volume

Volume = Mass /Density

Volume of Br2 = 2880/3.1

Volume of Br2 = 929mL

Finally, we shall convert 929mL to L.

1000mL = 1L

Therefore, 929mL = 929/1000 = 0.929L

Therefore, 0.929L of Br2 were used in the reaction.

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