Answer:
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.
