How many grams of bromine are required to react completely with 3704 grams aluminum chloride?
To determine the number of grams of bromine required to react completely with 3704 grams of aluminum chloride, we first need to determine the molar ratio between aluminum chloride (AlCl3) and bromine (Br2). The balanced chemical equation for the reaction between aluminum chloride and bromine is:
2AlCl3 + 3Br2 → 2AlBr3 + 3Cl2
From the balanced equation, we can see that 2 moles of aluminum chloride react with 3 moles of bromine.
Step 1: Convert grams of aluminum chloride to moles:
To do this, we need the molar mass of aluminum chloride (AlCl3). The molar mass of AlCl3 can be calculated by adding the atomic masses of aluminum (Al) and three times the atomic mass of chlorine (Cl).
Molar mass of Al = 26.98 g/mol
Molar mass of Cl = 35.45 g/mol
Molar mass of AlCl3 = (26.98 g/mol) + 3 × (35.45 g/mol) = 133.34 g/mol
Now, we can convert the grams of aluminum chloride to moles using the following equation:
moles = mass / molar mass
moles of AlCl3 = 3704 g / 133.34 g/mol ≈ 27.77 moles
Step 2: Determine the mole ratio between AlCl3 and Br2:
From the balanced chemical equation, we know that 2 moles of AlCl3 react with 3 moles of Br2.
Step 3: Convert moles of AlCl3 to moles of Br2:
Using the mole ratio, we can calculate the moles of bromine required:
moles of Br2 = (moles of AlCl3) × (3 moles of Br2 / 2 moles of AlCl3)
moles of Br2 ≈ 27.77 moles × (3/2)
moles of Br2 ≈ 41.65 moles
Step 4: Convert moles of Br2 to grams:
Finally, to convert moles of bromine to grams, we use the molar mass of bromine (Br2), which is equal to 159.808 g/mol.
grams of Br2 = (moles of Br2) × (molar mass of Br2)
grams of Br2 ≈ 41.65 moles × 159.808 g/mol
grams of Br2 ≈ 6652.34 grams
Therefore, approximately 6652.34 grams of bromine are required to react completely with 3704 grams of aluminum chloride.