How many grams of bromine are required to react completely with 37.4 grams aluminum chloride?
To determine the number of grams of bromine required to react completely with aluminum chloride, we need to balance the chemical equation of the reaction between bromine and aluminum chloride. Let's assume the formula for aluminum chloride is AlCl3.
The balanced chemical equation for the reaction is:
2 AlCl3 + 3 Br2 → 2 AlBr3 + 3 Cl2
From the balanced equation, we can see that 2 moles of aluminum chloride react with 3 moles of bromine to produce 2 moles of aluminum bromide and 3 moles of chlorine.
To find the number of moles of aluminum chloride, we divide its mass by its molar mass:
37.4 grams AlCl3 / (26.98 g/mol + (35.45 g/mol × 3)) = 37.4 g / 133.83 g/mol ≈ 0.2796 moles AlCl3
According to the stoichiometry of the balanced equation, 2 moles of aluminum chloride react with 3 moles of bromine. Therefore, we can set up a ratio:
2 moles AlCl3 : 3 moles Br2 = 0.2796 moles AlCl3 : x
By cross-multiplying, we get:
2 moles AlCl3 * x = 3 moles Br2 * 0.2796 moles AlCl3
2x = 0.8388
x = 0.8388 / 2
x ≈ 0.4194 moles Br2
Now we can determine the mass of bromine required by multiplying the number of moles by its molar mass:
Mass of bromine = 0.4194 moles Br2 × 159.81 g/mol = 66.98 grams
Therefore, approximately 66.98 grams of bromine are required to react completely with 37.4 grams of aluminum chloride.
To determine how many grams of bromine are required to react completely with 37.4 grams of aluminum chloride, you need to write a balanced chemical equation for the reaction. The balanced equation for the reaction between bromine and aluminum chloride is:
2 AlCl₃ + 3 Br₂ → 2 AlBr₃ + 3 Cl₂
From the balanced equation, we can determine the stoichiometric ratio between bromine and aluminum chloride, which is 3 moles of Br₂ to 2 moles of AlCl₃.
Now, we'll calculate the number of moles of aluminum chloride:
moles of AlCl₃ = mass / molar mass
moles of AlCl₃ = 37.4 g / (26.98 g/mol + (35.45 g/mol * 3))
moles of AlCl₃ = 37.4 g / 133.32 g/mol
moles of AlCl₃ = 0.2801 mol
According to the stoichiometric ratio, we need 3 moles of bromine for every 2 moles of aluminum chloride. So, we can set up a proportion to find the number of moles of bromine required:
2 mol AlCl₃ / 3 mol Br₂ = 0.2801 mol AlCl₃ / x mol Br₂
Cross-multiplying the proportion:
2x = 0.2801 * 3
2x = 0.8403
x = 0.8403 / 2
x = 0.4202 mol Br₂
Finally, we'll calculate the grams of bromine required:
mass of Br₂ = moles of Br₂ * molar mass
mass of Br₂ = 0.4202 mol * (79.904 g/mol)
mass of Br₂ = 33.61 g
Therefore, approximately 33.61 grams of bromine are required to react completely with 37.4 grams of aluminum chloride.
Actually, the reaction you're proposing will not occur; however, here is how you work the problem.
2AlCl3 + 3Br2 ==> 2AlBr3 + 3Cl2
mols AlCl3 = grams/molar mass
Convert mols AlCl3 to mols Br2 using the coefficients in the balanced equation.
Convert mols Br2 to g. g = mols x molar mass.