During the electrolysis of molten aluminum chloride in an electrolytic cell, 5.40 g of aluminum is produced at the cathode. Predict the mass of chlorine produced at the anode.
Solutions of this question
To predict the mass of chlorine produced at the anode during the electrolysis of molten aluminum chloride, we can use the principle of conservation of mass.
The balanced equation for the electrolysis of aluminum chloride is:
2AlCl3(l) -> 2Al(l) + 3Cl2(g)
From the balanced equation, we can see that for every 2 moles of aluminum produced at the cathode, 3 moles of chlorine gas will be produced at the anode.
Step 1: Calculate the moles of aluminum produced at the cathode.
Given mass of aluminum = 5.40 g
Molar mass of aluminum (Al) = 26.98 g/mol
Number of moles of aluminum = mass/molar mass
Number of moles of aluminum = 5.40 g / 26.98 g/mol
Step 2: Use the mole ratio to find the moles of chlorine.
According to the balanced equation, the mole ratio of aluminum to chlorine is 2:3.
Number of moles of chlorine = (Number of moles of aluminum) * (3/2)
Step 3: Calculate the mass of chlorine.
Molar mass of chlorine (Cl2) = 70.90 g/mol
Mass of chlorine = number of moles * molar mass of chlorine
Mass of chlorine = (Number of moles of chlorine) * 70.90 g/mol
Now we can substitute the values into the equation.
Number of moles of aluminum = 5.40 g / 26.98 g/mol
Number of moles of chlorine = (5.40 g / 26.98 g/mol) * (3/2)
Mass of chlorine = (5.40 g / 26.98 g/mol) * (3/2) * 70.90 g/mol
Calculating the equation above will give us the mass of chlorine produced at the anode during the electrolysis of molten aluminum chloride.
To predict the mass of chlorine produced at the anode during the electrolysis of molten aluminum chloride, we need to understand the stoichiometry of the reaction.
The balanced chemical equation for the electrolysis of aluminum chloride is:
2 AlCl3 → 2 Al + 3 Cl2
From the equation, we can see that for every 2 moles of aluminum chloride that react, 3 moles of chlorine are produced.
Now, let's calculate the moles of aluminum produced at the cathode using the given mass of 5.40 g.
First, we need to determine the molar mass of aluminum. The molar mass of aluminum is 26.98 g/mol.
Moles of aluminum = mass of aluminum / molar mass of aluminum
= 5.40 g / 26.98 g/mol
= 0.200 mol
According to the balanced equation, the ratio of moles of aluminum to moles of chlorine is 2:3. Therefore, the moles of chlorine produced will be:
Moles of chlorine = (2/3) × moles of aluminum
= (2/3) × 0.200 mol
= 0.133 mol
To calculate the mass of chlorine produced, we need to know the molar mass of chlorine. The molar mass of chlorine is 35.45 g/mol.
Mass of chlorine = moles of chlorine × molar mass of chlorine
= 0.133 mol × 35.45 g/mol
= 4.71 g
Therefore, the predicted mass of chlorine produced at the anode during the electrolysis of molten aluminum chloride is 4.71 grams.
This is just a stoichiometry problem. You don't need anything about electrolysis.
2AlCl3 ==> 2Al + 3Cl2
Convert g Al to mols.
Convert mols Al to mols Cl2.
Convert mols Cl2 to grams.
Check my thinking.