The balanced equation for the reaction of aluminum metal and chlorine gas is
2Al(s) + 3Cl2(g) → 2AlCl3(s)
Assume that 0.40 g Al is mixed with 0.25 g Cl2.
What is the maximum amount of AlCl3, in grams, that can be produced?
This is a limiting reagent (LR) problem and is worked like two stochiometry problems.
2Al(s) + 3Cl2(g) → 2AlCl3(s)
mols Al = 0.4/about 27 = approx 0.015 but that's an estimate. You should use better numbers throughout the problem because all of my numbers are estimates.
How many mols COULD that produce? That's
0.015 mols Al x (2 mols AlCl3/2 mols Al) = 0.015 mols Al
mols Cl2 gas = 0.25/71 = est 0.0035 mols. How many mols AlCl3 can we get from this. That's 0.0035 mols Cl2 x (2 mols AlCl3/3 mols Cl2) = est 0.002 mols AlCl3.
So you have two choices. You might get 0.002 mols AlCl3 OR you could get 0.015. Obviously you can get only the smaller amount which means you can produce about 0.002 mols AlC3 and Cl2 is the LR. Redo the numbers to get a more accurate figure and convert mols to grams. Post your work if you get stuck.
To find the maximum amount of AlCl3 that can be produced, we need to determine which reactant limits the reaction. This is done by calculating the number of moles of each reactant and then comparing the ratios between them.
First, let's calculate the number of moles of Al and Cl2 using their respective molar masses. The molar mass of Al is 26.98 g/mol, and the molar mass of Cl2 is 70.90 g/mol.
Number of moles of Al = mass of Al / molar mass of Al
Number of moles of Al = 0.40 g / 26.98 g/mol
Number of moles of Cl2 = mass of Cl2 / molar mass of Cl2
Number of moles of Cl2 = 0.25 g / 70.90 g/mol
Now we can compare the mole ratios of the reactants to the balanced equation to determine which reactant is limiting. According to the equation, the ratio of Al to Cl2 is 2:3, meaning 2 moles of Al react with 3 moles of Cl2.
Now let's calculate the moles of AlCl3 that can potentially be produced for each reactant:
Al:Cl2 mole ratio = 2:3
Moles of AlCl3 produced from Al = Moles of Al * (Mole ratio of AlCl3 to Al)
Moles of AlCl3 produced from Al = (0.40 g / 26.98 g/mol) * (2 mol AlCl3 / 2 mol Al)
Moles of AlCl3 produced from Cl2 = Moles of Cl2 * (Mole ratio of AlCl3 to Cl2)
Moles of AlCl3 produced from Cl2 = (0.25 g / 70.90 g/mol) * (2 mol AlCl3 / 3 mol Cl2)
Next, let's determine the limiting reactant. The reactant that produces the smaller amount of moles of AlCl3 will be the limiting reactant.
Finally, to find the maximum amount of AlCl3 produced, we multiply the number of moles of AlCl3 by its molar mass:
Maximum amount of AlCl3 produced = Moles of limiting reactant * Molar mass of AlCl3
This calculation will give us the answer, which is the maximum amount of AlCl3 that can be produced in grams.
To find the maximum amount of AlCl3 that can be produced, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed, thus limiting the amount of product that can be formed.
First, we need to calculate the number of moles of each reactant.
The molar mass of Al is 26.98 g/mol, so the number of moles of Al is:
moles of Al = mass of Al / molar mass of Al
moles of Al = 0.40 g / 26.98 g/mol
moles of Al ≈ 0.0148 mol
The molar mass of Cl2 is 70.90 g/mol, so the number of moles of Cl2 is:
moles of Cl2 = mass of Cl2 / molar mass of Cl2
moles of Cl2 = 0.25 g / 70.90 g/mol
moles of Cl2 ≈ 0.0035 mol
Now, we need to determine the limiting reactant by comparing the moles of each reactant to the stoichiometric ratio in the balanced equation.
From the balanced equation:
2 moles of Al reacts with 3 moles of Cl2 to produce 2 moles of AlCl3
So, the moles of AlCl3 that can be produced from the moles of Al is:
moles of AlCl3 = (moles of Al) x (2 moles of AlCl3 / 2 moles of Al)
moles of AlCl3 = 0.0148 mol x (2/2)
moles of AlCl3 = 0.0148 mol
The moles of AlCl3 that can be produced from the moles of Cl2 is:
moles of AlCl3 = (moles of Cl2) x (2 moles of AlCl3 / 3 moles of Cl2)
moles of AlCl3 = 0.0035 mol x (2/3)
moles of AlCl3 ≈ 0.0023 mol
Since the moles of AlCl3 that can be produced from Cl2 is less than the moles of AlCl3 that can be produced from Al, the limiting reactant is Cl2.
To calculate the mass of AlCl3 that can be produced, we need to calculate the moles of AlCl3 produced using the limiting reactant and then convert it to grams.
The molar mass of AlCl3 is 133.34 g/mol.
mass of AlCl3 = moles of AlCl3 x molar mass of AlCl3
mass of AlCl3 = 0.0023 mol x 133.34 g/mol
mass of AlCl3 ≈ 0.31 g
Therefore, the maximum amount of AlCl3 that can be produced is approximately 0.31 grams.