10g of alumium reats with 35g of chlorine gas to produce alumium chloride. Which reactant is limiting, which is in excess, and how much product is produced? (Al:27g/mol, AlCl3:166.5g/mol)
2Al + 3Cl2 -> 2AlCl3
alcl3:166.5g/mol (X) -> 133.5g/mol
2Al + 3Cl2 -> 2AlCl3
Here is how you work these limiting reagent problems (LR) but I must be honest and say I do them the long way.
I've estimated all calculations so you will need to recalculate all of them.
Step 1. Convert what you have to grams.
a. mols Al = 10/27 = about 0.4
b. mols Cl2 = 35/35.5 = about 0.5
Step 2. Convert mols Al and mols Cl2 to mols AlCl3 using the coefficients in the balanced equation.
a. 0.4 mols Al x (2 mols AlCl3/1 mols Al) = about 0.4
b. 0.5 mols Cl2 x (2 mols AlCl3/3 mols Cl2) = about 0.33
c. In LR problems, the small number always wins. Do you understand why? So Cl2 is the LR and Al is the ER (excess reagent)/
Step 3. mols AlCl3 x molar mass AlCl3 = grams AlCl3.
To determine which reactant is limiting and which is in excess, we need to compare the amount of each reactant to the stoichiometry of the balanced equation.
1. Convert the mass of aluminum (Al) and chlorine gas (Cl2) to moles:
- Aluminum (Al):
Mass of Al = 10g
Molar mass of Al = 27g/mol
Moles of Al = Mass of Al / Molar mass of Al
= 10g / 27g/mol
≈ 0.37 mol
- Chlorine gas (Cl2):
Mass of Cl2 = 35g
Molar mass of Cl2 = 2*(35.5g/mol)
= 71g/mol
Moles of Cl2 = Mass of Cl2 / Molar mass of Cl2
= 35g / 71g/mol
≈ 0.49 mol
2. Use the balanced equation to determine the stoichiometric ratio between Al and Cl2:
According to the balanced equation: 2Al + 3Cl2 -> 2AlCl3
The stoichiometric ratio is 2:3, meaning that 2 moles of Al react with 3 moles of Cl2 to produce 2 moles of AlCl3.
3. Determine the limiting reactant:
Since we have 0.37 mol of Al and 0.49 mol of Cl2, we need to determine which reactant is present in lower quantity, relative to the stoichiometric ratio.
By comparing the moles present, we see that we have fewer moles of Al (0.37 mol) than Cl2 (0.49 mol). From this, we can conclude that Al is the limiting reactant.
4. Determine the excess reactant:
Since Al is the limiting reactant, Cl2 is the excess reactant.
5. Calculate the moles of AlCl3 produced:
Based on the balanced equation, 2 moles of Al react to produce 2 moles of AlCl3. Therefore, the moles of AlCl3 produced will be equal to the moles of Al used.
Moles of AlCl3 = Moles of Al
= 0.37 mol
6. Calculate the mass of AlCl3 produced:
Mass of AlCl3 = Moles of AlCl3 * Molar mass of AlCl3
= 0.37 mol * 166.5 g/mol (given molar mass of AlCl3)
≈ 61.6 g
Therefore, the limiting reactant is aluminum (Al), chlorine gas (Cl2) is in excess, and approximately 61.6 grams of aluminum chloride (AlCl3) is produced.
To find out which reactant is limiting and which is in excess, we need to calculate the number of moles for each reactant.
First, let's calculate the number of moles of aluminum (Al):
Number of moles of Al = Mass of Al / Molar mass of Al
Number of moles of Al = 10g / 27g/mol ≈ 0.37 mol
Next, let's calculate the number of moles of chlorine gas (Cl2):
Number of moles of Cl2 = Mass of Cl2 / Molar mass of Cl2
Number of moles of Cl2 = 35g / 71g/mol ≈ 0.49 mol
Now, let's determine the mole ratio between Al and Cl2 from the balanced chemical equation:
2Al + 3Cl2 -> 2AlCl3
The ratio is 2:3, which means 2 moles of Al react with 3 moles of Cl2 to produce 2 moles of AlCl3.
Since the mole ratio is 2:3, the number of moles of Cl2 should be multiplied by 2/3 to calculate the expected moles of AlCl3:
Expected moles of AlCl3 = (Number of moles of Cl2) * (2/3)
Expected moles of AlCl3 = 0.49 mol * (2/3) ≈ 0.33 mol
Now let's compare the expected moles of AlCl3 with the actual moles of AlCl3 that can be produced. We can use the mole ratio between Al and AlCl3 to do this calculation:
Actual moles of AlCl3 = (Number of moles of Al) * (2/2)
Actual moles of AlCl3 = 0.37 mol * (2/2) = 0.37 mol
Based on these calculations:
- The limiting reactant is aluminum (Al) because its actual moles of AlCl3 produced (0.37 mol) are less than the expected moles of AlCl3 (0.33 mol).
- The excess reactant is chlorine gas (Cl2) because its actual moles of AlCl3 produced (0.37 mol) are greater than the expected moles of AlCl3 (0.33 mol).
To calculate the amount of product produced, we need to determine the moles of AlCl3 produced and then convert it to grams.
Moles of AlCl3 produced = Actual moles of AlCl3 = 0.37 mol
Now let's convert the moles of AlCl3 to grams:
Mass of AlCl3 = Moles of AlCl3 * Molar mass of AlCl3
Mass of AlCl3 = 0.37 mol * 166.5 g/mol ≈ 61.5 g
Therefore, the amount of aluminum chloride (AlCl3) produced is approximately 61.5 grams.