Aluminum reacts with sulfur gas to form aluminum sulfide. Initially 1.18 ml of aluminum and 2.25mol of sulfur are combined.
1. write a balanced equation?
2. what is the limiting reaction?
3. what is the theoretical yield of aluminum sulfide in moles?
How many moles of excess reactant remain unreacted?
I will take any help I can on this one please?
1. The balanced equation for the reaction between aluminum and sulfur gas to form aluminum sulfide is:
2 Al + 3 S₂ → 2 Al₂S₃
2. To determine the limiting reactant, we need to compare the moles of each reactant to their respective stoichiometric coefficients in the balanced equation.
1.18 ml of aluminum can be converted to moles using its molar volume:
1 ml = 1 cm³ → 1.18 ml = 1.18 cm³ = 1.18 cm³ × (1 cm / 10 mm) = 1.18 cm³ × (1 cm³ / 1,000 mm³) = 1.18 cm³ × (1 cm³ / 1 mL) = 1.18 cm³ × (1 mL / 1 cm³) = 1.18 mL
Using the density of aluminum (2.70 g/cm³) and molar mass (26.98 g/mol):
1.18 mL × (2.70 g/cm³) × (1 mol / 26.98 g) = 0.131 mol of aluminum
2.25 mol of sulfur are already given.
Comparing the moles of aluminum and sulfur, we have:
Aluminum: 0.131 mol
Sulfur: 2.25 mol
Since the stoichiometric coefficient for aluminum is 2 and for sulfur is 3, we can determine the limiting reactant by dividing the moles of each reactant by its respective stoichiometric coefficient:
Aluminum: 0.131 mol / 2 = 0.0655 mol
Sulfur: 2.25 mol / 3 = 0.75 mol
The limiting reactant is aluminum with 0.0655 mol.
3. The theoretical yield of aluminum sulfide can be calculated using the stoichiometry of the balanced equation.
From the balanced equation, we can see that 2 moles of aluminum react to form 2 moles of aluminum sulfide. Therefore, the theoretical yield of aluminum sulfide in moles is equal to the moles of aluminum.
Theoretical yield of aluminum sulfide = 0.0655 mol
To determine the moles of excess reactant remaining unreacted, we can subtract the moles of the limiting reactant from the initial moles of each reactant:
Excess moles of sulfur = Initial moles of sulfur - Moles of limiting reactant
Excess moles of sulfur = 2.25 mol - 0.0655 mol = 2.1845 mol
Sure! Let's break down each question step by step:
1. To write a balanced equation for the reaction between aluminum and sulfur gas, we need to determine the chemical formula for aluminum sulfide first. Aluminum is represented by the symbol Al, and sulfur is represented by the symbol S. Since both elements have a charge of 0 in their elemental form, we simply combine them together to form the compound:
Al + S → Al2S3
Note that the subscript 2 after Al indicates that there are two aluminum atoms, and the subscript 3 after S indicates that there are three sulfur atoms.
2. To determine the limiting reactant, we need to compare the moles of each reactant to the stoichiometry of the balanced equation. The stoichiometric ratio between aluminum and sulfur in this reaction is 2 moles of aluminum to 3 moles of sulfur. Let's calculate the moles of each reactant:
Moles of aluminum = volume of aluminum in liters × molarity of aluminum
= 1.18 mL × (1 L / 1000 mL) × (2.25 mol/L)
= 0.002655 mol
Moles of sulfur = 2.25 mol (given)
Now we can compare the stoichiometric ratio to determine the limiting reactant. Since the ratio is 2 moles of aluminum to 3 moles of sulfur, for every 3 moles of sulfur, we need 2/3 × 3 = 2 moles of aluminum. However, we only have 0.002655 moles of aluminum, which is less than what is required. Therefore, aluminum is the limiting reactant.
3. The theoretical yield is the maximum amount of product that can be formed based on the limiting reactant. From the balanced equation, we can see that for every 2 moles of aluminum, we produce 1 mole of aluminum sulfide (Al2S3). Therefore, since 0.002655 moles of aluminum is the limiting reactant, the theoretical yield of aluminum sulfide is:
Theoretical yield = 0.002655 mol of Al × (1 mol of Al2S3 / 2 mol of Al)
= 0.0013275 mol of Al2S3
4. To find the moles of excess reactant remaining unreacted, we need to determine how much sulfur is in excess. Since aluminum is the limiting reactant, we can consider the stoichiometry of the balanced equation to calculate the amount of sulfur that reacts with the available aluminum.
The stoichiometric ratio between aluminum and sulfur tells us that for every 2 moles of aluminum, 3 moles of sulfur are required. We already know that we have 0.002655 moles of aluminum, so we can calculate the moles of sulfur required:
Moles of sulfur required = (3 moles of S / 2 moles of Al) × 0.002655 moles of Al
= 0.0039825 moles of S
Since we initially had 2.25 moles of sulfur, the excess sulfur is equal to the initial moles of sulfur minus the moles of sulfur required:
Excess sulfur = 2.25 moles of S - 0.0039825 moles of S
= 2.2460175 moles of S
Therefore, the moles of excess reactant remaining unreacted is approximately 2.25 moles of sulfur - 0.0039825 moles of sulfur = 2.246 moles of sulfur.
I hope this helps! Let me know if you have any further questions.
ml of aluminum? Hmmmm. So get the denstiy of aluminum, and convert the ml to mg. This is odd, check your problem.
Now convert the mass of Al to moles.
2Al+ 3S>>>Al2S3
Notice for each mole of S, you need 2/3 mole of Al. If you do not have that much of Al, then Al is the limiting reageant.