disulfur dichloride is used to vulcanize rubber. it can be made by treating molten sulfur with gaseous chlorine: S8(1)+Cl2(g) -----> S2Cl2(1) starting with a mixture of 32.0g of sulfur and 71.0g of Cl2,
a.) which is the limiting reactant?
b.) what is the theoretical yield of S2Cl2?
c.) what mass of the excess reactant remains when the reaction is completed?
a.) Well, my dear inquirer, to determine the limiting reactant, we need to compare the number of moles of each reactant and use the stoichiometry of the balanced equation.
First, let's convert the mass of sulfur to moles. The molar mass of S8 is approximately 256 g/mol, so 32.0g of sulfur is equal to 32.0g / 256 g/mol = 0.125 mol.
Next, we need to convert the mass of Cl2 to moles. The molar mass of Cl2 is approximately 71 g/mol, so 71.0g of Cl2 is equal to 71.0g / 71 g/mol = 1 mol.
From the balanced equation, we can see that the mole ratio between S8 and Cl2 is 1:1. Therefore, the limiting reactant is sulfur, as it has fewer moles compared to Cl2.
b.) Ah, the theoretical yield! To find that, we can use the mole ratio from the balanced equation. Since sulfur is the limiting reactant, we know that the mole ratio of S8 to S2Cl2 is 1:1.
Hence, the theoretical yield of S2Cl2 would be the same as the number of moles of sulfur, which is 0.125 mol. To find the mass, we'll multiply the number of moles by the molar mass. The molar mass of S2Cl2 is approximately 135 g/mol.
The theoretical yield of S2Cl2 would be 0.125 mol * 135 g/mol = 16.875 g.
c.) Oh, the excess reactant is like that one friend who just never knows when to leave the party! To determine the mass of the excess reactant, we'll need to compare the moles of Cl2 used with the moles of Cl2 required.
From our previous calculations, we know that 71.0g of Cl2 is equal to 1 mol. Since sulfur is the limiting reactant, we only need 0.125 mol of Cl2 to react completely.
Therefore, the excess amount of Cl2 is 1 mol - 0.125 mol = 0.875 mol.
To find the mass of the excess reactant, we need to multiply the number of moles by the molar mass. The molar mass of Cl2 is approximately 71 g/mol.
Hence, the mass of the remaining excess Cl2 would be 0.875 mol * 71 g/mol = 62.125 g.
To determine the limiting reactant, we need to compare the moles of sulfur and chlorine to see which one will be completely consumed in the reaction.
a) Calculate the number of moles for sulfur (S8) and chlorine (Cl2):
Molar mass of sulfur (S8) = 8 × 32.06 g/mol = 256.48 g/mol
Molar mass of chlorine (Cl2) = 2 × 35.45 g/mol = 70.90 g/mol
Moles of sulfur = mass of sulfur / molar mass of sulfur = 32.0 g / 256.48 g/mol ≈ 0.125 mol
Moles of chlorine = mass of chlorine / molar mass of chlorine = 71.0 g / 70.90 g/mol ≈ 1.001 mol
Since the stoichiometric ratio of S8 to Cl2 is 1:1, it is evident that sulfur is the limiting reactant because it has fewer moles than chlorine.
b) The balanced equation indicates that 1 mole of S8 reacts with 1 mole of Cl2 to produce 1 mole of S2Cl2. Therefore, the theoretical yield of S2Cl2 will be equal to the number of moles of the limiting reactant.
Theoretical yield of S2Cl2 = moles of sulfur = 0.125 mol
c) To calculate the mass of the excess reactant remaining, we need to determine how much of the excess reactant is consumed in the reaction. This can be done by using the stoichiometric ratio between the limiting reactant (sulfur) and the excess reactant (chlorine).
From the reaction equation, 1 mole of S8 reacts with 1 mole of Cl2. Since sulfur is the limiting reactant, it will react completely, consuming 0.125 moles of Cl2.
Moles of Cl2 remaining = moles of Cl2 initially - moles of Cl2 consumed = 1.001 mol - 0.125 mol ≈ 0.876 mol
Mass of excess Cl2 remaining = moles of Cl2 remaining × molar mass of Cl2 = 0.876 mol × 70.90 g/mol ≈ 62.09 g
Therefore, approximately 62.09 grams of excess chlorine will remain when the reaction is completed.
To determine the limiting reactant, theoretical yield, and mass of excess reactant, we need to compare the amount of product formed by each reactant.
First, we need to determine the molar masses of sulfur and Cl2:
- The molar mass of sulfur (S8) is 8 * atomic mass of sulfur (32.06 g/mol) = 256.48 g/mol.
- The molar mass of Cl2 is 2 * atomic mass of chlorine (35.45 g/mol) = 70.90 g/mol.
Now we can calculate the number of moles of each reactant:
- Moles of sulfur: mass / molar mass = 32.0 g / 256.48 g/mol = 0.1250 mol.
- Moles of Cl2: mass / molar mass = 71.0 g / 70.90 g/mol = 1.00 mol.
The balanced equation tells us that the stoichiometric ratio between S8 and Cl2 is 1:1. Therefore, for every mol of S2Cl2 formed, we need 1 mol of sulfur and 1 mol of Cl2.
a.) To determine the limiting reactant, we compare the moles of each reactant to the stoichiometric ratio. In this case, we see that sulfur has fewer moles (0.1250 mol) than Cl2 (1.00 mol). Therefore, sulfur is the limiting reactant.
b.) Theoretical yield is the maximum amount of product that can be formed based on the limiting reactant. Since sulfur is the limiting reactant, we use the moles of sulfur (0.1250 mol) to calculate the moles of S2Cl2. Since the stoichiometric ratio between S8 and S2Cl2 is 1:1, we have 0.1250 mol of S2Cl2 as the theoretical yield.
To calculate the mass of S2Cl2, we multiply the moles by the molar mass:
Mass = moles * molar mass = 0.1250 mol * (2 * atomic mass of sulfur (32.06 g/mol) + 2 * atomic mass of chlorine (35.45 g/mol)) = 37.77 g.
c.) To determine the mass of the excess reactant, we subtract the moles of the limiting reactant from the moles of the excess reactant. Since sulfur is the limiting reactant, we need to find the moles of Cl2 in excess. From the balanced equation, the stoichiometric ratio is 1:1 between S8 and Cl2. We started with 0.1250 mol of sulfur, so the moles of Cl2 used will also be 0.1250 mol.
Moles of Cl2 in excess = moles of Cl2 initially - moles of Cl2 used
= 1.00 mol - 0.1250 mol
= 0.8750 mol.
Finally, we can calculate the mass of the excess reactant:
Mass = moles * molar mass = 0.8750 mol * (2 * atomic mass of chlorine (35.45 g/mol))
= 61.24 g.
Therefore, when the reaction is completed, there will be 61.24 g of excess Cl2 remaining.
1. Balance the equation.
2. Convert 32.0 g S8 to moles. moles = grams/molar mass.
3. Convert 71.0 g Cl2 to moles.
4a. Convert moles S8 to moles S2Cl2 using the coefficients in the balanced equation.
4b. Same for moles Cl2 to moles S2Cl2.
4c. The SMALLER number of moles S2Cl2 will be the moles S2Cl2 formed and the reagent from which the smaller amount arises will be the limiting reagent.
5. Theoretical yield is found by converting moles S2Cl2 to grams. Grams = moles x molar mass.
6. Go through the same kind of procedure to determine the grams of the excess reagent used to form the theoretical yield of S2Cl2, then subtract from the amount with which you started. Post your work if you get stuck.