Disulfur dichloride, S2Cl2, is used to vulcanize rubber. It can be made by treating molten sulfur with gaseous chlorine:
S8(l) + 4 Cl2(g) --> 4 S2Cl2(l)
[Molar masses: 256.6 70.91 135.0]
Starting with a mixture of 32.0 g of sulfur and 71.0 g of Cl2, which is the limiting reactant? What mass of S2Cl2 (in grams) can be produced? What mass of the excess reactant remains when the limiting reactant is consumed?
Worked the same way as as the post by Rachel on the same computer.
To determine the limiting reactant and calculate the mass of S2Cl2 produced, we need to compare the amount of product that can be formed from each reactant.
Step 1: Convert the given masses of sulfur (S8) and chlorine (Cl2) to moles.
Molar mass of sulfur (S8) = 256.6 g/mol
Molar mass of chlorine (Cl2) = 70.91 g/mol
Moles of sulfur = mass of sulfur / molar mass of sulfur
Moles of sulfur = 32.0 g / 256.6 g/mol
Moles of sulfur = 0.1249 mol (approximately 0.125 mol)
Moles of chlorine = mass of chlorine / molar mass of chlorine
Moles of chlorine = 71.0 g / 70.91 g/mol
Moles of chlorine = 1.002 mol (approximately 1.00 mol)
Step 2: Use the balanced equation to determine the stoichiometry and find the limiting reactant.
From the balanced equation, we can see that:
1 mol of S8 reacts with 4 mol of Cl2 to produce 4 mol of S2Cl2.
However, we have 0.125 mol of sulfur and 1.002 mol of chlorine.
So, for every 1 mol of sulfur, we would need 4/0.125 = 32 moles of chlorine to completely react.
Since we only have 1.002 mol of chlorine, it is the limiting reactant because it is not in excess.
Step 3: Calculate the mass of S2Cl2 produced using the stoichiometry from the balanced equation.
Molar mass of S2Cl2 = 135.0 g/mol
To find the mass of S2Cl2 produced:
Moles of limiting reactant (chlorine) * (4 mol S2Cl2 / 1 mol Cl2) * (molar mass of S2Cl2)
Mass of S2Cl2 produced = 1.002 mol * 4 * 135.0 g/mol
Mass of S2Cl2 produced = 541.08 g
Step 4: Calculate the mass of the excess reactant (sulfur) remaining when the limiting reactant is consumed.
To determine the excess reactant remaining, we first need to find the amount of sulfur consumed. Since we know that 1 mol of S8 reacts with 4 mol of Cl2, we can use the stoichiometry to find the moles of S8 that react with 1.002 mol of Cl2.
Moles of sulfur consumed = moles of limiting reactant (chlorine) * (1 mol S8 / 4 mol Cl2)
Moles of sulfur consumed = 1.002 mol * (1 mol S8 / 4 mol Cl2)
Moles of sulfur consumed = 0.2505 mol
To find the mass of the excess sulfur remaining, we subtract the moles of sulfur consumed from the initial moles of sulfur.
Moles of sulfur remaining = moles of sulfur - moles of sulfur consumed
Moles of sulfur remaining = 0.125 mol - 0.2505 mol
Moles of sulfur remaining = -0.1255 mol
Since we cannot have a negative amount of a substance, we conclude that all the sulfur is consumed, and thus there is no excess sulfur remaining.
Summary:
The limiting reactant is chlorine (Cl2).
The mass of S2Cl2 produced is 541.08 grams.
There is no excess sulfur remaining.
To determine the limiting reactant and calculate the amounts of products and remaining reactant, we need to follow a step-by-step approach:
Step 1: Convert the given masses of reactants (sulfur and chlorine) to moles using their molar masses.
Molar mass of sulfur (S8) = 256.6 g/mol
Molar mass of chlorine (Cl2) = 70.91 g/mol
Moles of sulfur = mass of sulfur / molar mass of sulfur
Moles of sulfur = 32.0 g / 256.6 g/mol
Moles of chlorine = mass of chlorine / molar mass of chlorine
Moles of chlorine = 71.0 g / 70.91 g/mol
Step 2: Determine the molar ratio between the reactants and the product using the balanced equation.
From the balanced equation: 1 mole of S8 reacts with 4 moles of Cl2 to produce 4 moles of S2Cl2.
Step 3: Determine the limiting reactant.
To determine the limiting reactant, compare the moles of each reactant with the stoichiometric ratio.
For sulfur:
Moles of S8 = 32.0 g / 256.6 g/mol ≈ 0.125 mol
For chlorine:
Moles of Cl2 = 71.0 g / 70.91 g/mol ≈ 1.002 mol
By comparing the moles of each reactant, it is clear that sulfur is the limiting reactant since it has fewer moles.
Step 4: Calculate the theoretical yield of S2Cl2.
Using the stoichiometric ratio, we can determine the moles of S2Cl2 produced when all the sulfur reacts.
Moles of S2Cl2 = Moles of S8 (limiting reactant) × (4 moles of S2Cl2 / 1 mole of S8)
Moles of S2Cl2 = 0.125 mol × (4 mol / 1 mol)
Step 5: Calculate the mass of S2Cl2 produced.
Mass of S2Cl2 = Moles of S2Cl2 × Molar mass of S2Cl2
Mass of S2Cl2 = 0.125 mol × 135.0 g/mol
Step 6: Calculate the remaining reactant.
To find the mass of the excess reactant remaining, we need to determine the moles of the reactant that is in excess and then convert it to mass.
Moles of chlorine in excess = Moles of Cl2 (initial) - Moles of Cl2 used
Moles of chlorine in excess = 1.002 mol - (0.125 mol × (4 mol / 1 mol))
Step 7: Calculate the mass of the excess reactant remaining.
Mass of the excess reactant = Moles of chlorine in excess × Molar mass of Cl2
Now, you can perform the calculations using the values obtained in the above steps to find the limiting reactant, the mass of S2Cl2 produced, and the mass of the excess reactant remaining.