S8+4Cl2 -----> 4S2Cl2
A) what mass of S2Cl2 can be produced?
B) what mass of the excess reactant remains
When the limiting reactant is consumed?
This problem can't be solved with no quantities listed.
To determine the answers, we need the molar masses of the reactants.
The molar mass of S8 is 256.52 g/mol.
The molar mass of Cl2 is 70.906 g/mol.
Now, let's calculate the answers step-by-step.
A) To find the mass of S2Cl2 produced, we need to determine the limiting reactant first.
1) Calculate the number of moles of each reactant:
Moles of S8 = Mass / Molar mass = 256.52 g / 256.52 g/mol = 1 mol
Moles of Cl2 = Mass / Molar mass = 4Cl2 * 70.906 g/mol = 283.624 g / 70.906 g/mol = 4 mol
2) Determine the molar ratio between the reactants:
From the balanced equation: S8 + 4Cl2 -> 4S2Cl2
The ratio of S8 to Cl2 is 1:4.
3) Compare the moles of the reactants with the stoichiometric ratio:
Since there are equal moles of S8 and Cl2 (1:4 ratio), Cl2 is the limiting reactant. Therefore, the amount of S8 is in excess.
4) Calculate the moles of S2Cl2 produced using the limiting reactant:
Moles of S2Cl2 = Moles of limiting reactant = 4 mol
5) Calculate the mass of S2Cl2 produced:
Mass of S2Cl2 = Moles of S2Cl2 * Molar mass = 4 mol * (2S + 2Cl) = 4 mol * (32.06 g/mol) = 128.24 g
Therefore, the mass of S2Cl2 produced is 128.24 g.
B) To find the mass of the excess reactant remaining when the limiting reactant is consumed, we will calculate the moles and mass of the excess reactant.
1) Calculate the moles of Cl2 used:
Moles of Cl2 used = Moles of S2Cl2 produced * (4 mol Cl2 / 4 mol S2Cl2) = 4 mol * (4 mol Cl2 / 4 mol S2Cl2) = 4 mol
2) Calculate the moles of Cl2 remaining:
Moles of Cl2 remaining = Moles of Cl2 initial - Moles of Cl2 used = 4 mol - 4 mol = 0 mol
3) Calculate the mass of Cl2 remaining:
Mass of Cl2 remaining = Moles of Cl2 remaining * Molar mass of Cl2 = 0 mol * 70.906 g/mol = 0 g
Therefore, the mass of the excess reactant (Cl2) remaining when the limiting reactant is consumed is 0 g.
To answer these questions, we need to understand the concept of limiting reactant and how to calculate the mass of a compound in a chemical reaction.
A) To determine the mass of S2Cl2 that can be produced, we first need to identify the limiting reactant. The limiting reactant is the reactant that will be completely consumed first, limiting the amount of product that can be formed.
In this reaction, we have S8 and Cl2 as the reactants. To find the limiting reactant, we need to compare the moles of each reactant to the stoichiometric ratio in the balanced equation.
First, determine the moles of S8 and Cl2:
Moles of S8 = mass of S8 / molar mass of S8
Moles of Cl2 = volume of Cl2 (if given) / molar volume at STP (22.7 L/mol)
Then, compare the mole ratios between S8 and Cl2 in the balanced equation:
S8:Cl2 = 1:4 (from the balanced equation)
Whichever reactant has a smaller mole ratio compared to the stoichiometric ratio is the limiting reactant. The other reactant is in excess.
Once we determine the limiting reactant, we can calculate the moles of S2Cl2 produced using the stoichiometric ratio from the balanced equation:
S8: S2Cl2 = 1:4 (from the balanced equation)
Finally, calculate the mass of S2Cl2 produced using the calculated moles of S2Cl2 and the molar mass of S2Cl2:
Mass of S2Cl2 = moles of S2Cl2 * molar mass of S2Cl2
B) To find the mass of the excess reactant remaining when the limiting reactant is consumed, we can use the stoichiometry of the balanced equation and the moles of the limiting reactant.
First, calculate the moles of the limiting reactant used using the stoichiometric ratio from the balanced equation:
Moles of limiting reactant used = moles of limiting reactant * stoichiometric ratio of limiting reactant in the balanced equation
Next, calculate the moles of the excess reactant remaining by subtracting the moles of the limiting reactant used from the initial moles of excess reactant:
Moles of excess reactant remaining = initial moles of excess reactant - moles of limiting reactant used
Finally, calculate the mass of the excess reactant remaining using the moles of the excess reactant and its molar mass:
Mass of excess reactant remaining = moles of excess reactant remaining * molar mass of excess reactant
By following these steps, you can determine the mass of S2Cl2 produced and the mass of the excess reactant remaining when the limiting reactant is consumed.