find composition of post-reaction mixture if 10.0g of sulfur and 20.0g of iron were reacted in a closed container producing iron (II) sulfide
I don't know what the problem is asking. The word composition seems out of place with the rest of the problem.
that is exactly how it is worded on the pretest we have to study by
Find composition of post-reaction mixture if 10.0 g of sulfur and 20.0 g of iron were reacted in a closed container producing iron (II) sulfide.
To find the composition of the post-reaction mixture, we need to determine the amount of each component in the mixture.
First, let's write the balanced chemical equation for the reaction:
Fe + S → FeS
According to the equation, 1 mole of iron reacts with 1 mole of sulfur to produce 1 mole of iron (II) sulfide.
Step 1: Calculate the moles of sulfur and iron used in the reaction.
The molar mass of sulfur (S) is 32.06 g/mol, and the molar mass of iron (Fe) is 55.85 g/mol.
Moles of sulfur = mass of sulfur / molar mass
Moles of sulfur = 10.0 g / 32.06 g/mol ≈ 0.312 mol
Moles of iron = mass of iron / molar mass
Moles of iron = 20.0 g / 55.85 g/mol ≈ 0.358 mol
Step 2: Determine the limiting reactant.
To find the limiting reactant, we compare the moles of each reactant to the stoichiometry of the balanced equation. The reactant that produces the smallest amount of product is the limiting reactant.
From the balanced equation, the 1:1 ratio shows that both reactants are in a 1:1 ratio. So, the limiting reactant will be the one that is present in a lesser amount.
In this case, sulfur (S) is the limiting reactant since we have 0.312 moles, while we have an excess of iron (0.358 moles). So, all of the sulfur will be consumed in the reaction, and there will be some unreacted iron.
Step 3: Calculate the moles of iron (II) sulfide formed.
From the balanced equation, we know that 1 mole of sulfur reacts with 1 mole of iron to produce 1 mole of iron (II) sulfide.
Moles of iron (II) sulfide = moles of sulfur ≈ 0.312 mol.
Step 4: Calculate the mass of iron (II) sulfide formed.
The molar mass of iron (II) sulfide (FeS) is 87.91 g/mol.
Mass of iron (II) sulfide formed = moles of iron (II) sulfide × molar mass of iron (II) sulfide
Mass of iron (II) sulfide formed = 0.312 mol × 87.91 g/mol ≈ 27.41 g
Step 5: Calculate the remaining amount of iron.
Since iron is in excess, some of it will remain unreacted. We subtract the moles of sulfur from the moles of iron to find the excess.
Excess moles of iron = moles of iron - moles of sulfur
Excess moles of iron = 0.358 mol - 0.312 mol ≈ 0.046 mol
Step 6: Calculate the mass of unreacted iron.
Mass of unreacted iron = excess moles of iron × molar mass of iron
Mass of unreacted iron = 0.046 mol × 55.85 g/mol ≈ 2.57 g
Step 7: Calculate the composition of the post-reaction mixture.
The post-reaction mixture will consist of iron (II) sulfide (FeS) and unreacted iron (Fe).
Mass of iron (II) sulfide = 27.41 g (calculated in Step 4)
Mass of unreacted iron = 2.57 g (calculated in Step 6)
Therefore, the composition of the post-reaction mixture is approximately:
- Iron (II) sulfide (FeS): 27.41 g
- Unreacted iron (Fe): 2.57 g