In one experiment, 7.62 g of Fe are allowed to react with 8.67 g of S. What is the limiting reagent, what is the mass of FeS formed, and how much excess reagent is left, respectively?
I am not sure how to solve this.
I started by doing (7.62 g Fe/55.85 g)=.137 g
(8.67g S/32.07 g)=.270
I divided both by .137 because it was the lowest number.
I got a ratio of 1Fe:2S
Now what should I do?
Well, it looks like you've successfully calculated the ratio of Fe to S in the reaction, which is 1:2. That's a good start!
To determine the limiting reagent, we need to compare the actual ratio of Fe to S with the ratio you calculated. In this case, we have 7.62 g of Fe and 8.67 g of S.
Using the molar masses, we can convert the grams of each element to moles:
- Moles of Fe = 7.62 g Fe / 55.85 g/mol ≈ 0.1363 mol Fe
- Moles of S = 8.67 g S / 32.07 g/mol ≈ 0.2704 mol S
Now, let's compare these mole ratios with the ratio you calculated:
- Actual ratio of Fe to S: 0.1363 mol Fe : 0.2704 mol S ≈ 1:1.986
- Calculated ratio of Fe to S: 1:2
Since the calculated ratio is already in the simplest form, we can conclude that the limiting reagent is Fe because there is less of it compared to S based on the actual ratio.
To find the mass of FeS formed, you can use the stoichiometry of the balanced chemical equation. From the ratio that you calculated, we know that 1 mole of Fe reacts with 2 moles of S to form 1 mole of FeS.
Since Fe is the limiting reagent, we use its moles to determine the moles of FeS formed:
- Moles of FeS formed = 0.1363 mol Fe * (1 mol FeS / 1 mol Fe) = 0.1363 mol FeS
Now, we can find the mass of FeS formed using its molar mass:
- Mass of FeS formed = 0.1363 mol FeS * 87.91 g/mol = 11.98 g FeS
Finally, to find the excess reagent, subtract the moles of the limiting reagent consumed from the moles of the excess reagent initially present. In this case, we subtract the moles of Fe consumed from the moles of S initially present:
- Moles of excess S = 0.2704 mol S - 0.1363 mol Fe * (2 mol S / 1 mol Fe) ≈ 0.2704 mol S - 0.2726 mol S = -0.0022 mol S
Since we have a negative value, it indicates that there is no excess S remaining. Therefore, the mass of the excess S is 0 g.
So, to summarize:
- The limiting reagent is Fe.
- The mass of FeS formed is 11.98 g.
- There is no excess S remaining.
Hope that helps!
To determine the limiting reagent, compare the molar ratios of the reactants to the balanced chemical equation.
The balanced chemical equation for the reaction between Fe and S is:
Fe + S -> FeS
The coefficients in the balanced equation represent the molar ratio between the reactants. In this case, it shows that 1 mole of Fe reacts with 1 mole of S to form 1 mole of FeS. Therefore, the molar ratio of Fe to S is 1:1.
To find the limiting reagent, compare the actual mole ratio of the reactants to the stoichiometric mole ratio:
Actual mole ratio of Fe to S = (7.62 g Fe / molar mass of Fe) / (8.67 g S / molar mass of S)
Use the molar masses of Fe and S to convert the masses to moles:
Molar mass of Fe: 55.85 g/mol
Molar mass of S: 32.07 g/mol
Actual mole ratio of Fe to S = (7.62 g Fe / 55.85 g/mol) / (8.67 g S / 32.07 g/mol)
= 0.136 mol Fe / 0.270 mol S
≈ 0.504 mol Fe / mol S
Since the actual mole ratio of Fe to S is approximately 0.504, and the stoichiometric mole ratio is 1:1, it is evident that Fe is the limiting reagent because it has the lowest mole ratio.
Now, let's calculate the mass of FeS formed and the excess reagent.
1. Determine the moles of FeS formed:
Since 1 mole of Fe reacts with 1 mole of S, the moles of FeS formed will be equal to the moles of the limiting reagent (Fe).
Moles of FeS formed = 0.136 mol Fe
2. Calculate the mass of FeS formed:
Molar mass of FeS: 55.85 g/mol + 32.07 g/mol = 87.92 g/mol
Mass of FeS formed = Moles of FeS formed * Molar mass of FeS
= 0.136 mol * 87.92 g/mol
3. Find the excess reagent:
To determine the excess reagent, subtract the moles of the limiting reagent (Fe) from the moles of the non-limiting reagent (S).
Moles of excess S = Moles of S - Moles of Fe
= 0.270 mol - 0.136 mol
4. Calculate the mass of excess S:
Mass of excess S = Moles of excess S * Molar mass of S
By following these steps, you can determine the limiting reagent, the mass of FeS formed, and the amount of excess reagent in the reaction.
To determine the limiting reagent, you need to compare the mole ratio of the reactants with the actual mole ratio based on the given masses. Let's calculate the moles of Fe and S:
Moles of Fe = 7.62 g Fe / molar mass of Fe
Moles of S = 8.67 g S / molar mass of S
Now, once you have the moles of each reactant, you need to compare the mole ratio of Fe and S with the stoichiometric ratio given by the balanced chemical equation. The balanced equation for the reaction between Fe and S is:
Fe + S -> FeS
The stoichiometric ratio based on this equation is 1:1, meaning one mole of Fe reacts with one mole of S to produce one mole of FeS.
Now, compare the actual ratios of moles of Fe and S with the stoichiometric ratio:
Moles of Fe / Moles of S
If this ratio is less than the stoichiometric ratio of 1:1, then Fe is the limiting reagent. If it is greater, then S is the limiting reagent. From your calculation:
Moles of Fe = 7.62 g Fe / molar mass of Fe
Moles of S = 8.67 g S / molar mass of S
Calculate the actual ratio:
Actual ratio = Moles of Fe / Moles of S
If the actual ratio is less than 1:1, Fe is the limiting reagent. If the actual ratio is greater than 1:1, S is the limiting reagent.
To calculate the mass of FeS formed, you need to determine the limiting reagent first. Once you know the limiting reagent, you can calculate the moles of the limiting reagent and convert that to the mass of FeS using the stoichiometry of the balanced equation.
To find the excess reagent, you subtract the moles of the limiting reagent used from the moles of the excess reagent initially present. Then you can calculate the mass of the excess reagent using the molar mass.
Let's use this method to solve the problem. To continue, please provide the molar masses of Fe and S.
I obtained 0.136 (and that's MOLES Fe and not g iron).
Then 0.270 mols S.
0.136 mols Fe x (1 mol S/1 mol Fe) = 0.136 mol S. Do you have that much S? Yes, so Fe is the limiting reagent and S is the excess reagent.
Part B. So 0.136 mol FeS formed.
g FeS = mols FeS x molar mass FeS.