5.6g of iron metal Fe are heated strong and mixed with 8g of sulfur S to form 4.4g of iron(ll) sulfide FeS. a, identify the limiting and excess reactants. b, calculate the mass of FeS formed. C,calculet the percent yield of FeS.
Fe + S --> FeS
a. moles Fe = g/atomic mass = 5.6/56 = 0.1 mol
moles S = 8/32 = 0.25
0.1 mole Fe will require 0.1 mol S and you have that much; therefore, Fe is the limiting regent (LR) and S is the excess reagent (ER).
b. I think you meant how much FeS COULD BE FORMED since you know from the problem that 4.4 g was formed; i.e., ou want to know the theoretical yield. 0.1 mol Fe will produce 0.1 mol FeS. grams FeS produced = mols x molar mass = 0.1 mole x 88 = 8.8 grams FeS. 8.8 g FeS is the theoretical yield.
c. % yield = (actual yield/theoretical yield) *100 = ?
a) Well, well, looks like we have ourselves a chemical reaction party! To identify the limiting and excess reagents, we'll have to do a little math. Let's start by comparing the moles of iron (Fe) and sulfur (S) in the ratio given by the balanced equation.
The balanced equation is:
2Fe + S → FeS
First, let's find the moles of iron (Fe):
Moles of Fe = mass of Fe (g) / molar mass of Fe (g/mol)
Moles of Fe = 5.6g Fe / 55.845 g/mol = 0.100 mol Fe
Next up, we'll find the moles of sulfur (S):
Moles of S = mass of S (g) / molar mass of S (g/mol)
Moles of S = 8g S / 32.06 g/mol = 0.249 mol S
Now, let's compare the moles of Fe and S. The balanced equation tells us that 1 mol of Fe reacts with 1 mol of S to produce 1 mol of FeS. So, we'll use the moles of S as our reference.
Looking at our calculations, we can see that there are 0.249 moles of S, which is greater than the 0.100 moles of Fe. This means that sulfur is the excess reactant, and iron is the limiting reactant.
b) Alrighty, time to calculate the mass of FeS formed. Since we know the iron is the limiting reactant, we can use the stoichiometry of the balanced equation to find the theoretical yield of FeS.
From the balanced equation, we know that:
1 mol Fe reacts with 1 mol S to produce 1 mol FeS
So, the moles of FeS formed would be equal to the moles of Fe used since it's the limiting reactant. In this case, that's 0.100 mol Fe. Now, we just need to calculate the mass of FeS:
Mass of FeS = moles of FeS × molar mass of FeS
Mass of FeS = 0.100 mol × (55.845 g/mol + 32.06 g/mol)
Mass of FeS = 0.100 mol × 87.905 g/mol
Mass of FeS = 8.7905 g FeS
Therefore, the mass of FeS formed is 8.7905 grams.
c) To calculate the percent yield of FeS, we'll need the actual yield and the theoretical yield. Lucky for us, we already know the theoretical yield, which is 8.7905 g from the previous calculation.
Since the question tells us that 4.4g of FeS was formed, we can use that as the actual yield. Now it's just plug and chug into the percent yield formula:
Percent yield = (actual yield / theoretical yield) × 100
Percent yield = (4.4g / 8.7905g) × 100
After a quick calculation, the percent yield of FeS is approximately 50.07%.
Remember, chemists have a sense of humor too. So let's not take these calculations too seriously—we're just clowning around!
To solve this problem, we first need to calculate the moles of each reactant involved.
Given:
Mass of iron (Fe) = 5.6 g
Mass of sulfur (S) = 8 g
Mass of iron(II) sulfide (FeS) formed = 4.4 g
a. To identify the limiting and excess reactants, we need to compare the moles of each reactant.
Step 1: Calculate the moles of each reactant.
Molar mass of Fe = 55.85 g/mol
Molar mass of S = 32.06 g/mol
Moles of Fe = mass of Fe / molar mass of Fe
Moles of Fe = 5.6 g / 55.85 g/mol
Moles of Fe ≈ 0.1 mol
Moles of S = mass of S / molar mass of S
Moles of S = 8 g / 32.06 g/mol
Moles of S ≈ 0.25 mol
The balanced equation for the reaction is:
Fe + S → FeS
Looking at the stoichiometric ratio between Fe and S, we see that 1 mole of Fe reacts with 1 mole of S to produce 1 mole of FeS.
Comparing the moles of Fe and S, we can see that we have more moles of S than Fe. Therefore, Fe is the limiting reactant, and S is the excess reactant.
b. To calculate the mass of FeS formed, we can use the stoichiometric ratio from the balanced equation.
Moles of FeS = moles of Fe (since 1 mole of Fe reacts to form 1 mole of FeS)
Mass of FeS = Moles of FeS * molar mass of FeS
Molar mass of FeS = molar mass of Fe + molar mass of S
Molar mass of FeS = 55.85 g/mol + 32.06 g/mol
Molar mass of FeS ≈ 87.91 g/mol
Mass of FeS = 0.1 mol * 87.91 g/mol
Mass of FeS ≈ 8.79 g
c. To calculate the percent yield of FeS, we need to compare the actual yield (mass of FeS formed) to the theoretical yield (mass of FeS calculated).
Percent Yield = (Actual Yield / Theoretical Yield) * 100
Actual Yield = Mass of FeS formed = 4.4 g
Theoretical Yield = Mass of FeS calculated = 8.79 g
Percent Yield = (4.4 g / 8.79 g) * 100
Percent Yield ≈ 50.1%
Therefore, the answers to the questions are:
a. The limiting reactant is Fe, and the excess reactant is S.
b. The mass of FeS formed is approximately 8.79 g.
c. The percent yield of FeS is approximately 50.1%.
To identify the limiting and excess reactants, we need to compare the amount of each reactant with the stoichiometric ratio from the balanced chemical equation.
The balanced chemical equation for the reaction between iron and sulfur to form iron(II) sulfide is:
Fe + S → FeS
a) Calculating the amount of each reactant:
- Iron (Fe): 5.6g
- Sulfur (S): 8g
Now, we need to convert the mass of each reactant to moles by dividing by their molar masses.
- Molar mass of Fe: 55.85 g/mol
Number of moles of Fe = 5.6g / 55.85 g/mol = 0.100 moles of Fe
- Molar mass of S: 32.06 g/mol
Number of moles of S = 8g / 32.06 g/mol = 0.250 moles of S
b) Identifying the limiting and excess reactants:
To determine the limiting reactant, we need to compare the stoichiometric ratio of the reactants to find which one would require more moles to react with the other.
From the balanced equation, we can see that the stoichiometric ratio of Fe:S is 1:1. This means that 1 mole of Fe requires 1 mole of S to react completely.
Since we have 0.100 moles of Fe and 0.250 moles of S, we can see that Fe is the limiting reactant because it requires fewer moles of S for a complete reaction. This also tells us that S is the excess reactant.
c) Calculating the mass of FeS formed:
To calculate the mass of FeS formed, we first need to determine the limiting reactant, which is Fe. Then, we use its stoichiometric ratio from the balanced equation to find the amount of FeS that can be formed.
From the balanced equation, we know that the stoichiometric ratio of FeS:Fe is 1:1. This means that 1 mole of Fe reacts to produce 1 mole of FeS.
Since we have 0.100 moles of Fe, we can conclude that 0.100 moles of FeS can be formed.
To determine the mass of FeS formed, we multiply the number of moles of FeS by its molar mass.
- Molar mass of FeS: 87.91 g/mol
Mass of FeS = 0.100 moles × 87.91 g/mol = 8.791g
Therefore, the mass of FeS formed is 8.791 grams.
c) Calculating the percent yield of FeS:
The percent yield is a measure of how much product is obtained compared to the theoretically calculated yield. It is given by the formula:
Percent Yield = (Actual Yield / Theoretical Yield) × 100
In this case, the actual yield is given as 4.4g, and the theoretical yield (calculated earlier) is 8.791g.
- Actual Yield: 4.4g
- Theoretical Yield: 8.791g
Now we can calculate the percent yield:
Percent Yield = (4.4g / 8.791g) × 100 = 50.03%
Therefore, the percent yield of FeS is approximately 50.03%.