Question
Given the equation: 4 Fe(s) + 3 O2(g) ? 2 Fe2O3(s), what is the theoretical yield of Fe2O3 from the reaction of 4.86 g Fe with excess O2? b. If 6.76 g Fe2O3 are actually obtained from 4.86 g Fe, what is the percent yield? c. Under other conditions, the percent yield is 75.6%. How many grams of Fe2O3 are formed from 4.86 g Fe under these conditions?
mols Fe = 4.86 g Fe/atomic mass Fe = ?
Using the coefficients in the balcned equation, convert mols Fe to mols Fe2O3.
Now convert mols Fe2O3 to grams Fe2O3. g = mols x molar mass = ? and this is the theoretical yield (TY) The actual yield (AY) is 6.76.
%yield = (AY/TY)*100 = ?
Just rearrange the %yield equation above from
%yield = (AY/TY)*100
75.6 = (AY/TY)*100 and solve for AY.
a. Well, let's solve this equation! According to the balanced equation, we know that 4 moles of Fe react with 3 moles of O2 to produce 2 moles of Fe2O3. So, first we need to convert 4.86 g of Fe to moles.
Using the molar mass of Fe (55.85 g/mol), we can calculate the number of moles:
4.86 g Fe * (1 mol Fe / 55.85 g Fe) = 0.087 moles of Fe
Since the ratio of Fe2O3 to Fe is 2:4, we can find the theoretical yield of Fe2O3:
0.087 moles of Fe * (2 moles of Fe2O3 / 4 moles of Fe) = 0.0435 moles of Fe2O3
Now, we can convert moles of Fe2O3 to grams using the molar mass of Fe2O3 (159.69 g/mol):
0.0435 moles of Fe2O3 * (159.69 g Fe2O3 / 1 mol Fe2O3) = 6.94 g of Fe2O3
So, the theoretical yield of Fe2O3 is 6.94 g.
b. If 6.76 g of Fe2O3 are actually obtained, we can calculate the percent yield using the formula:
Percent yield = (actual yield / theoretical yield) * 100
Plugging in the values:
Percent yield = (6.76 g / 6.94 g) * 100 = 97.4%
c. If the percent yield is 75.6%, we can rearrange the formula to find the actual yield:
Actual yield = (percent yield / 100) * theoretical yield
Plugging in the values:
Actual yield = (75.6 / 100) * 6.94 g = 5.24 g
So, under these conditions, 5.24 g of Fe2O3 are formed from 4.86 g of Fe.
To find the theoretical yield of Fe2O3, we need to use the stoichiometry of the balanced equation.
a. The equation is balanced as follows:
4 Fe(s) + 3 O2(g) -> 2 Fe2O3(s)
The molar mass of Fe is 55.845 g/mol, and the molar mass of Fe2O3 is 159.6882 g/mol.
First, convert the mass of Fe to moles:
moles of Fe = mass of Fe / molar mass of Fe
moles of Fe = 4.86 g / 55.845 g/mol
moles of Fe = 0.087 moles
Using the balanced equation, we can determine the moles of Fe2O3 produced:
From the balanced equation, the molar ratio of Fe to Fe2O3 is 4:2, which simplifies to 2:1.
moles of Fe2O3 = moles of Fe * (2 moles of Fe2O3 / 4 moles of Fe)
moles of Fe2O3 = 0.087 moles * (2 / 4)
moles of Fe2O3 = 0.0435 moles
To find the theoretical yield of Fe2O3, convert moles to grams:
mass of Fe2O3 = moles of Fe2O3 * molar mass of Fe2O3
mass of Fe2O3 = 0.0435 moles * 159.6882 g/mol
mass of Fe2O3 = 6.95 g
b. To calculate the percent yield, use the formula:
percent yield = (actual yield / theoretical yield) * 100
Given that the actual yield is 6.76 g Fe2O3, the percent yield is:
percent yield = (6.76 g / 6.95 g) * 100
percent yield = 97.83%
c. Now, let's find the mass of Fe2O3 formed under conditions with a percent yield of 75.6%.
Using the formula for percent yield, rearrange the formula to find the actual yield:
actual yield = (percent yield / 100) * theoretical yield
percent yield = 75.6%
theoretical yield = 6.95 g (from part a)
actual yield = (75.6 / 100) * 6.95 g
actual yield = 5.25 g
Therefore, 5.25 g of Fe2O3 would be formed from 4.86 g of Fe under these conditions.
To find the theoretical yield of Fe2O3 from the reaction of 4.86 g Fe with excess O2, we need to use stoichiometry. Here are the step-by-step calculations:
Step 1: Determine the molar mass of Fe (iron) and Fe2O3 (iron(III) oxide).
- The molar mass of Fe is 55.845 g/mol.
- The molar mass of Fe2O3 is (2 * 55.845) + (3 * 16.00) = 159.69 g/mol.
Step 2: Convert the mass of Fe to moles using its molar mass.
- Moles of Fe = mass of Fe / molar mass of Fe
= 4.86 g / 55.845 g/mol
≈ 0.087 mol Fe
Step 3: Use the balanced chemical equation to determine the stoichiometry between Fe and Fe2O3.
- From the balanced equation, we can see that 4 moles of Fe react to form 2 moles of Fe2O3. Therefore, the molar ratio is 4:2.
Step 4: Calculate the moles of Fe2O3 produced.
- Moles of Fe2O3 = (moles of Fe * moles of Fe2O3) / moles of Fe
= (0.087 mol * 2 mol) / 4 mol
= 0.0435 mol Fe2O3
Step 5: Convert the moles of Fe2O3 to grams using its molar mass.
- Mass of Fe2O3 = moles of Fe2O3 * molar mass of Fe2O3
= 0.0435 mol * 159.69 g/mol
≈ 6.94 g Fe2O3
Therefore, the theoretical yield of Fe2O3 from the reaction of 4.86 g Fe is approximately 6.94 g Fe2O3.
b. To find the percent yield, we need to compare the actual yield (6.76 g Fe2O3) to the theoretical yield (6.94 g Fe2O3).
Percent yield = (actual yield / theoretical yield) * 100
= (6.76 g / 6.94 g) * 100
≈ 97.4%
Therefore, the percent yield is approximately 97.4%.
c. To find how many grams of Fe2O3 are formed from 4.86 g Fe under the condition of a 75.6% yield, we will reverse the percent yield calculation.
Percent yield = (actual yield / theoretical yield) * 100
Solving for the actual yield:
Actual yield = (percent yield / 100) * theoretical yield
= (75.6 / 100) * 6.94 g
≈ 5.25 g Fe2O3
Therefore, under these conditions, approximately 5.25 g of Fe2O3 are formed from 4.86 g Fe.