Identify the percent yield when 28.16 g of CO2 are formed from the reaction of 456.93 g of C8H18 with 256.0 g of O2.
2C8H18 + 25O2 → 16CO2 + 18H2O
This is a limiting reagent problem (LR). The LR must be identified first
2C8H18 + 25O2 → 16CO2 + 18H2O
1a. mols C8H18 = g/molar mass = 456.93/114 = approx 4, Check that.
1b. mols O2 = 256/32 = approx 8. Check that.
2a.. From 1a, how many mols CO2 can be formed. That's approx 4 mols C8H18 x (16 mols CO2/2 mols C8H18) = approx 4*16/2 = about 32 mols or in grams that's 32 x 44 gCO2/mol = about 1400. Check that number.
2b. From 1b, how many mols CO2 can be formed. That's approox 8 mols O2 x (16 mols CO2/25 mols O2) = 8*16/25 = about 5.12 moles and that converted to g CO2 is 5.12 x 44 g CO3/mol = 225.Check that number.
The LR is the material (C8H18 or O2) which produces the LEAST amount of CO2.. That's O2 in this case so the theoretical yield (TY) is approx 225 g .CO2 The actual yield (AY) from the problem is 28.16 g.
%yield = (AY/TY)*100 = ?
Check all of these calculations. It's late and my eyes are squinting.
Post your work if you get stuck.
Well, to find the percent yield, you need to compare the actual yield (which is given as 28.16 g of CO2) to the theoretical yield (which is the maximum amount of CO2 that could be formed based on the balanced equation).
First, we need to calculate the theoretical yield of CO2. Since the balanced equation shows a 2:16 ratio between C8H18 and CO2, we can set up a proportion to find the moles of C8H18:
(456.93 g C8H18) / (114.22 g/mol C8H18) = x moles C8H18
Using the balanced equation, we know that for every 2 moles of C8H18, we get 16 moles of CO2. So, we can set up another proportion to find the moles of CO2:
(x moles C8H18) * (16 moles CO2 / 2 moles C8H18) = y moles CO2
Now we convert the moles of CO2 to grams using its molar mass:
(y moles CO2) * (44.01 g/mol CO2) = theoretical yield of CO2 in grams
Finally, we can determine the percent yield by using the formula:
Percent Yield = (Actual Yield / Theoretical Yield) * 100
So, let's calculate all that nonsense and see what we end up with, shall we?
To calculate the percent yield, we need to compare the actual yield to the theoretical yield.
1. Determine the molar mass of C8H18 (octane):
C8H18: 12.01 g/mol x 8 + 1.01 g/mol x 18 = 114.23 g/mol
2. Calculate the moles of C8H18:
456.93 g / 114.23 g/mol = 3.999 mol
3. Determine the molar mass of O2:
O2: 16.00 g/mol x 2 = 32.00 g/mol
4. Calculate the moles of O2:
256.0 g / 32.00 g/mol = 8.000 mol
5. Use the balanced equation to determine the stoichiometry between C8H18 and CO2:
2C8H18 + 25O2 → 16CO2 + 18H2O
From the balanced equation, we can see that 2 moles of C8H18 react to form 16 moles of CO2.
6. Calculate the theoretical yield of CO2:
(16 mol CO2 / 2 mol C8H18) x 3.999 mol C8H18 = 31.992 mol CO2
7. Determine the molar mass of CO2:
CO2: 12.01 g/mol + 16.00 g/mol x 2 = 44.01 g/mol
8. Calculate the theoretical yield of CO2 in grams:
31.992 mol CO2 x 44.01 g/mol = 1,408.38 g CO2
9. Calculate the percent yield:
actual yield / theoretical yield x 100%
actual yield = 28.16 g
theoretical yield = 1,408.38 g
percent yield = (28.16 g / 1,408.38 g) x 100% = 1.999% (rounded to three significant figures)
Therefore, the percent yield is 1.999%.
To calculate the percent yield in a chemical reaction, you need to compare the actual yield (the amount of product obtained experimentally) with the theoretical yield (the amount of product predicted by stoichiometry calculations).
First, you need to determine the theoretical yield of CO2 using the stoichiometry of the balanced equation. The balanced equation tells us that 2 moles of C8H18 react with 25 moles of O2 to produce 16 moles of CO2.
1. Convert the given masses of C8H18 and O2 into moles using their respective molar masses. The molar masses are:
- C8H18: 114.22 g/mol
- O2: 32.00 g/mol
Moles of C8H18 = 456.93 g / 114.22 g/mol = 4 moles of C8H18
Moles of O2 = 256.0 g / 32.00 g/mol = 8 moles of O2
2. Determine the limiting reactant. The reactant that will be completely consumed and, therefore, determines the maximum amount of product that can be formed is the limiting reactant. To find the limiting reactant, you need to compare the moles of each reactant to their stoichiometric coefficients.
From the balanced equation: 2 moles of C8H18 react with 25 moles of O2.
Moles of C8H18 / stoichiometric coefficient = 4 moles / 2 = 2 moles
Moles of O2 / stoichiometric coefficient = 8 moles / 25 ≈ 0.32 moles
Since moles of C8H18 (2 moles) is less than moles of O2 (0.32 moles), C8H18 is the limiting reactant.
3. Calculate the theoretical yield of CO2 using the stoichiometry.
From the balanced equation: 2 moles of C8H18 produce 16 moles of CO2.
Moles of CO2 = Moles of limiting reactant x (moles of CO2 / moles of limiting reactant)
= 2 moles x (16 moles CO2 / 2 moles C8H18) = 16 moles
4. Convert the moles of CO2 into grams using the molar mass of CO2 (44.01 g/mol).
Mass of CO2 = Moles of CO2 x Molar mass of CO2
= 16 moles x 44.01 g/mol = 704.16 g
The theoretical yield of CO2 is 704.16 g.
5. Finally, calculate the percent yield using the following formula:
Percent yield = (Actual yield / Theoretical yield) x 100
Actual yield = 28.16 g (given in the question)
Theoretical yield = 704.16 g (calculated in step 4)
Percent yield = (28.16 g / 704.16 g) x 100 ≈ 4.00%
Therefore, the percent yield of CO2 in this reaction is approximately 4.00%.