Iron metal is formed during the reduction of iron(III) oxide:
Fe2O3 + 3CO 2Fe + 3CO2
When 24.3 kg of Fe2O3 react with 14.6 kg of CO and 7.8 kg of Fe are recovered, what is the percent yield for the reaction?
Pls.. help I am having a hard time finding the answer. The wordings are getting me confused in weather i have to choose 24.3 or 7.8 :S
You use both. This is a limiting reagent type problem; I know that because the problem gives BOTH reactants. That means that one of them is used completely while the other one will have some remaing after the reaction has occurred. I work these problem by working TWO simple stoichiometry problems (simple meaning they are not limiting reagent problems). Use either reagent and determine how much of the product is formed. Then use the other one and determine the amount of the same product. The answers probably will be different; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Here is an example of a simple stoichiometry problem. Just follow the steps. Post your work if you get stuck.
oops.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the percent yield of the reaction, we need to compare the actual yield (7.8 kg of Fe) with the theoretical yield that can be calculated from the balanced equation.
First, we need to find the amount of Fe2O3 consumed based on the provided information. From the balanced equation, we can see that the stoichiometric ratio between Fe2O3 and Fe is 1:2. Therefore, if 7.8 kg of Fe is formed, the amount of Fe2O3 consumed can be calculated as:
7.8 kg Fe x (1 mol Fe2O3 / 2 mol Fe) x (159.688 g Fe2O3 / 1 mol Fe2O3) = X g Fe2O3
Next, we calculate the theoretical yield of Fe, which can be obtained from the stoichiometric ratio between Fe2O3 and Fe:
X g Fe2O3 x (2 mol Fe / 1 mol Fe2O3) x (55.845 g Fe / 1 mol Fe) = Y g Fe
Now, we can calculate the percent yield:
Percent Yield = (Actual Yield / Theoretical Yield) x 100
Percent Yield = (7.8 kg / Y g) x 100
To find the weight Y g, we need to determine whether the 24.3 kg of Fe2O3 or the 7.8 kg of Fe is the limiting reactant. The limiting reactant is the one that is completely consumed and limits the amount of product that can be formed. In this case, we can compare the amounts of Fe2O3 used in the reaction and the amounts of Fe formed.
Using the molar mass of Fe2O3 (159.688 g/mol) and CO (28.010 g/mol), we can calculate the number of moles of each substance:
Moles of Fe2O3 = 24.3 kg x (1000 g/kg) / (159.688 g/mol)
Moles of CO = 14.6 kg x (1000 g/kg) / (28.010 g/mol)
Next, using the coefficients from the balanced equation, we can calculate the stoichiometric ratios:
Fe2O3 : CO = 1 : 3
So, the ratio of moles of Fe2O3 to CO is 1:3. We can compare the moles of Fe2O3 and CO to determine the limiting reactant. If the moles of Fe2O3 are smaller, it is the limiting reactant. If the moles of CO are smaller, it is the limiting reactant.
Compare (moles of Fe2O3) / 1 = X
Compare (moles of CO) / 3 = Y
If X is smaller, we can use the balanced equation and calculate:
Moles of Fe formed = (moles of Fe2O3) × 2
If Y is smaller, we can use the balanced equation and calculate:
Moles of Fe formed = (moles of CO) × 2
Finally, we can compare the actual yield (7.8 kg) with the theoretical yield (calculated moles of Fe formed) to determine the percent yield.
I apologize for the complexity of the calculations involved. Understanding the concept of limiting reactants and stoichiometry is essential to solve this problem accurately.