Hello, how would I solve this question? I tried solving it, but I got 109% as my answer which seems incorrect. This is the question:
4FeCr2O4(s) + 8K2CO3(aq) + 7O2(g) → 2Fe2O3(s) + 8K2CrO4(aq) + 8CO2(g) .
4.0 g of FeCr2O4 and 6.0 g of K2CO3 were reacted in excess O2. If 0.51 g of Fe2O3 was obtained, what was the percent yield?
Thank you!
This is a limiting reagent (LR) problem.
1. Convert grams FeCr2O4 to mols. mols = g/molar mass = ?
2. Convert grams K2CO3 to mols the same way.
3a. Using the coefficients in the balanced equation, convert mols FeCr2O4 to mols Fe2O3.
3b. Do the same and convert mols K2CO3 to mols Fe2O3.
3c. It is likely that the mols Fe2O3 from 3a and 3b will NOT be the same. The correct answer in LR problems is ALWAYS the smaller number of mols.
4, Using the smaller number, convert mols Fe2O3 to grams Fe2O3. grams = mols x molar mass = ? This is the theoretical yield (TY). The actual yield (AY) in the problem is 0.51 g.
5. % yield = (AY/TY)*100 = ?
To solve this question, we need to calculate the percent yield by comparing the actual yield (0.51 g of Fe2O3) with the theoretical yield. The theoretical yield is the amount of product that would be obtained if the reaction proceeded perfectly.
To calculate the theoretical yield, we first need to determine the limiting reagent. The limiting reagent is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
To find the limiting reagent, we compare the number of moles of FeCr2O4 and K2CO3 using their molar masses.
The molar mass of FeCr2O4 is calculated as follows:
(1 × atomic mass of Fe) + (2 × atomic mass of Cr) + (4 × atomic mass of O)
= (1 × 55.845 g/mol) + (2 × 51.9961 g/mol) + (4 × 15.9994 g/mol)
= 223.854 g/mol
The molar mass of K2CO3 is calculated as follows:
(2 × atomic mass of K) + (1 × atomic mass of C) + (3 × atomic mass of O)
= (2 × 39.0983 g/mol) + (1 × 12.0107 g/mol) + (3 × 15.9994 g/mol)
= 138.204 g/mol
Now, we can calculate the number of moles for each reactant:
moles of FeCr2O4 = (mass of FeCr2O4) / (molar mass of FeCr2O4)
= 4.0 g / 223.854 g/mol
= 0.017863 mol
moles of K2CO3 = (mass of K2CO3) / (molar mass of K2CO3)
= 6.0 g / 138.204 g/mol
= 0.043409 mol
Next, we need to determine the mole ratio of Fe2O3 to FeCr2O4. From the balanced equation, we can see that 4 moles of FeCr2O4 produce 2 moles of Fe2O3.
The number of moles of Fe2O3 produced can be calculated as:
moles of Fe2O3 = (moles of FeCr2O4) × (2 moles of Fe2O3 / 4 moles of FeCr2O4)
= 0.017863 mol × (2/4)
= 0.008931 mol
Now, we can calculate the theoretical yield of Fe2O3:
mass of Fe2O3 = (moles of Fe2O3) × (molar mass of Fe2O3)
= 0.008931 mol × (2 × 55.845 g/mol)
= 0.99661 g
The calculated theoretical yield of Fe2O3 is 0.99661 g.
Finally, we can calculate the percent yield:
percent yield = (actual yield / theoretical yield) × 100
= (0.51 g / 0.99661 g) × 100
= 51.2%
Therefore, the percent yield of Fe2O3 in this reaction is approximately 51.2%.