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 = ?
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!
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