Question
please answer the rest of the question
and show work
first reaction is: Fe(NH4)2(SO4)2 6H2O + H2C2O4 rightarrow FeC2O4 2H2O + (NH4)2SO4 + H2SO4 + 4H2O (intermediate) 2FeC2O4 2H2O + H2C2O4 + H2O2 + 3K2C2O4 rightarrow 2K3Fe(C2O4)3 3H2O objective is to prepare as great a yield as possible. To measure the efficiency are you will calculate the percentage yield. Percent yield = actual yield/theoretical yield times 100% Mass of Fe(NH_4)2(SO4)2 6H2O used: 5.039 molar mass = 392.140 g/mol Yield of K3Fe(C2O4)3 3H2O: 2.33 g molar mass = 491.23 g/mol Show all calculations for the following questions: What is the theoretical yield of K3Fe(C2O4)3 3H2O? What is the percent yield of K3Fe(C2O4)3 3H2O? Give three reasons why a yield could beless than 100% for this experiment. Give one reason why a yield could be greater than 100%.
Explanation / Answer
2)
percent yield = 2.33/2.54 *100=91.7%
2.33 actul yield...(given)
theritical yield = 2.54...(you calculated it)
