I just need help with the final questions at the bottom. My attached answers to

Question

I just need help with the final questions at the bottom. My attached answers to everything previous are also there. I would greatly appreciate the full process shown. Thank you!

Part D The synthesis and Analysis of an Iron Oxalato Transition Metal Complex L With the completion of the determinations of % potassium, % iron, and % oxalate in the crystals, you may calculate the % water. The can now be calculated from the the percentage composition. Once the formula is know it is then possible to calculate the percent yield o You have entered the following values: From Part A: Mass of KxFe(C204)y zH20 prepared: 4.444 g Mass of FeCl3 1.60 g From Part B: 90 Potassium in compound : 21.90 % % Iron (from ion exchange & titration vs. NaOH) : 10.70 % From Part C: 96 Oxlate : 56.50 % Now,_let's finish the calculation and the determination of the formula of the iron compound: Calculate the % water of hydration : 10.90 You are correct. Your receipt no. is 163-8835 Calculate the following for Fe3+: g in 100 g sample mol in 100 g sample 10.70 mol/mol Fe (3 sig figs) (whole number) 1.00 mol/mol Fe 0.191 You are correct. Your receipt no. is 163-1778? previous ines Calculate the following for K+: g in 100 g sample 21.90 Submit Answer Answer Submitted: Your final submission will be graded after the due date. Tries 2/3 Previous Tries Calculate the following for C204 mol/mol Fe (3 sig figs) 2.93 mol/mol Fe (whole number) 3 mol in 100 g sample 0.560 mol/mol Fe mol/mol Fe q in 100 q sample mol in 100 q sample

Explanation / Answer

Theoretical moles of K3Fe(C2O4)3.3H2O is nothing but the theoretical yield in mol and the theoretical yield is the maximum amount of the product that can be produced in the reaction at given condition.

In this reaction, one mol of FeCl3 can produce the one mole of product that is K3Fe(C2O4)3.3H2O

So the 9.86E-3 (is nothing but 9.86 * 10-3) mol of the FeCl3 can produce 9.86 * 10-3 mol of the K3Fe(C2O4)3.3H2O

hence the Theoretical moles of K3Fe(C2O4)3.3H2O =  9.86 * 10-3 mol

An actual mole of K3Fe(C2O4)3.3H2O = mass of product prepared / molecular weight

=  (4.444 g) / ( 491.25 g/mol) = 9.046 * 10-3 mol

percent yield = (actual mole of K3Fe(C2O4)3.3H2O / theorotical mol of K3Fe(C2O4)3.3H2O) * 100

= (9.046 * 10-3 mol / 9.86 * 10-3 mol) * 100 = 91.75 %

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