Given the following data, determine the mole ratio of iron (III) to oxalate (Fe

Question

Given the following data, determine the mole ratio of iron (III) to oxalate (Fe3+:C2O42-) in a sample of an iron oxalate complex with the general formula Kz[Fex(C2O4)y].wH2O.

Mw(Fe3+) = 55.85 g/mol

Mw(C2O42-) = 88.02 g/mol

Step 1: Assume you have 100 g complex.

Enter your answers in the boxes provided to 3 significant figures. Do not enter units.

%m(Fe3+) = 11.0%

= 11.0 g in 100 g

%m(C2O42-) = 52.0%

= 52.0 g in 100 g

n(Fe3+) = m(Fe3+)/Mw(Fe3+)

= Answer __________ mol

n(C2O42-)   = m(C2O42-)/Mw(C2O42-)

= Answer __________ mol

Step 2: Divide by smallest no. of moles (in this case Fe3+) to determine the ratios

Round your answer to the nearest whole number.

Ratio (C2O42-) = n(C2O42-)/ n(Fe3+) = Answer _________ = y         

Ratio (Fe3+) = n(Fe3+)/ n(Fe3+) = Answer __________ = x

Therefore molar ratio of Fe3+:C2O42- = Answer _________ : Answer __________ = x : y

Explanation / Answer

%m(Fe3+) = 11.0%

= 11.0 g in 100 g

%m(C2O42-) = 52.0%

= 52.0 g in 100 g

n(Fe3+) = m(Fe3+)/Mw(Fe3+) = 11/55.85

= Answer 0.197 mol

n(C2O42-)   = m(C2O42-)/Mw(C2O42-) = 52/88

= Answer 0.59 mol

Ratio (C2O42-) = n(C2O42-)/ n(Fe3+) = 0.59/0.197 Answer 3 = y         

Ratio (Fe3+) = n(Fe3+)/ n(C2O42-) = Answer 0.33 = x

Therefore molar ratio of Fe3+:C2O42- = Ratio (Fe3+) = n(Fe3+)/ n(Fe3+) = Answer 1 = x

so molar ratio is Fe : C2O4 0.197 : 0.59 which is 0.197/0.197 : 0.59/0.197 = 1:3

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