sample of an Iron Oxalato complex salt weighting 0.145 grams requires 31.13 mL o

Question

sample of an Iron Oxalato complex salt weighting 0.145 grams requires 31.13 mL of 0.015 M KMnO4 to turn the solution a very light pink color at the quivalence point.

Calculate the number of moles of KMnO4 added.

Calculate the number of moles of C2O42 in the sample of the oxalate salt.

Calculate the percent weight of C2O42 in the original Iron Oxalato Complex salt sample

Calculate the number of grams of C2O42 in the sample.Calculate the number of moles of C2O42 in the sample of the oxalate salt.Calculate the number of grams of C2O42 in the sample.

Explanation / Answer

The full equation:

5 C2O42- (aq) +2 MnO4- (aq) + 16 H+ (aq) =10 CO2 (g) + 8 H2O (l) + 2 Mn2+(aq)

ratio is 5 mol of oxalate per 2 mol of permanganate

so...

mol of KMnO4 used = MV = 31.13*0.015= 0.46695 mmol of KMnO4

a) mol of KMnO4 added --> 0.46695 *10^-3 mol = 0.00046695 mol of KMnO4

b)

since ratio is: 5:2 then

mol of C2O4-2 --> 5/2*0.00046695 = 0.001167375 mmol of C2O4-2

c)

% mass of C2O4-2 in sample = mass of C2O4 / total mass *100%

MW of C2O4 = 80.2 g/mol

so

mass = 80.2*0.001167375 = 0.09362 g of oxalate

% mass of C2O4-2 in sample = mass of C2O4 / total mass *100% = 0.09362 / 0.145 * 100 = 64.56551%

d)

mass of C2O4 in sample --> 0.09362

mol in sample --> 0.001167375 mol of C2O4-2

88.019

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