Question

In: Chemistry

sample of an Iron Oxalato complex salt weighting 0.145 grams requires 31.13 mL of 0.015 M...

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.

Solutions

Expert Solution

a) Millimols of KMnO4 added = (31.13 mL)*(0.015 M) = 0.46695 mmol.

Mol(s) of KMnO4 added = (0.46695 mmol)*(1 mol)/(1000 mmol) = 4.6695*10-4 mol,

b) The balanced chemical equation for the reaction between oxalate and MnO4- is given as below.

MnO4- (aq) + 8 H+ (aq) + 5 e- --------> Mn2+ (aq) + 4 H2O (l)

C2O42- (aq) ---------> 2 CO2 (g) + 2 e-

Multiply the first equation by 2 and the second by 5 to give the balanced equation as

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

As per the stoichiometric equation,

2 mols MnO4- = 5 mols C2O42-.

Mols C2O42- in the sample = (4.6695*10-4 mol)*(5 mols C2O42-)/(2 mols MnO4-)

= 1.167375*10-3 mol.

c) The atomic masses are

C: 12.011 u

O: 15.999 u

The MW of C2O42- = (2*12.011 + 4*15.999) g/mol

= 88.018 g/mol.

Mass of oxalate in the sample = (1.167375*10-3 mol)*(88.018 g/mol)

= 0.10275 g ≈ 0.103 g (ans, correct to 3 sig. figs).

Mass percent oxalate in the sample = (0.103 g)/(0.145 g)*100

= 71.03% ≈ 71.0% (ans, correct to 3 sig. figs).

d) Number of grams of C2O42- in the sample = 71.0% (ans).


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