# AP chem

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A 0.345g sample of anhydrous BeC2O4, which contains an inert impurity, was dissolved in a sufficient water to produce 100ml of solution. A 20.0ml portion of the solution was titrated w/ KMnO4(aq). The balanced equation for the reaction that occurred is as follows:
16H^+(aq) + 2MnO4^-(aq) + 5C2O4^2-(aq) ---> 2Mn^2+(aq) + 10CO2(g) + 8H2O(l)

the volume of 0.0150M KMnO4(aq) required to reach the equivalence point was 17.80ml.

a) identify the reducing agent in the reaction
b) for the titration at the equivalence point, calculate the number of moles of each of the following that reacted:
MnO4^-(aq)
C2O4^2-(aq)
c) calculate the total # of moles of C2O4^2-(aq) that were present in the 100ml of prepared solution.
d)calculate the mass percent of BeC2O4(s) in the impure 0.345g sample.

can u give me the steps to solving this please? thanx

A 0.345g sample of anhydrous BeC2O4, which contains an inert impurity, was dissolved in a sufficient water to produce 100ml of solution. A 20.0ml portion of the solution was titrated w/ KMnO4(aq). The balanced equation for the reaction that occurred is as follows:
16H^+(aq) + 2MnO4^-(aq) + 5C2O4^2-(aq) ---> 2Mn^2+(aq) + 10CO2(g) + 8H2O(l)

the volume of 0.0150M KMnO4(aq) required to reach the equivalence point was 17.80ml.

a) identify the reducing agent in the reaction is the substance that is oxidized.
b) for the titration at the equivalence point, calculate the number of moles of each of the following that reacted:
MnO4^-(aq) mols=L x M =??
C2O4^2-(aq) = ?? mols MnO4^- *(5 mols C2O4^-2)/2 mols MnO4^-). Use the coefficients in the balanced equation. Note units of mols permangante cancel leaving mols oxalate.
c) calculate the total # of moles of C2O4^2-(aq) that were present in the 100ml of prepared solution. b gives mols present in 20 mL. That was 1/5 the initial sample; therefore, multiply mols found in the titration by 5.
d)calculate the mass percent of BeC2O4(s) in the impure 0.345g sample.

mols x molar mass = grams.
%= (grams/weight sample)x100

are these right::
a) Mn
b) 3x10-4 and 7.5x10-4
c) 1.125x10-6
d) ??

• AP chem -

i. (C2O4)-2 Loss of electrons....

ii. (MnO4)-1:
0.0150M*.0178L = mol(MnO4)-1
=>2.67*10^-4 mol

(C2O4)-2:
2.67*10^-4 mol * 5(C2O4)/2(MnO4) (ratio)
=>6.68*10^-4 mol

iii. 6.68*10^-4 mol * 5 (100.ml/20.ml=5)
=>3.34*10^-3 mol

iv. 3.34*10^-3 mol * 97.032gfm = 0.324g
(0.324g / 0.345g) * 100.
=>93.9%

• AP chem -

i. The reducing agent is the one that undergoes oxidation (the one that loses electrons). Mn goes from being -1 to +2 meaning that Mn will be the reducing agent. But since the reducing agent is the whole compound, MnO4- is the R.A>

• AP/IB chem -

i) Oxidation numbers are not the overall charge of an ion. The Oxidation numbers are for each atom. The reducing agent in this case is C2O4 because the Carbon is Oxidized because it's oxidation number goes from +3 to +4. MnO4 is the Oxidizing Agent because Mn's oxidation number goes from +7 to +2.

• AP chem -

iv. The question is asking for the mass percent of BeC2O4 in the impure sample. The impure sample is the BeC2O4 with the water. So the question is wanting to know the percentage of BeC2O4 out of the entire sample. Therefore you must divide the mass of the BeC2O4(97.0302g) by the mass of the BeC2O4 in addition to the H2O(total of 115.045). Giving you an answer of 84.3%

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