A 0.345 g sample of anhydrous BeC2O4, containing an inert impurity, was dissolved in water to produce 100 mL of solution. A 20.0 mL portion of the solution was titrated with KMnO4 (aq). The balanced equation for the reaction is as follows.
16H[+] (aq) + 2MnO4[-] (aq) + 5C2O4[2-] (aq) -> 2Mn[2+] (aq) + 10CO2 (g) + 8H2O (l)
The volume of 0.0150 M KMnO4 (aq) required to reach equivalence point was 17.80 mL.
1) Identify the reducing agent in the titration reaction.
I've been googling everything, and I've NOW learned about OILRIG, LEOGER, reduction = reduction in charge.
But I don't understand where all the numbers come from like "Mn goes from +7 (in MnO4^-1) to +2; Mn is reduced so MnO4^-1is the oxidizing agent.C goes from +3 (in C2O4^-2) to +4 (in CO2); C is oxidized so C2O4^-2 is the reducing agent."
16H[+] (aq) + 2MnO4[-] (aq) + 5C2O4[2-] (aq) -> 2Mn[2+] (aq) + 10CO2 (g) + 8H2O (l)
The volume of 0.0150 M KMnO4 (aq) required to reach equivalence point was 17.80 mL.
1) Identify the reducing agent in the titration reaction.
I've been googling everything, and I've NOW learned about OILRIG, LEOGER, reduction = reduction in charge.
But I don't understand where all the numbers come from like "Mn goes from +7 (in MnO4^-1) to +2; Mn is reduced so MnO4^-1is the oxidizing agent.C goes from +3 (in C2O4^-2) to +4 (in CO2); C is oxidized so C2O4^-2 is the reducing agent."
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the reducing agent get oxidized
C2O4-2 each O is -2 which is -8 for all 4
thus C2 is +3 so +6 -8 = -2
in CO2 each O is -2 so C is +4
thus C is oxidized from +3 to +4
screw LEO says GER that is for babies
the oxidization goes up - it is oxidized
since C is oxidized C2O4-2 is the reducing agent
C2O4-2 each O is -2 which is -8 for all 4
thus C2 is +3 so +6 -8 = -2
in CO2 each O is -2 so C is +4
thus C is oxidized from +3 to +4
screw LEO says GER that is for babies
the oxidization goes up - it is oxidized
since C is oxidized C2O4-2 is the reducing agent