a) Calculate the moles of oxalate, C2O42‐, in each titrated sample
of the iron oxalate

compound. Show one sample calculation and give the values of the
other trial.

b) Calculate the average moles of oxalate, C2O42‐, per gram of the iron oxalate compound.

UPADATED VALUE AS PER A COMMENT:

The molarity of KMnO_{4} in a solution of 0.0908 g
of KMnO_{4} in 0.500 L of solution is
1.149×10^{‒}^{3} M.

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To determine the moles of oxalate we will first convert the molarity to normality of the solution. Oxalate has valency of two or 2 acidic hydrogen. Molarity of oxalate = normality / 2. In case of KMnO4 , the oxidation number change or reaction electrons involved is 5, Molarity of KMnO4 = normality / 5 Fe_(2)(C_(2)O_(4))_(3)longrightarrowFe^(3+)+C_(2)O_(4)^(2-)For C_(2)O_(4)^(2^(-)),2^(6)+ve^(') is required for neutralisation. That's why the conversion factor from morarity from norenality is 2 .KnnO_(4)^(+7)(" with ")/(" oxalate ")rarrMn^(2+)+7 to +2rarr5e^(-)are required.So, conversion factor io 5 .   Fon neutralisation the requered equation iov_(1)S_(1)=v_(2)S_(2)Where V_(1) and V_(2) are volume of KMnO_(4) and onalate in LitreS_(1) and S_(2) are strength of KunO_(4) and oralate in nonmalityGiven{:[V_(1)=34.84mL=34.84 xx10^(-3)L],[S_(1)=1.149 xx10^(-3)N//5]:} {:[" volume (L) "xx" molarity "=L xx" mole "//L=" moles "], ... See the full answer