Home/Natural Sciences & Medicine/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 KMnO4 in a solution of 0.0908 g of KMnO4 in 0.500 L of solution is 1.149×10‒3 M. Table 1: Analysis of Iron Oxalate Product for Oxalate Trial 1 Trial 2 Mass of weigh boat + iron oxalate product /g 0.8128 0.7665 Mass of "emptied" weigh boat /g 0.6643 0.6149 Mass of iron oxalate product /g 0.1485 0.1516 Initial Buret Reading /mL 2.23 1.06 Final Buret Reading /mL 37.07 37.75 Volume KMnO4 dispensed /mL 34.84 36.69 a) Calculate the moles of oxalate, C2042, in each titrated sample of the iron oxalate compound. Show one sample calculation and give the values of the other trial.

Question Solved1 Answer 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 KMnO4 in a solution of 0.0908 g of KMnO4 in 0.500 L of solution is 1.149×10‒3 M. Table 1: Analysis of Iron Oxalate Product for Oxalate Trial 1 Trial 2 Mass of weigh boat + iron oxalate product /g 0.8128 0.7665 Mass of "emptied" weigh boat /g 0.6643 0.6149 Mass of iron oxalate product /g 0.1485 0.1516 Initial Buret Reading /mL 2.23 1.06 Final Buret Reading /mL 37.07 37.75 Volume KMnO4 dispensed /mL 34.84 36.69 a) Calculate the moles of oxalate, C2042, in each titrated sample of the iron oxalate compound. Show one sample calculation and give the values of the other trial.

PNBPYK The Asker · Chemistry


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 KMnO4 in a solution of 0.0908 g of KMnO4 in 0.500 L of solution is 1.149×10‒3 M.

Transcribed Image Text: Table 1: Analysis of Iron Oxalate Product for Oxalate Trial 1 Trial 2 Mass of weigh boat + iron oxalate product /g 0.8128 0.7665 Mass of "emptied" weigh boat /g 0.6643 0.6149 Mass of iron oxalate product /g 0.1485 0.1516 Initial Buret Reading /mL 2.23 1.06 Final Buret Reading /mL 37.07 37.75 Volume KMnO4 dispensed /mL 34.84 36.69 a) Calculate the moles of oxalate, C2042, in each titrated sample of the iron oxalate compound. Show one sample calculation and give the values of the other trial.
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Transcribed Image Text: Table 1: Analysis of Iron Oxalate Product for Oxalate Trial 1 Trial 2 Mass of weigh boat + iron oxalate product /g 0.8128 0.7665 Mass of "emptied" weigh boat /g 0.6643 0.6149 Mass of iron oxalate product /g 0.1485 0.1516 Initial Buret Reading /mL 2.23 1.06 Final Buret Reading /mL 37.07 37.75 Volume KMnO4 dispensed /mL 34.84 36.69 a) Calculate the moles of oxalate, C2042, in each titrated sample of the iron oxalate compound. Show one sample calculation and give the values of the other trial.
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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

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