Asked by sara
The titration of 5.40 mL of a saturated solution of sodium oxalate, Na2C2O4, at 25∘C requires 29.4 mL of 2.200×10−2 M KMnO4 in acidic solution.
What mass of Na2C2O4 in grams would be present in 1.00L of this saturated solution?
What mass of Na2C2O4 in grams would be present in 1.00L of this saturated solution?
Answers
Answered by
DrBob222
You can balance the entire equation if you wish but the only part that matter is the redox part. Here is that part.
5C2O4^2- + 2MnO4^-==>2Mn^2+ + 10CO2
mols KMnO4 = M x L = ?
Using the coefficients in the balanced equation, convert mols KMnO4 to molsl Na2C2O4.
Then grams Na2C2O4 = mols Na2C2O4 x molar mass Na2C2O4 and that is grams in a 5.40 mL sample of the Na2C2O4. Convert to grams in 1 L.
5C2O4^2- + 2MnO4^-==>2Mn^2+ + 10CO2
mols KMnO4 = M x L = ?
Using the coefficients in the balanced equation, convert mols KMnO4 to molsl Na2C2O4.
Then grams Na2C2O4 = mols Na2C2O4 x molar mass Na2C2O4 and that is grams in a 5.40 mL sample of the Na2C2O4. Convert to grams in 1 L.
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