Asked by Aaron
A 0.750 g sample of an unknown solid is dissolved in 100 mL of water and acidfied with 25 mL of 3 M H2SO4 then titrated with a 0.0200 M KMnO4 solution. If the unknown solid requires 12.5 mL of the KMnO4 solution to reach the endpoint, what is the % sodium oxalate in the unknown solid? Assume the only compound in the unknown solid that reacts with the KMnO4 solution is sodium oxalate. Report you answer as a percentage, but do not include the percent sign, %, in your answer
Answers
Answered by
DrBob222
You can finish balancing the equation but the following is all you need.
2MnO4^- + 5C2O4^2- ==> 10CO2 + 2Mn^2+
mols MnO4^- = M x L = ?
Using the coefficients in the balanced equation, convert mols MnO4^- to mols C2O4^2-.
Then convert mols C2O4^2- to mols Na2C2O4 (that's the same value). Convert mols Na2C2O4 to g. g Na2C2O4 = mols Na2C2O4 x molar mass Na2C2O4.
%Na2C2O4 = (grams Na2C2O4/0.750)*100 = ?
2MnO4^- + 5C2O4^2- ==> 10CO2 + 2Mn^2+
mols MnO4^- = M x L = ?
Using the coefficients in the balanced equation, convert mols MnO4^- to mols C2O4^2-.
Then convert mols C2O4^2- to mols Na2C2O4 (that's the same value). Convert mols Na2C2O4 to g. g Na2C2O4 = mols Na2C2O4 x molar mass Na2C2O4.
%Na2C2O4 = (grams Na2C2O4/0.750)*100 = ?
Answered by
Bob
14.5
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