Asked by George
3.392 g of unknown oxalate product is titrated with 0.201 M KMnO4. The initial burette reading is 11.23 mL, and the final burette reading is 46.88 mL. How many moles of MnO4- were added?
Once again any helpwould be greatly appreciated.
Once again any helpwould be greatly appreciated.
Answers
Answered by
GK
The balanced chemical equation is:
2MnO4^-(aq) + 16H^+(aq) + 5C2O4^-2 ---> 2Mn^+2 + 8H2O + 10CO2
46.88mL - 11.23mL = 35.65mL MnO4- solution
•Moles of MnO4^- = (0.03565L)(0.201mol/L)
2MnO4^-(aq) + 16H^+(aq) + 5C2O4^-2 ---> 2Mn^+2 + 8H2O + 10CO2
46.88mL - 11.23mL = 35.65mL MnO4- solution
•Moles of MnO4^- = (0.03565L)(0.201mol/L)
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