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.

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)

To find the number of moles of MnO4- that were added, we need to determine the difference between the initial and final volumes of the KMnO4 solution used in the titration.

Step 1: Calculate the volume of KMnO4 solution used in the titration:
Volume used = Final volume - Initial volume
= 46.88 mL - 11.23 mL
= 35.65 mL

Step 2: Convert the volume used from milliliters to liters:
Volume used = 35.65 mL * (1 L / 1000 mL)
= 0.03565 L

Step 3: Calculate the number of moles of MnO4- using the equation:
moles of MnO4- = Molarity * Volume (in liters)

Given:
Molarity (M) = 0.201 M
Volume (in liters) = 0.03565 L

moles of MnO4- = 0.201 M * 0.03565 L
= 0.00718365 moles

Therefore, approximately 0.00718365 moles of MnO4- were added.