Question

(I showed some work below. Please show step-by-step process as I'm confused)

What is the:
a) Number of moles of KMnO₄ required for reaction?
b) Volume of KMnO₄, M

	Trial 1
Mass of sodium oxalate, g	0.2575 g
Moles of sodium oxalate*	0.001922 mol
a) Moles of KMnO4 required for reaction	see below
Initial burette reading, mL	2.15 mL
Final burette reading, mL	17.90 mL
b) Volume of KMnO4, mL	see below

Calculation for *:
0.2575 g Na₂C₂O₄ x (1 mol Na₂C₂O₄ / 133.996 g Na₂C₂O₄)
= 0.0019216991 mol Na₂C₂O₄
= 0.001922 mol Na₂C₂O₄

a) Does this mean I calculate the moles of KMnO_{4 (permanganante)} as I did for the Na₂C₂O_{4 (sodium oxalate)}?

Step 1: Finding molar mass
K = 39.098
Mn = 54.938 x 1 = 54.938
O₄ = 15.999 x 4 = 63.996
K + Mn + O₄ = 158.032 g

Step 2: Convert grams to mol
0.2575 g KMnO₄ x (1 mol KMnO₄ / 158.032 g KMnO₄)
= 0.0016294168 mol KMnO₄
= 0.001630 mol KMnO₄

Is this correct?
__________________________________________
b) Is it: final burette reading - initial burette reading = volume of KMnO₄ solution?

17.90 - 2.15 = 15.75 mL
Volume of KMnO₄ solution is 15.75 mL (?)

Answer 1