Question

Molarity of your standardized KMnO₄ solution : 0.0200 M

Experiment Number	#1	#2	#3	#4
Mass of sample (g)	0.142	0.136	0.155	0.140
Volume of KMnO4 required (mL) :	14.70	16.80	12.10	13.40
Volume of blank titration (mL) :	0.0500	0.0500	0.0500	0.0500

1) Balance the equation below for the reaction between MnO₄⁻ and C₂O₄²⁻.

2) Calculate the corrected volumes for each experiment for KMnO₄:

3)Calculate the moles for each experiment of KMnO₄ added :

4)Calculate the moles in each experiment of C₂O₄²⁻ in the sample :

5)Calculate the mass in each experiment of C₂O₄²⁻ in the sample :

6)Calculate the percent in each experiment of C₂O₄²⁻ in the sample :

7) Now, using the statistical treatment of data, calculate the percent of C₂O₄²⁻ in your sample :

Answer 1

1)Balanced equation for the reaction between Mno, and co- 2Mno; (aq)+5C0(aq) +16H* (aq) +2Mn++ (aq)+10C02(g)+8H20(1) 2)Corrected volumes of all the experiments for KMnO4 Experiment 1, V2 = 14.70 -0.0500 mL = 14.65 mL Experiment 2, V, = 16.80 -0.0500 mL = 16.75 mL Experiment 3, V, = 12.10 -0.0500 mL = 12.05 mL Experiment 4, V. = 13.40 -0.0500 mL =13.35 mL

3)Moles of KMnO, in each experiment IL Experiment 1, n = 0.0200M X14.65 mL X- 1000 mL = 0.000293 mol IL Experiment 2, n = 0.0200M X16.75 mL X- 1000 mL = 0.000335 mol IL Experiment 3, nz = 0.0200 Mx12.05 mL X- 1000 mL = 0.000241 mol IL Experiment 4, n = 0.0200 M x 13.35 mL X- 1000 mL = 0.000267 mol 4)Moles of CO2 in each experiment Experiment 1, n = 0.000293 mol KMnO, x 3 mol C20, 2 mol KMno = 0.0007325 mol 5 mol CO2- Experiment 2, no = 0.000335 mol KMnO, 2 mol KMnO = 0.0008375 mol 5 molco- Experiment 3, ncc = 0.000241mol KMnO 2 mol KMnO = 0.0006025 mol 5 molc,O2- Experiment 4, n.o = 0.000267 mol KMnO,x= 2 mol KMno = 0.0006675 mol