Question

Redox CHM 111 Titration Name Date Lab Partner Data Sheet Molarity of KMnO: 1.234M Balanced chemical equation: Trial 1 Trial Trial Trial Trial Trial 1. Mass of H, O, solution _2419_ _2.899_ 1.989 _2.329 _2.569 1.999 2. Initial burette reading 0.00mL._ _27.4mL_ _25.23mL_ _0.00ml. 23.5ml 26.52mL 3. Final burette reading 27.4mL._ _49.6ml._ _48.58mL.__25.31mL.__41.25mL._ 42.89mL_ 4. Volume of KMnO, used 5. Moles of KMnO, 6. Moles of Ho 7. Grams of 1,0 8. Q-calculated value 9. Q-table value 10. Average grams of но 11. Standard deviation for grams of H,O,

Answer 1

Answer -

Balanced Equation of Reaction -

1st half-reaction is

i) Basic Equation

KMnO4  $\rightarrow$ K+ + Mn2+

ii) Balancing the molecules of 'O' by adding H2O in RHS, Now

KMnO4  $\rightarrow$ K+ + Mn2+ + 4 H2O

iii) Now balancing charge and molecules of 'H'

8H+ + 5e- + KMnO4  $\rightarrow$ K+ + Mn2+ + 4 H2O $\leftarrow$ Equation (1)

2nd half-reaction is

i) Basic equation

H2O2 $\rightarrow$ O2

ii) Balancing 'H' & 'O' molecules and charge, Now

H2O2 $\rightarrow$ O2 + 2H+ + 2e-    $\leftarrow$ Equation (2)

Now, multiply equation (1) to 2 and equation (2) to 5 separately and Add Equation (1) and Equation (2)

Balanced Chemical Equation/Reaction -

6H+ + 5H2O2 + 2KMnO4 $\rightarrow$ 5O2 + 2K+ + 2Mn2+ + 8 H2O

Given -

Molarity of KMnO4 = 1.234

From the balanced reaction,

Moles of H2O2 = 2.5 x Moles of KMnO4 used

Title	Trial 1	Trial 2	Trial 3	Trial 4	Trial 5	Trial 6
The volume of KMnO4 used(in Litre)	0.0274	0.0222	0.02333	0.02531	0.01775	0.01637
Moles of the KMnO4 (Molarity x Volume(L))	0.033812	0.027395	0.028789	0.031233	0.021904	0.020201
Moles of H2O2	0.084529	0.068487	0.071973	0.078081	0.054759	0.050501
Grams of H2O2(Mass x Moles)	0.203715	0.197927	0.142507	0.181149	0.140182	0.100498

Average Grams of  H2O2 = 0.160996

Standard Deviation = 0.1574

Qexp = (X2 - X1) / (Xn - X1)

x1 is the smallest (suspect) value,

x2 is the second smallest value,

and xn is the largest value.

Applying this formula

Qexp = 0.3845

Q crt = 0.560 for Alpha = 0.1 which is greater than Qexp so it is not an outer of 10%.

Thank you