Question

conc of oxalic acid: 0.7550M

conc of MnO4=0.13M

3H₂C₂O₄(aq) + 2MnO^-₄(aq)-------6CO₂ (g)+2MnO₂₍s)+2OH^- (aq)+2H₂O(l)

Experiment #	H20	Oxalic Acid	KMnO4
1	6	5	1
2	1	10	1
3	5	5	2

Base on the equation and the reactants shown I need to determine:

a)The initial concentration of each reactant. Show calculations

b)final concentration of each reactance, considering:

-which compound is the limiting reactant?Show calculations

-which compound is in excess?Show calculations

Answer 1

For the given reaction,

Table

a) Initial concentrations

Expt.         Oxalic acid (initial) (M)                   KMnO4 (initial) (M)          

   1         0.755 M x 5 ml/12 ml = 0.314         0.13 M x 1 ml/12 ml = 0.011

   2         0.755 M x 10 ml/12 ml = 0.63         0.13 M x 1 ml/12 ml = 0.011

   3         0.755 M x 5 ml/12 ml = 0.314         0.13 M x 2 ml/12 ml = 0.022

b) Final concentrations

For complete consuption of oxalic acid we would need = 0.314 x 2/3 = 0.209 M KMnO4

Since KMnO4 available is less, KMnO4 is limiting reagent

Expt.         Oxalic acid (final) (M)                   KMnO4 (final) (M)          

   1           0.314 M - 0.0165 = 0.2975                         0

   2           0.2975 - 0.0165 - 0.2810                            0

   3           0.314 M - 0.033 = 0.2810                           0