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15 mL of 0.5 M H2O2 reacts with 10 mL of 0.5 M KMnO4 in excess of sulphuric acid, according to reaction 2KMnO4 + 3H2SO4 + 5H2O2 → K2SO4 + 2MnSO4 + 8H2O + 5O2 The volume of O2 produced at STP is

15 mL of 0.5 M H2O2 reacts with 10 mL of 0.5 M KMnO4 in excess of sulphuric acid, according to reaction 2KMnO4 + 3H2SO4 + 5H2O2 → K2SO4 + 2MnSO4 + 8H2O + 5O2 The volume of O2 produced at STP is

Grade:11

1 Answers

JAMES SELVA RAJ
13 Points
4 years ago
 2KMnO4 + 3 H2SO4 + 5 H2SO4 \rightarrow K2SO4 + 2MnSO4 +8H2O + 5O2
M= n/v(L)                                                 n2= 0.5x15x10-3
n1= 0.50x 10x10-3                                          n2=0.0075 moles
n1=0.005 moles
2 moles(KMnO4) reacts with 5 moles(H2O2)
0.005 moles reacts with?
                         =5x0.005/2
                          =0.0125 (limiting reagent)
 
    5moles gives 5x22.4 L of O2
      0.0075moles gives?
                      =5x22.4x0.0075/5
                   =0.168 litre
                    =168 ml

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