Question

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

Answer 1

Answer:

$2KMnO_{4} +3H_{2} SO_{4} +5H_{2} SO_{4}$--->k2so4+2MnSO4+8H2O+5O2

M=n/v(L)

n1=$0.50*10*10^{-3}$

n1=0.005moles

n2=$0.5*15*10^{-3}$

n2=0.0075moles

2 moles(KMnO4) reacts with 5 moles(H2O2)

0.005 moles reacts with =5*0.005/2

                                         =0.0125

5moles gives 5*22.4L of O2

=5*22.4*0.0075/5

=0.168litre

=168ml