Question

In the chemical reaction, 5S + 6KNO3 + 2CaCO3 -->3K2SO4 + 2CaSO4 + 2CO2 + 3N2, a. Give the oxidation number (ON) of each atom on each reactant and product b. Indicate which atom is reduced and which one is oxidized c. Indicate the oxidizing agent and the reducing agent d. Indicate how many electrons are lost/gained(transferred)

Answer 1

5S + 6 KNO₃ + 2 CaCO₃ -----> 3 K₂SO₄+ 2 CaSO₄ + 2 CO₂ + 3N₂

a) Here oxidation no. Of reactants:

S =0 , K = +1, N= +5 , O= -2, Ca= +2 , C= +4

Oxidation no. Of products:

K = +1 , S= +6, O= -2, Ca= +2, N=0, C= +4

b)here ON of S changes from 0 to +6 so it gets oxidised in the given reaction.

ON of N changes from +5 to 0 , so it gets reduced in the given reaction.

c) oxidising agent is the substance which undergo reduction, So KNO₃ is oxidising agent, in this N undergo reduction from +5 to zero.

Reducing agent is the substance which undergo oxidation, So here S is reducing agent, undergo oxidation from zero to +6.

d) S looses 6 e- in the given reaction, here 5 S are involved so total loss will be 5*6 =30e-

N gains 5e- in the given reaction, here 6 N are involved so total gain will be 6*5= 30e- .