The unbalanced reaction is aKNO3+5C12H22O11→bN2 +cCO2+dH2O+eK2CO3 Number of H-atom on reactant side = Number of H − atom on product side 5×22=d×2 d=55 Number of N-atom on R= Number of N-atom on P a=2b ....(i) Number of O on R= Number of O on P 3a+55=2c+55+3e 3a=2c+3e..(ii) Number of K on R= Number of ⋅K on P a=2e.....(iii) Number of C on R= Number of C on P 60=c+e.....(iv) Putting Eq. (iii) in Eq. (ii) 6e=2c+3e− 2c−3e=0....(v) Adding Eqs. (v) and (iii) × (iv) 2c−3e=0 3c−3e=180 5c=180 c=
180
5
=36 c+e=60 ⇒e=60−36=24⇒a=2e ⇒a=2×24=48⇒a=2b ⇒b=
48
2
=24 So, values of a,b,c,d and e are 48,24,36,55 and 24 respectively.