1.

Balance the following chemical equations by the oxidation number method (i) CuO+NH_(3)rarrCu+N_(23)+H_(2)O (ii) C_(6)H_(6)+O_(2)rarrCO_(2)+H_(2)O (iii)SnO_(2)+CrarrSn+CO (iv) Fe_(2)O_(3)rarrFe+CO (v) P+HNO_(3)rarrHPO_(3)+NO+H_(2)O (vi) FeS_(2)+O_(2)rarrFe_(2)O_(3)+SO_(2)

Answer»

Solution :`(i) CuO+NH_(3)rarrCu+N_(2)+H_(2)O`
The balancing is done in the folowing STEPS:
1. write the O.N of each atom in the skeleton equation.
`overset(+2)Cuoverset(-2)O+overset(-3)Noverset(+1)H_(3)rarrCoverset(0)u+overset(0)N_(2)+overset(+1)H_(2)overset(-2)O`
2.Identify the atoms which undergo change in O.N.
`overset(+2)CuO+overset(-3)NH_(3)rarroverset(0)Cu+overset(0)Cu+overset(N_(2))+H_(2)O`
3.Calculate the increaseand decrease in O.N w.r.t reactant atoms

4.Equate the increase and decreasein O.N on the reactant side.
`3CuO+2NH_(3)rarrCu+N_(2)+H_(2)O`
5. Balance the number of Cu and N atoms on both sides of the equation
`3CuO+2NH_(3)rarr3Cu+N_(2)+H_(2)O`
6. Now balacne H and O atoms byu HIT and trial method
`3CuO +2NH_(3)rarr3Cu+N_(2)+3H_(2)O`
`(ii) C_(6)H_(6)+O_(2)rarrO_(2)rarrCO_(2)+H_(2)O`
The balancing is done in the folowing steps:
1.Write the O.N of each atom in the skeleton equation
`overset(-1)C_(6)overset(+1)H_(6)+overset(0)_(2)rarroverset(+4)CO_(2)+H_(2)overset(-2)O_(2)`
3. Calculate the total increase and decrease in O.N w.r.t reactant atoms

4. Equate the increase and decrease in O.N onthe reactatn side after taking out a commonfactor of 2
`2C_(6)H_(6)+15O_(2)rarrCO_(2)+H_(2)O`
5. Balance the number of C an d O atoms on both sides of the equation
`2C_(6)H_(6)+154O_(2)rarr12CO_(2)+6H_(2)O`
The H atoms are already balanced in the above equation
`(iii) SnO_(2) +Crarrsn+CO`
The balancing is done int the following steps:
1. Write the O.N of each atom in the skeleton equation
`overset(+4)SnO_(2)+overset(0)CrarrSoverset(0)n+overset(+2)CO`
3. Calculate the increase and decrease in O.N w.r.t reactant atoms.

4. Equate the increase and decrease in O.N on the reacttant side after taking out a common factor of 2 `SnO_(2)+2CrarrSn+CO`
5. Balance the number of Snand C atoms onboth sides of the equation
`SnO_(2)+2CrarrSn+2CO`
The O atoms are already balanced in the equation
`(IV) Fe_(2)O_(3)+CrarrFerarrCO`
The balancing is done in the following steps:
1. Write the O.N of each atyom in the skeleton equation
`overset(+3)Fe_(2)overset(-2)O_(3)+overset(0)Crarroverset(0)Fe+overset(0)Fe+overset(+2)CO`
3. Calculate the increase and decreasein O.N w.r.t reactant atoms
4.Equate the inccrease and decrease in O.N on the reacttnt side.
`3P+5HNO_(3)rarrHPO_(3)+NO+H_(2)O`
5. Balance the P amd N atoms on both sides of the equation
`3P+5HNO_(3)rarr3HPO_(3)+5NO+H_(2)O`
The O and H atoms are already blanced in the equation.
`(vi) FeS_(2)+O_(2)rarrFe_(2)O_(3)+SO_(2)`
The balacning is done in the following steps:
1. Write the O.N of each atom in the skeleton equation
`overset(+2)Feoverset(-1)S_(2)+overset(0)O_(2)rarroverset(+3)Fe_(2) overset(-2)O_(3)+overset(+4)S Ooverset(-2)O_(2)`
2.Identify the atoms which undergo change in O.N
`overset(+3)Fe_(2)overset(-2)O_(3)+overset(0)Crarroverset(0)Fe+overset(0)Fe+overset(+2)CO`
3. Calculate the increase and decreasein O.N w.r.t reactant atoms
4.Equate the inccrease and decrease in O.N on the reacttnt side.
`3P+5HNO_(3)rarrHPO_(3)+NO+H_(2)O`
5. Balance the P amd N atoms on both sides of the equation
`3P+5HNO_(3)rarr3HPO_(3)+5NO+H_(2)O`
The O and H atoms are already blanced in the equation.
`(vi) FeS_(2)+O_(2)rarrFe_(2)O_(3)+SO_(2)`
The balacning is done in the following steps:
1. Write the O.N of each atom in the skeleton equation
`overset(+2)Feoverset(-1)S_(2)+overset(0)O_(2)rarroverset(+3)Fe_(2) overset(-2)O_(3)+overset(+4)S Ooverset(-2)O_(2)`
3.Calculate the increase and decrease in O.N w.r.t reactant atoms.

4. Equate the increase and decrease in O.Nin the reactant side.
`2[Cr(OH)_(4)]^(-)+3H_(2)O_(2)rarr(CrO_(4))^(2-)+H_(2)`
5.Balance the number of Cr atoms in the equation
`2[Cr(OH)_(4)]^(-)+3H_(2)O_(2)rarr(CrO_(4))^(2-)+H_(2)`
6.In order to blalnce the number of oxygen atoms, ADD five `H_(2)O` molecules on the product side
`2[Cr(OH)_(4)]^(-)+3H_(2)O_(2)rarr2(CrO)_(4)^(2)+6H_(2)O`
7.As the reaction is carried in the basic medium in order to blance the number of negative charges add two OH ions on the reatant sides and two `H_(2)O` molecules on the product side.
`2[Cr(OH)_(4)]^(-)+3H_(2)O_(2)rarr2(CrO)_(4)^(2)+6H_(2)O+2H_(2)O`
or `2[Cr(OH)_(4)]^(-)+3H_(2)O_(2)rarr2(CrO)_(4)^(2)+8H_(2)O`
(iii) `MnO_(4)^(-)+Fe^(2)rarrMn^(2)+Fe^(3)+H_(2)O`(Acidic medium)
The balancing is done in the following steps:
1. Write the O.N of each atom in the skeleton equation
`(overset(+7)(Mn)overset(-2)O_(4)) + (overset(+2)Fe)^(2+)rarr(overset(+2)Mn)^(2+)+(overset(+3)Fe)^(3+)+overset(+1)H_(2)overset(-2)O`
2. Identify the atoms which undergo change in O.N.
`(overset(+7)(Mn)O_(4))^(-) + (overset(+2)Fe)^(2+)rarr(overset(+2)Mn)^(2+)+(overset(+3)Fe)^(3+)+H_(2)O`
3. Calculate the increase and decrease in O.N w.r.t reactant atoms.
4.Equate the increase and decrease in O.N on the reactant side.
`MnO_(4)^(-) + (5Fe)^(2+)rarrMn^(2+)+Fe^(3+)+H_(2)O`
5. Balance Mn and Fe atoms on both sides of the equation
`MnO_(4)^(-) (3+)+H_(2)O`
6.As the reaction is carried in the acidic mediium to balace O atoms, add three `H_(2)O` molecules on the product side
`MnO_(4)^(-) + (5Fe)^(2+)rarrMn^(2+)+Fe^(3+)+H_(2)O`
7.In order to balance H atoms, add `8H^(+)` on the reactant side.
`MnO_(4)^(-) + (5Fe)^(2+)rarrMn^(2+)+Fe^(3+)+H_(2)O`
The final equation is balaced w.r.t charge also.


Discussion

No Comment Found

Related InterviewSolutions