Balancing Chemical Reactions with Examples
Balancing Chemical Reactions with Examples
Conservation of Mass Theorem:
In a chemical reaction, mass is conserved, it is not lost or created. Thus;
- Number of atoms of elements are conserved. In other words, sum of the atoms in reactants part is equal to sum of the atoms of products.
- Mass of elements is conserved. Masses of reactants are equal to masses of products.
- Charges of elements/compounds are conserved. Total charges of reactants are equal to total charges of products.
- In a chemical reaction, number of molecules is not conserved always. For example;
H2 + Cl2 → 2HCl
In this reaction; 1 H2 molecule and 1Cl2 molecule reacts and 2 molecule HCl is produced. So, number of molecules is conserved. On the contrary,
2H2 + O2→ 2H2O
In this reaction, 2 molecule H2 react with 1 molecule O2 and 2 molecule H2O is produced. Thus, number of molecules is not conserved.
Example: 4 g substance A reacts with 2,5 g substance B and 1,4 liters C gas and 3,5 g D are produced. Find molar mass of C.
Solution:
In chemical reactions mass is always conserved. So;
Amass + Bmass = Cmass + Dmass
4 + 2,5 = Cmass+ 3,5
Cmass=3 g
In standard conditions, 1mol gas is 22,4 liters, number of moles of C;
nC=1,4/22,4=1/16moles
1/16moles C is 3g
1mol C is X
___________________
X=48 g
Molar mass of C=48 g
- Balancing chemical equations, you should balance H and O after balancing other elements.
- You can multiply coefficients of reaction with fractions like 3/2.
Example: Balance following chemical equation.
C3H4 + O2→ CO2 + H2O
Reactants:
3 C atoms, 4 H atoms, 2 O atoms
Products:
1 C atom, 3 O atoms, 2 H atoms,
To balance reaction, we multiply CO2 with 3 and reaction becomes;
C3H4 + O2→ 3CO2 + H2O
Now, we have 4 H atoms in reactants and 2 H atom in products, so we multiply H2O with 2 and reaction becomes;
C3H4 + O2 → 3CO2 + 2H2O
Now, we have 2 O atoms in reactants and 6+2=8 O atoms in products, to balance O we multiply O2 in reactants with 4.
C3H4+ 4O2 → 3CO2 + 2H2O
Now our reaction is balanced.
Balancing Redox (Oxidation-Reduction) Reactions:
In balancing redox reactions, you should balance number of atoms and charges of matters in reaction. Thus, you should know oxidation states of atoms of elements. We give some examples and try to explain this subject on them.
Example: Balance following reaction;
HNO3 + H2S → NO + H2O + S
Solution:
We first write oxidation states of all elements;
H+N+5O3-2 + H+2S-2 → N+2O-2+ H+2O-2 + S0
Then we write half reactions that show oxidation and reduction (electron transfer) of elements.
Reduction: N+5 +3e- → N+2
Oxidation: S-2 → S0 + 2e-
To balance number of electrons gained and lost, we multiply reduction reaction with 2 and oxidation reaction with 3.
To balance number of N atoms in both sides, we add 2 in front of molecules including N, and to balance number of S atoms we write 3 in front of all matters including S. Now reaction becomes;
2HNO3 + 3H2S → 2NO + H2O + 3S
Now we balance number of H atoms in both sides, there are 8 H atoms in left hand side and 2 H atoms in right hand side. If we multiply H2O in products we balance number of H atoms in reaction. Final reaction becomes;
2HNO3+ 3H2S → 2NO + 4H2O + 3S
You can check number of O in both sides, it is also balanced. We have 6 O atoms in left side and 6 O atoms in right side.
Be Careful!
Balancing (+) charge in ionic reactions taking place in acidic medium, add H+, and to balance number of H and O atoms add H2O.
Example: Balance following reaction in acidic medium;
ClO3- + Cr+3 → ClO2 + Cr2O7-2
We write oxidation states of all elements, be careful, do not confuse charge of ion and oxidation states of elements.
Cl+5O-23- + Cr+3 → Cl+4O-22 + Cr+62O-27-2
Oxidation and reduction half reactions;
Reduction: Cl+5 + e- → Cl+4
Oxidation: Cr+3 → Cr+6 + 3e-
To balance number of electrons in both sides we multiply reduction half reaction with 6 and oxidation half reaction with 2. Now reaction becomes;
6ClO3-+2Cr+3→ 6ClO2 + Cr2O7-2
We calculate electric charges of both sides and balance them.
Left side: ClO3-: 6x(ion charge=-1)=-6
Cr+3: 2x(+3)=+6
-6+6=0
Total electric charge in left side is zero.
Right side: Cr2O7-2: -2
Total charge is -2, balancing electric charge we add H+ ions in acidic mediums. Thus we add 2H+ ion to right side of reaction and make reaction balanced.
6ClO3- +2Cr+3 → 6ClO2 + Cr2O7-2 + 2H+
Now, we must balance number of H atoms in both side of reaction. There is no H atom in lefts side but there are 2H atoms in right side. To balance reaction, we add 1 H2O molecule to left side of reaction and balance it. Final balanced reaction is;
6ClO3- +2Cr+3 + H2O → 6ClO2 + Cr2O7-2 + 2H+
Be Careful!
Balancing (-) charge in ionic reactions taking place in basic medium, add OH-, and to balance number of H and O atoms add OH- ion.
Example: Balance following reaction in basic medium;
Br2→ Br- +BrO3-
We write oxidation states of all elements, be careful, do not confuse charge of ion and oxidation states of elements.
Br20 → Br- +Br+5O-23-
Oxidation and reduction half reactions;
Reduction: Br0 + e- → Br-1
Oxidation: Br0 → Br+5 + 5e-
To balance number of electrons in both sides we multiply reduction half reaction with 5 and oxidation half reaction with 1.To balance number of Br in both sides we add 3 in front of Br in reactants. Now reaction becomes;
3Br2 → 5Br- +BrO3-
We calculate electric charges of both sides and balance them.
Left side: Br2: 0
Total electric charge in left side is zero.
Right side: 5Br- +BrO3-: -5+(-1)=-6
Total charge is -6, balancing electric charge we add OH- ions to left side in basic mediums. Thus we add 6OH- ion to left side of reaction and make reaction balanced.
3Br2 +6OH- → 5Br- + BrO3-
Now, we must balance number of H atoms in both side of reaction. There is no H atom in right side but there are 6H atoms in left side. To balance reaction, we add 3 H2O molecule to right side of reaction and balance it. Final balanced reaction is;
3Br2 +6OH- → 5Br- + BrO3- + 3H2O
Now, all charges and number of atoms are balanced.