Balance the following chemical equation: Mastering Chemical Equation Balancing
More than 30 important balanced reactions with solutions
Introduction
Chemical equations are the language of chemistry, representing chemical reactions in a concise and standardized way. Balancing these equations is a fundamental skill that ensures we’re correctly accounting for all atoms in a reaction, adhering to the law of conservation of mass.
This comprehensive guide will walk you through the principles of balancing chemical equations, provide a step-by-step methodology, and offer over 30 practice problems with detailed solutions to help you master this essential skill.
Why Balance Chemical Equations?
Conservation of Mass
According to the law of conservation of mass, matter cannot be created or destroyed in a chemical reaction. Balancing ensures we account for every atom, maintaining this fundamental principle.
Reaction Stoichiometry
Balanced equations allow chemists to determine the exact quantities of reactants needed and products formed, essential for laboratory work and industrial processes.
Scientific Communication
Properly balanced equations provide a standard way to communicate chemical reactions across the scientific community, ensuring clarity and precision.
Problem Solving
Balancing is the first step in solving many chemistry problems, from calculating reaction yields to determining limiting reagents.
Basic Principles of Balancing
Before diving into practice problems, let’s understand the key principles and strategies for balancing chemical equations:
General Steps for Balancing Chemical Equations:
- Write the unbalanced equation with correct formulas for all reactants and products.
- Count the atoms of each element on both sides of the equation.
- Begin balancing with elements that appear in only one compound on each side.
- Balance polyatomic ions (if present) as a unit when possible.
- Leave hydrogen and oxygen for last (unless they appear in only one compound).
- Use fractional coefficients if needed during the process, but convert to whole numbers at the end.
- Verify that all atoms are balanced by recounting.
Important Note:
Always adjust the coefficients (numbers in front of compounds) to balance an equation, never change the subscripts (numbers within chemical formulas), as this would change the actual compounds involved.
Practice Problems
Below are 30+ chemical equations for you to practice balancing. Click the “Show Solution” button to reveal a detailed step-by-step solution for each equation.
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 2 H atoms and 2 O atoms
Products: 2 H atoms and 1 O atom
Step 2: The O atoms are not balanced (2 on the left, 1 on the right).
To balance O atoms, add a coefficient of 2 before H₂O:
H₂ + O₂ → 2H₂O
Step 3: Recount the atoms.
Reactants: 2 H atoms and 2 O atoms
Products: 4 H atoms and 2 O atoms
Step 4: The H atoms are not balanced (2 on the left, 4 on the right).
To balance H atoms, add a coefficient of 2 before H₂:
2H₂ + O₂ → 2H₂O
Step 5: Final verification
Reactants: 4 H atoms and 2 O atoms
Products: 4 H atoms and 2 O atoms
The equation is now balanced!
Balanced equation: 2H₂ + O₂ → 2H₂O
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 3 C atoms, 8 H atoms, and 2 O atoms
Products: 1 C atom, 2 H atoms, and 3 O atoms
Step 2: Balance C atoms first.
Add a coefficient of 3 before CO₂:
C₃H₈ + O₂ → 3CO₂ + H₂O
Step 3: Balance H atoms next.
There are 8 H atoms on the left and 2 on the right.
Add a coefficient of 4 before H₂O:
C₃H₈ + O₂ → 3CO₂ + 4H₂O
Step 4: Balance O atoms last.
Count O atoms: 2 on the left, (3×2) + (4×1) = 10 on the right
Add a coefficient of 5 before O₂:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Step 5: Final verification
Reactants: 3 C atoms, 8 H atoms, and 10 O atoms
Products: 3 C atoms, 8 H atoms, and 10 O atoms
The equation is balanced!
Balanced equation: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 1 Fe atom and 2 O atoms
Products: 2 Fe atoms and 3 O atoms
Step 2: Balance Fe atoms first.
Add a coefficient of 2 before Fe:
2Fe + O₂ → Fe₂O₃
Step 3: Check O atoms.
The O atoms are not yet balanced (2 on the left, 3 on the right).
Step 4: To balance O atoms, try using fractions temporarily.
We need 3/2 O₂ molecules to get 3 O atoms:
2Fe + 3/2O₂ → Fe₂O₃
Step 5: Multiply all coefficients by 2 to eliminate fractions:
4Fe + 3O₂ → 2Fe₂O₃
Step 6: Final verification
Reactants: 4 Fe atoms and 6 O atoms
Products: 4 Fe atoms and 6 O atoms
The equation is balanced!
Balanced equation: 4Fe + 3O₂ → 2Fe₂O₃
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 1 H atom, 1 Cl atom, 1 Na atom, 1 O atom, 1 H atom
Products: 1 Na atom, 1 Cl atom, 2 H atoms, 1 O atom
Step 2: All atoms are already balanced!
| Element | Reactants | Products | Status |
|---|---|---|---|
| H | 2 (1 from HCl + 1 from NaOH) | 2 (in H₂O) | Balanced ✓ |
| Cl | 1 | 1 | Balanced ✓ |
| Na | 1 | 1 | Balanced ✓ |
| O | 1 | 1 | Balanced ✓ |
Balanced equation: HCl + NaOH → NaCl + H₂O
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 1 Al atom, 1 H atom, 1 Cl atom
Products: 1 Al atom, 3 Cl atoms, 2 H atoms
Step 2: Balance Cl atoms first.
Add a coefficient of 3 before HCl:
Al + 3HCl → AlCl₃ + H₂
Step 3: Check H atoms.
Reactants: 3 H atoms (from 3HCl)
Products: 2 H atoms (from H₂)
The H atoms are not balanced yet.
Step 4: Balance H atoms.
We need 3/2 H₂ molecules to get 3 H atoms.
Al + 3HCl → AlCl₃ + 3/2H₂
Step 5: Multiply all coefficients by 2 to eliminate fractions:
2Al + 6HCl → 2AlCl₃ + 3H₂
Step 6: Final verification
Reactants: 2 Al atoms, 6 H atoms, 6 Cl atoms
Products: 2 Al atoms, 6 H atoms, 6 Cl atoms
The equation is balanced!
Balanced equation: 2Al + 6HCl → 2AlCl₃ + 3H₂
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 1 C atom, 4 H atoms, 2 O atoms
Products: 1 C atom, 2 H atoms, 3 O atoms
Step 2: C is already balanced (1 on each side).
Step 3: Balance H atoms.
Add a coefficient of 2 before H₂O:
CH₄ + O₂ → CO₂ + 2H₂O
Step 4: Check O atoms.
Reactants: 2 O atoms
Products: 4 O atoms (2 from CO₂ + 2 from 2H₂O)
The O atoms are not balanced yet.
Step 5: Balance O atoms.
Add a coefficient of 2 before O₂:
CH₄ + 2O₂ → CO₂ + 2H₂O
Step 6: Final verification
Reactants: 1 C atom, 4 H atoms, 4 O atoms
Products: 1 C atom, 4 H atoms, 4 O atoms
The equation is balanced!
Balanced equation: CH₄ + 2O₂ → CO₂ + 2H₂O
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 1 K atom, 1 Cl atom, 3 O atoms
Products: 1 K atom, 1 Cl atom, 2 O atoms
Step 2: K and Cl are already balanced (1 of each on both sides).
Step 3: The O atoms are not balanced (3 on the left, 2 on the right).
We need to ensure the total number of O atoms is equal on both sides.
Step 4: Since we have 3 O atoms on the left, we need multiples of 3 on both sides.
Let’s try with 2 KClO₃ molecules:
2KClO₃ → 2KCl + ?O₂
Now we have 6 O atoms on the left, so we need 6 on the right.
Each O₂ has 2 O atoms, so we need 3 O₂ molecules:
2KClO₃ → 2KCl + 3O₂
Step 5: Final verification
Reactants: 2 K atoms, 2 Cl atoms, 6 O atoms
Products: 2 K atoms, 2 Cl atoms, 6 O atoms
The equation is balanced!
Balanced equation: 2KClO₃ → 2KCl + 3O₂
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 1 Zn atom, 1 H atom, 1 Cl atom
Products: 1 Zn atom, 2 Cl atoms, 2 H atoms
Step 2: Zn is already balanced (1 on each side).
Step 3: Balance Cl atoms.
Add a coefficient of 2 before HCl:
Zn + 2HCl → ZnCl₂ + H₂
Step 4: Check H atoms.
Reactants: 2 H atoms (from 2HCl)
Products: 2 H atoms (from H₂)
H atoms are now balanced.
Step 5: Final verification
Reactants: 1 Zn atom, 2 H atoms, 2 Cl atoms
Products: 1 Zn atom, 2 H atoms, 2 Cl atoms
The equation is balanced!
Balanced equation: Zn + 2HCl → ZnCl₂ + H₂
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 3 Na atoms, 1 P atom, 4 O atoms, 1 Ca atom, 2 Cl atoms
Products: 3 Ca atoms, 2 P atoms, 8 O atoms, 1 Na atom, 1 Cl atom
Step 2: Let’s start with balancing P atoms.
There’s 1 P on the left and 2 P on the right.
Add a coefficient of 2 before Na₃PO₄:
2Na₃PO₄ + CaCl₂ → Ca₃(PO₄)₂ + NaCl
Step 3: Balance Ca atoms.
There’s 1 Ca on the left and 3 Ca on the right.
Add a coefficient of 3 before CaCl₂:
2Na₃PO₄ + 3CaCl₂ → Ca₃(PO₄)₂ + NaCl
Step 4: Now check Na atoms.
There are 6 Na atoms on the left (2 × 3) and 1 Na atom on the right.
Change the coefficient for NaCl to 6:
2Na₃PO₄ + 3CaCl₂ → Ca₃(PO₄)₂ + 6NaCl
Step 5: Check Cl atoms.
There are 6 Cl atoms on the left (3 × 2) and 6 Cl atoms on the right (6 × 1).
Cl atoms are balanced.
Step 6: Final verification
Reactants: 6 Na atoms, 2 P atoms, 8 O atoms, 3 Ca atoms, 6 Cl atoms
Products: 3 Ca atoms, 2 P atoms, 8 O atoms, 6 Na atoms, 6 Cl atoms
The equation is balanced!
Balanced equation: 2Na₃PO₄ + 3CaCl₂ → Ca₃(PO₄)₂ + 6NaCl
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 6 C atoms, 12 H atoms, 6 O atoms, 2 O atoms
Products: 1 C atom, 2 H atoms, 3 O atoms
Step 2: Balance C atoms first.
Add a coefficient of 6 before CO₂:
C₆H₁₂O₆ + O₂ → 6CO₂ + H₂O
Step 3: Balance H atoms next.
There are 12 H atoms on the left and 2 on the right.
Add a coefficient of 6 before H₂O:
C₆H₁₂O₆ + O₂ → 6CO₂ + 6H₂O
Step 4: Finally, balance O atoms.
Count O atoms: (6 + 2) = 8 on the left, (6×2) + (6×1) = 18 on the right
We need 10 more O atoms on the left.
Since each O₂ provides 2 oxygen atoms, we need 5 more O₂ molecules.
Total O₂ needed: 1 + 5 = 6
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O
Step 5: Final verification
Reactants: 6 C atoms, 12 H atoms, (6 + 12) = 18 O atoms
Products: 6 C atoms, 12 H atoms, (12 + 6) = 18 O atoms
The equation is balanced!
Balanced equation: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 2 Fe atoms, 3 O atoms, 1 C atom
Products: 1 Fe atom, 1 C atom, 2 O atoms
Step 2: Balance Fe atoms first.
Add a coefficient of 2 before Fe:
Fe₂O₃ + C → 2Fe + CO₂
Step 3: Now check O atoms.
There are 3 O atoms on the left and 2 on the right.
To balance O atoms, we need to ensure there’s a common multiple. The LCM of 2 and 3 is 6.
Step 4: Let’s try with 2 Fe₂O₃ molecules:
2Fe₂O₃ + C → 4Fe + CO₂
Now we have 6 O atoms on the left and 2 on the right.
We need 3 CO₂ molecules to get 6 O atoms on the right:
2Fe₂O₃ + C → 4Fe + 3CO₂
Step 5: But now C atoms are not balanced (1 on the left, 3 on the right).
Add a coefficient of 3 before C:
2Fe₂O₃ + 3C → 4Fe + 3CO₂
Step 6: Final verification
Reactants: 4 Fe atoms, 6 O atoms, 3 C atoms
Products: 4 Fe atoms, 6 O atoms, 3 C atoms
The equation is balanced!
Balanced equation: 2Fe₂O₃ + 3C → 4Fe + 3CO₂
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 1 Cu atom, 1 H atom, 1 N atom, 3 O atoms
Products: 1 Cu atom, 2 N atoms, 6 O atoms, 1 N atom, 1 O atom, 2 H atoms, 1 O atom
Step 2: This is a redox reaction, which makes balancing more complex.
Let’s start by balancing Cu.
Cu is already balanced (1 on each side).
Step 3: Let’s balance N atoms.
There is 1 N on the left and 3 N on the right (2 in Cu(NO₃)₂ + 1 in NO).
Add a coefficient of 3 before HNO₃:
Cu + 3HNO₃ → Cu(NO₃)₂ + NO + H₂O
Step 4: Check H atoms.
There are 3 H atoms on the left and 2 on the right.
We need to add a coefficient before H₂O to balance hydrogen:
Since we need 3 H atoms total, and H₂O has 2 H atoms, we need 3/2 H₂O.
Let’s modify our approach to avoid fractions.
Step 5: Let’s try with 3 Cu atoms and scale everything:
3Cu + ?HNO₃ → 3Cu(NO₃)₂ + ?NO + ?H₂O
For N atoms to balance, we need 9 N atoms on the left (for 6 in Cu(NO₃)₂ + 3 in NO).
So we need 9 HNO₃:
3Cu + 9HNO₃ → 3Cu(NO₃)₂ + 3NO + ?H₂O
Step 6: Now balance H atoms.
We have 9 H atoms on the left, so we need 9/2 H₂O molecules.
Multiply everything by 2 to eliminate fractions:
6Cu + 18HNO₃ → 6Cu(NO₃)₂ + 6NO + 9H₂O
Step 7: Check O atoms.
Left side: 18 × 3 = 54 O atoms
Right side: (6 × 6) + (6 × 1) + (9 × 1) = 36 + 6 + 9 = 51 O atoms
The O atoms are not balanced yet. We need 3 more O atoms on the right.
Step 8: To balance O atoms, one approach is to modify the coefficient of one of the oxygen-containing products.
Let’s try adding water:
6Cu + 18HNO₃ → 6Cu(NO₃)₂ + 6NO + 12H₂O
This would give us 54 O atoms on the right, but would add 6 more H atoms.
Another approach is to add more NO molecules:
6Cu + 18HNO₃ → 6Cu(NO₃)₂ + 9NO + 9H₂O
This gives us: 36 + 9 + 9 = 54 O atoms on the right side.
Step 9: Final verification
Reactants: 6 Cu atoms, 18 H atoms, 18 N atoms, 54 O atoms
Products: 6 Cu atoms, 18 H atoms, (12 + 9) = 21 N atoms, 54 O atoms
N atoms are still not balanced. Let’s revise.
Step 10: Let’s start over with the correct approach for this redox reaction:
3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O
Step 11: Final verification
Reactants: 3 Cu atoms, 8 H atoms, 8 N atoms, 24 O atoms
Products: 3 Cu atoms, 8 H atoms, (6 + 2) = 8 N atoms, (18 + 2 + 4) = 24 O atoms
The equation is balanced!
Balanced equation: 3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O
Step-by-Step Solution:
Step 1: This is a complex redox reaction. Let’s count atoms on both sides.
Reactants: 2 K atoms, 2 Cr atoms, 7 O atoms, 1 H atom, 1 Cl atom
Products: 1 K atom, 1 Cl atom, 1 Cr atom, 3 Cl atoms, 2 H atoms, 1 O atom, 2 Cl atoms
Step 2: Balance K atoms first.
Add a coefficient of 2 before KCl:
K₂Cr₂O₇ + HCl → 2KCl + CrCl₃ + H₂O + Cl₂
Step 3: Balance Cr atoms.
Add a coefficient of 2 before CrCl₃:
K₂Cr₂O₇ + HCl → 2KCl + 2CrCl₃ + H₂O + Cl₂
Step 4: Now let’s focus on O atoms.
There are 7 O atoms on the left and 1 on the right.
Add a coefficient of 7 before H₂O:
K₂Cr₂O₇ + HCl → 2KCl + 2CrCl₃ + 7H₂O + Cl₂
Step 5: Check H atoms.
There are 1 H atoms on the left and 14 H atoms on the right (7 × 2).
Add a coefficient of 14 before HCl:
K₂Cr₂O₇ + 14HCl → 2KCl + 2CrCl₃ + 7H₂O + Cl₂
Step 6: Finally, check Cl atoms.
Left side: 14 Cl atoms
Right side: 2 + 6 + 2 = 10 Cl atoms
We’re missing 4 Cl atoms on the right.
Adjust the coefficient for Cl₂ to 3:
K₂Cr₂O₇ + 14HCl → 2KCl + 2CrCl₃ + 7H₂O + 3Cl₂
Step 7: Final verification
Reactants: 2 K atoms, 2 Cr atoms, 7 O atoms, 14 H atoms, 14 Cl atoms
Products: 2 K atoms, 2 Cr atoms, 7 O atoms, 14 H atoms, (2 + 6 + 6) = 14 Cl atoms
The equation is balanced!
Balanced equation: K₂Cr₂O₇ + 14HCl → 2KCl + 2CrCl₃ + 7H₂O + 3Cl₂
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 1 Pb atom, 2 N atoms, 6 O atoms, 1 K atom, 1 I atom
Products: 1 Pb atom, 2 I atoms, 1 K atom, 1 N atom, 3 O atoms
Step 2: Balance I atoms first.
Add a coefficient of 2 before KI:
Pb(NO₃)₂ + 2KI → PbI₂ + KNO₃
Step 3: Balance K atoms.
There are 2 K atoms on the left and 1 on the right.
Add a coefficient of 2 before KNO₃:
Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃
Step 4: Check N and O atoms.
N atoms: 2 on the left, 2 on the right ✓
O atoms: 6 on the left, 6 on the right (2 × 3) ✓
Step 5: Final verification
Reactants: 1 Pb atom, 2 N atoms, 6 O atoms, 2 K atoms, 2 I atoms
Products: 1 Pb atom, 2 I atoms, 2 K atoms, 2 N atoms, 6 O atoms
The equation is balanced!
Balanced equation: Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 1 Ca atom, 2 O atoms, 2 H atoms, 3 H atoms, 1 P atom, 4 O atoms
Products: 3 Ca atoms, 2 P atoms, 8 O atoms, 2 H atoms, 1 O atom
Step 2: Balance Ca atoms first.
There’s 1 Ca on the left and 3 on the right.
Add a coefficient of 3 before Ca(OH)₂:
3Ca(OH)₂ + H₃PO₄ → Ca₃(PO₄)₂ + H₂O
Step 3: Balance P atoms next.
There’s 1 P on the left and 2 on the right.
Add a coefficient of 2 before H₃PO₄:
3Ca(OH)₂ + 2H₃PO₄ → Ca₃(PO₄)₂ + H₂O
Step 4: Check H atoms.
Left side: (3 × 2) + (2 × 3) = 6 + 6 = 12 H atoms
Right side: 2 H atoms
We need 5 more H₂O molecules on the right:
3Ca(OH)₂ + 2H₃PO₄ → Ca₃(PO₄)₂ + 6H₂O
Step 5: Check O atoms.
Left side: (3 × 2) + (2 × 4) = 6 + 8 = 14 O atoms
Right side: 8 + 6 = 14 O atoms
O atoms are balanced.
Step 6: Final verification
Reactants: 3 Ca atoms, 12 H atoms, 2 P atoms, 14 O atoms
Products: 3 Ca atoms, 12 H atoms, 2 P atoms, 14 O atoms
The equation is balanced!
Balanced equation: 3Ca(OH)₂ + 2H₃PO₄ → Ca₃(PO₄)₂ + 6H₂O
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 1 Fe atom, 3 Cl atoms, 1 Na atom, 1 O atom, 1 H atom
Products: 1 Fe atom, 3 O atoms, 3 H atoms, 1 Na atom, 1 Cl atom
Step 2: Fe is already balanced (1 on each side).
Step 3: Balance OH groups.
We need 3 OH groups on the right, so we need 3 NaOH on the left:
FeCl₃ + 3NaOH → Fe(OH)₃ + NaCl
Step 4: Balance Na and Cl atoms.
Now we have 3 Na atoms on the left and 1 on the right.
Add a coefficient of 3 before NaCl:
FeCl₃ + 3NaOH → Fe(OH)₃ + 3NaCl
Step 5: Final verification
Reactants: 1 Fe atom, 3 Cl atoms, 3 Na atoms, 3 O atoms, 3 H atoms
Products: 1 Fe atom, 3 O atoms, 3 H atoms, 3 Na atoms, 3 Cl atoms
The equation is balanced!
Balanced equation: FeCl₃ + 3NaOH → Fe(OH)₃ + 3NaCl
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 1 N atom, 3 H atoms, 2 O atoms
Products: 1 N atom, 1 O atom, 2 H atoms, 1 O atom
Step 2: N is already balanced (1 on each side).
Step 3: Let’s balance H atoms.
There are 3 H atoms on the left and 2 on the right.
Since we can’t have fractional molecules, let’s try with multiple molecules.
With 4 NH₃ molecules, we’d have 12 H atoms, requiring 6 H₂O molecules:
4NH₃ + O₂ → NO + 6H₂O
Step 4: Now we need to balance N atoms.
We have 4 N atoms on the left, so we need 4 NO molecules on the right:
4NH₃ + O₂ → 4NO + 6H₂O
Step 5: Check O atoms.
Left side: 2 O atoms
Right side: 4 + 6 = 10 O atoms
We need 4 more O₂ molecules on the left:
4NH₃ + 5O₂ → 4NO + 6H₂O
Step 6: Final verification
Reactants: 4 N atoms, 12 H atoms, 10 O atoms
Products: 4 N atoms, 12 H atoms, 10 O atoms
The equation is balanced!
Balanced equation: 4NH₃ + 5O₂ → 4NO + 6H₂O
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 1 Al atom, 2 H atoms, 1 S atom, 4 O atoms
Products: 2 Al atoms, 3 S atoms, 12 O atoms, 2 H atoms
Step 2: Balance Al atoms first.
There’s 1 Al on the left and 2 on the right.
Add a coefficient of 2 before Al:
2Al + H₂SO₄ → Al₂(SO₄)₃ + H₂
Step 3: Balance S atoms next.
There’s 1 S on the left and 3 on the right.
Add a coefficient of 3 before H₂SO₄:
2Al + 3H₂SO₄ → Al₂(SO₄)₃ + H₂
Step 4: Check H atoms.
Left side: 3 × 2 = 6 H atoms
Right side: 2 H atoms
We need 2 more H₂ molecules on the right:
2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂
Step 5: Check O atoms.
Left side: 3 × 4 = 12 O atoms
Right side: 12 O atoms
O atoms are balanced.
Step 6: Final verification
Reactants: 2 Al atoms, 6 H atoms, 3 S atoms, 12 O atoms
Products: 2 Al atoms, 6 H atoms, 3 S atoms, 12 O atoms
The equation is balanced!
Balanced equation: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂
Step-by-Step Solution:
Step 1: This is a complex redox reaction. Let’s count atoms on both sides.
Reactants: 1 K atom, 1 Mn atom, 4 O atoms, 1 H atom, 1 Cl atom
Products: 1 K atom, 1 Cl atom, 1 Mn atom, 2 Cl atoms, 2 H atoms, 1 O atom, 2 Cl atoms
Step 2: K and Mn are already balanced (1 of each on both sides).
Step 3: Let’s balance O atoms.
There are 4 O atoms on the left and 1 on the right.
Add a coefficient of 4 before H₂O:
KMnO₄ + HCl → KCl + MnCl₂ + 4H₂O + Cl₂
Step 4: Balance H atoms.
Right side needs 8 H atoms (4 × 2).
Add a coefficient of 8 before HCl:
KMnO₄ + 8HCl → KCl + MnCl₂ + 4H₂O + Cl₂
Step 5: Check Cl atoms.
Left side: 8 Cl atoms
Right side: 1 + 2 + 2 = 5 Cl atoms
Add a coefficient of 5/2 before Cl₂:
KMnO₄ + 8HCl → KCl + MnCl₂ + 4H₂O + 5/2Cl₂
Step 6: To eliminate fractions, multiply all coefficients by 2:
2KMnO₄ + 16HCl → 2KCl + 2MnCl₂ + 8H₂O + 5Cl₂
Step 7: Final verification
Reactants: 2 K atoms, 2 Mn atoms, 8 O atoms, 16 H atoms, 16 Cl atoms
Products: 2 K atoms, 2 Mn atoms, 8 O atoms, 16 H atoms, (2 + 4 + 10) = 16 Cl atoms
The equation is balanced!
Balanced equation: 2KMnO₄ + 16HCl → 2KCl + 2MnCl₂ + 8H₂O + 5Cl₂
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 2 C atoms, 6 H atoms, 1 O atom, 2 O atoms
Products: 1 C atom, 2 O atoms, 2 H atoms, 1 O atom
Step 2: Balance C atoms first.
Add a coefficient of 2 before CO₂:
C₂H₅OH + O₂ → 2CO₂ + H₂O
Step 3: Balance H atoms next.
There are 6 H atoms on the left and 2 on the right.
Add a coefficient of 3 before H₂O:
C₂H₅OH + O₂ → 2CO₂ + 3H₂O
Step 4: Check O atoms.
Left side: 1 + 2 = 3 O atoms
Right side: 4 + 3 = 7 O atoms
Add a coefficient of 3 before O₂:
C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
Step 5: Final verification
Reactants: 2 C atoms, 6 H atoms, 7 O atoms
Products: 2 C atoms, 6 H atoms, 7 O atoms
The equation is balanced!
Balanced equation: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 1 P atom, 5 Cl atoms, 2 H atoms, 1 O atom
Products: 3 H atoms, 1 P atom, 4 O atoms, 1 H atom, 1 Cl atom
Step 2: P is already balanced (1 on each side).
Step 3: Balance Cl atoms.
There are 5 Cl atoms on the left and 1 on the right.
Add a coefficient of 5 before HCl:
PCl₅ + H₂O → H₃PO₄ + 5HCl
Step 4: Check H atoms.
Left side: 2 H atoms
Right side: 3 + 5 = 8 H atoms
Add a coefficient of 4 before H₂O:
PCl₅ + 4H₂O → H₃PO₄ + 5HCl
Step 5: Check O atoms.
Left side: 4 O atoms
Right side: 4 O atoms
O atoms are balanced.
Step 6: Final verification
Reactants: 1 P atom, 5 Cl atoms, 8 H atoms, 4 O atoms
Products: 1 P atom, 5 Cl atoms, 8 H atoms, 4 O atoms
The equation is balanced!
Balanced equation: PCl₅ + 4H₂O → H₃PO₄ + 5HCl
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 1 K atom, 1 O atom, 1 H atom, 3 H atoms, 1 P atom, 4 O atoms
Products: 3 K atoms, 1 P atom, 4 O atoms, 2 H atoms, 1 O atom
Step 2: Balance K atoms first.
There’s 1 K on the left and 3 on the right.
Add a coefficient of 3 before KOH:
3KOH + H₃PO₄ → K₃PO₄ + H₂O
Step 3: Check H atoms.
Left side: 3 + 3 = 6 H atoms
Right side: 2 H atoms
Add a coefficient of 3 before H₂O:
3KOH + H₃PO₄ → K₃PO₄ + 3H₂O
Step 4: Check O atoms.
Left side: 3 + 4 = 7 O atoms
Right side: 4 + 3 = 7 O atoms
O atoms are balanced.
Step 5: Final verification
Reactants: 3 K atoms, 6 H atoms, 1 P atom, 7 O atoms
Products: 3 K atoms, 6 H atoms, 1 P atom, 7 O atoms
The equation is balanced!
Balanced equation: 3KOH + H₃PO₄ → K₃PO₄ + 3H₂O
Step-by-Step Solution:
Step 1: This is a redox reaction. Let’s count atoms on both sides.
Reactants: 1 Mg atom, 1 H atom, 1 N atom, 3 O atoms
Products: 1 Mg atom, 2 N atoms, 6 O atoms, 2 H atoms, 1 N atom, 1 O atom
Step 2: Mg is already balanced (1 on each side).
Step 3: Let’s balance N atoms.
There’s 1 N on the left and 3 N on the right.
Add a coefficient of 3 before HNO₃:
Mg + 3HNO₃ → Mg(NO₃)₂ + H₂ + NO
Step 4: Check H atoms.
Left side: 3 H atoms
Right side: 2 H atoms
The H atoms are not balanced yet. Let’s set this aside for now.
Step 5: Let’s start with a different approach for this redox reaction.
The balanced equation is:
4Mg + 10HNO₃ → 4Mg(NO₃)₂ + 5H₂ + NO
Step 6: Final verification
Reactants: 4 Mg atoms, 10 H atoms, 10 N atoms, 30 O atoms
Products: 4 Mg atoms, 10 H atoms, (8 + 1) = 9 N atoms, (24 + 1) = 25 O atoms
There’s a problem. Let’s rebalance carefully.
Step 7: The correct balanced equation is:
4Mg + 10HNO₃ → 4Mg(NO₃)₂ + 5H₂ + N₂O
Step 8: Final verification
Reactants: 4 Mg atoms, 10 H atoms, 10 N atoms, 30 O atoms
Products: 4 Mg atoms, 10 H atoms, (8 + 2) = 10 N atoms, (24 + 1) = 25 O atoms
The O atoms are still not balanced. Let’s try again.
Step 9: The correct balanced equation for the given products is:
Mg + 4HNO₃ → Mg(NO₃)₂ + 2H₂O + 2NO
Step 10: Final verification
Reactants: 1 Mg atom, 4 H atoms, 4 N atoms, 12 O atoms
Products: 1 Mg atom, 4 H atoms, 4 N atoms, 12 O atoms
The equation is balanced!
Balanced equation: Mg + 4HNO₃ → Mg(NO₃)₂ + 2H₂O + 2NO
Note: The original equation had H₂ as a product, but this reaction typically produces H₂O instead of H₂. The balanced equation shown includes the corrected products.
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 1 S atom, 2 O atoms, 2 O atoms
Products: 1 S atom, 3 O atoms
Step 2: S is already balanced (1 on each side).
Step 3: Check O atoms.
Left side: 2 + 2 = 4 O atoms
Right side: 3 O atoms
The O atoms don’t balance in a 1:1 ratio.
Step 4: Let’s try with 2 SO₂ molecules:
2SO₂ + O₂ → 2SO₃
Step 5: Final verification
Reactants: 2 S atoms, (4 + 2) = 6 O atoms
Products: 2 S atoms, 6 O atoms
The equation is balanced!
Balanced equation: 2SO₂ + O₂ → 2SO₃
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 6 C atoms, 12 H atoms, 6 O atoms
Products: 2 C atoms, 6 H atoms, 1 O atom, 1 C atom, 2 O atoms
Step 2: Let’s check C atoms first.
Left side: 6 C atoms
Right side: 2 + 1 = 3 C atoms
The C atoms are not balanced.
Step 3: This is a fermentation reaction, which is typically written as:
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
Step 4: Final verification
Reactants: 6 C atoms, 12 H atoms, 6 O atoms
Products: (4 + 2) = 6 C atoms, 12 H atoms, (2 + 4) = 6 O atoms
The equation is balanced!
Balanced equation: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 1 Fe atom, 2 S atoms, 2 O atoms
Products: 2 Fe atoms, 3 O atoms, 1 S atom, 2 O atoms
Step 2: Balance Fe atoms first.
There’s 1 Fe on the left and 2 on the right.
Add a coefficient of 2 before FeS₂:
2FeS₂ + O₂ → Fe₂O₃ + SO₂
Step 3: Now check S atoms.
Left side: 2 × 2 = 4 S atoms
Right side: 1 S atom
Add a coefficient of 4 before SO₂:
2FeS₂ + O₂ → Fe₂O₃ + 4SO₂
Step 4: Check O atoms.
Left side: 2 O atoms
Right side: 3 + (4 × 2) = 11 O atoms
Add a coefficient of 11/2 before O₂:
2FeS₂ + 11/2O₂ → Fe₂O₃ + 4SO₂
Step 5: Multiply all coefficients by 2 to eliminate fractions:
4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
Step 6: Final verification
Reactants: 4 Fe atoms, 8 S atoms, 22 O atoms
Products: 4 Fe atoms, 8 S atoms, (6 + 16) = 22 O atoms
The equation is balanced!
Balanced equation: 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 6 C atoms, 6 H atoms, 2 O atoms
Products: 1 C atom, 2 O atoms, 2 H atoms, 1 O atom
Step 2: Balance C atoms first.
Add a coefficient of 6 before CO₂:
C₆H₆ + O₂ → 6CO₂ + H₂O
Step 3: Balance H atoms next.
There are 6 H atoms on the left and 2 on the right.
Add a coefficient of 3 before H₂O:
C₆H₆ + O₂ → 6CO₂ + 3H₂O
Step 4: Check O atoms.
Left side: 2 O atoms
Right side: (6 × 2) + 3 = 15 O atoms
Add a coefficient of 15/2 before O₂:
C₆H₆ + 15/2O₂ → 6CO₂ + 3H₂O
Step 5: Multiply all coefficients by 2 to eliminate fractions:
2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O
Step 6: Final verification
Reactants: 12 C atoms, 12 H atoms, 30 O atoms
Products: 12 C atoms, 12 H atoms, (24 + 6) = 30 O atoms
The equation is balanced!
Balanced equation: 2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 1 Ca atom, 2 O atoms, 2 H atoms, 1 C atom, 2 O atoms
Products: 1 Ca atom, 1 C atom, 3 O atoms, 2 H atoms, 1 O atom
Step 2: Check each element.
- Ca: 1 on both sides ✓
- C: 1 on both sides ✓
- H: 2 on both sides ✓
- O: 4 on the left (2 + 2), 4 on the right (3 + 1) ✓
Step 3: Final verification
All atoms are already balanced!
Balanced equation: Ca(OH)₂ + CO₂ → CaCO₃ + H₂O
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 2 Al atoms, 3 O atoms, 1 C atom
Products: 1 Al atom, 1 C atom, 1 O atom
Step 2: Balance Al atoms first.
Add a coefficient of 2 before Al:
Al₂O₃ + C → 2Al + CO
Step 3: Check O atoms.
Left side: 3 O atoms
Right side: 1 O atom
We need 2 more CO molecules on the right:
Al₂O₃ + C → 2Al + 3CO
Step 4: But now C atoms are not balanced (1 on the left, 3 on the right).
Add a coefficient of 3 before C:
Al₂O₃ + 3C → 2Al + 3CO
Step 5: Final verification
Reactants: 2 Al atoms, 3 O atoms, 3 C atoms
Products: 2 Al atoms, 3 O atoms, 3 C atoms
The equation is balanced!
Balanced equation: Al₂O₃ + 3C → 2Al + 3CO
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 4 C atoms, 10 H atoms, 2 O atoms
Products: 1 C atom, 2 O atoms, 2 H atoms, 1 O atom
Step 2: Balance C atoms first.
Add a coefficient of 4 before CO₂:
C₄H₁₀ + O₂ → 4CO₂ + H₂O
Step 3: Balance H atoms next.
There are 10 H atoms on the left and 2 on the right.
Add a coefficient of 5 before H₂O:
C₄H₁₀ + O₂ → 4CO₂ + 5H₂O
Step 4: Check O atoms.
Left side: 2 O atoms
Right side: (4 × 2) + 5 = 13 O atoms
Add a coefficient of 13/2 before O₂:
C₄H₁₀ + 13/2O₂ → 4CO₂ + 5H₂O
Step 5: Multiply all coefficients by 2 to eliminate fractions:
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
Step 6: Final verification
Reactants: 8 C atoms, 20 H atoms, 26 O atoms
Products: 8 C atoms, 20 H atoms, (16 + 10) = 26 O atoms
The equation is balanced!
Balanced equation: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 1 Zn atom, 1 Ag atom, 1 N atom, 3 O atoms
Products: 1 Zn atom, 2 N atoms, 6 O atoms, 1 Ag atom
Step 2: Zn and Ag are already balanced.
Step 3: Balance N and O atoms.
There’s 1 N on the left and 2 on the right.
Add a coefficient of 2 before AgNO₃:
Zn + 2AgNO₃ → Zn(NO₃)₂ + Ag
Step 4: But now Ag atoms are not balanced (2 on the left, 1 on the right).
Add a coefficient of 2 before Ag:
Zn + 2AgNO₃ → Zn(NO₃)₂ + 2Ag
Step 5: Final verification
Reactants: 1 Zn atom, 2 Ag atoms, 2 N atoms, 6 O atoms
Products: 1 Zn atom, 2 Ag atoms, 2 N atoms, 6 O atoms
The equation is balanced!
Balanced equation: Zn + 2AgNO₃ → Zn(NO₃)₂ + 2Ag
Step-by-Step Solution:
Step 1: Count the atoms on both sides.
Reactants: 2 H atoms, 1 S atom, 4 O atoms, 1 Na atom, 1 Cl atom
Products: 2 Na atoms, 1 S atom, 4 O atoms, 1 H atom, 1 Cl atom
Step 2: Balance Na atoms first.
There’s 1 Na on the left and 2 on the right.
Add a coefficient of 2 before NaCl:
H₂SO₄ + 2NaCl → Na₂SO₄ + HCl
Step 3: Now check Cl atoms.
There are 2 Cl atoms on the left and 1 on the right.
Add a coefficient of 2 before HCl:
H₂SO₄ + 2NaCl → Na₂SO₄ + 2HCl
Step 4: Check remaining atoms.
- H: 2 on the left, 2 on the right ✓
- S: 1 on both sides ✓
- O: 4 on both sides ✓
Step 5: Final verification
Reactants: 2 H atoms, 1 S atom, 4 O atoms, 2 Na atoms, 2 Cl atoms
Products: 2 Na atoms, 1 S atom, 4 O atoms, 2 H atoms, 2 Cl atoms
The equation is balanced!
Balanced equation: H₂SO₄ + 2NaCl → Na₂SO₄ + 2HCl
Common Challenges and Tips
Handling Redox Reactions
Redox reactions can be particularly challenging to balance due to electron transfers. For complex redox reactions, consider using the half-reaction method, which involves:
- Separating the reaction into oxidation and reduction half-reactions
- Balancing each half-reaction separately
- Equalizing the electrons transferred
- Combining the half-reactions
Polyatomic Ions
When balancing equations with polyatomic ions (like NO₃⁻, SO₄²⁻, PO₄³⁻), it’s often easier to:
- Treat the polyatomic ion as a single unit
- Balance these ions first before individual elements
- Verify that all atoms within the polyatomic ions are balanced at the end
Fractional Coefficients
Sometimes, using fractional coefficients temporarily can simplify the balancing process:
- Use fractions to balance atoms if needed
- Once the equation is balanced with fractions, multiply all coefficients by the least common denominator
- This eliminates fractions while maintaining balance
Systematic Approach
Follow a systematic order for balancing most chemical equations:
- Balance metals (except H)
- Balance non-metals (except O and H)
- Balance oxygen
- Balance hydrogen
- Verify the final equation
Conclusion
Balancing chemical equations is a fundamental skill in chemistry that requires practice and systematic thinking. By following the principles outlined in this guide and working through the practice problems, you’ll develop the ability to balance even complex chemical equations.
Remember that balanced equations are essential for understanding reaction stoichiometry, predicting reaction outcomes, and performing quantitative calculations in chemistry. The law of conservation of mass must always be maintained, and properly balanced equations ensure this principle is upheld.
Continue practicing with different types of reactions to strengthen your skills and confidence in this critical area of chemistry.
