Balancing chemical equations worksheet with ten problems to solve.
Worksheet titled "Balancing Equations" with ten chemical equations to balance, including reactants and products like CH₄, O₂, CO₂, H₂O, Na⁺, Cl⁻, NaCl, Al, O₂, Al₂O₃, N₂⁺, H₂, NH₃, CO(g), C₈H₁₈(l), H₂O, FeO₃(s), CO(g), Fe(l), CO₂(g), H₂SO₄, Pb(OH)₄, Pb(SO₄)₂, H₂O, Al, HCl, AlCl₃, H₂, Ca₃(PO₄)₂, H₂SO₄, CaSO₄, Ca(H₂PO₄)₂, H₃PO₄, HCl, PCl₅, H₂O, with spaces for coefficients and a "Name:" and "Date:" field at the top, and "About Chemistry" and a URL at the bottom.
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Step-by-step solution for: 49 Balancing Chemical Equations Worksheets [with Answers]
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Show Answer Key & Explanations
Step-by-step solution for: 49 Balancing Chemical Equations Worksheets [with Answers]
Here are the balanced chemical equations for each problem. I have worked through them step-by-step to ensure the number of atoms for each element is the same on both sides.
1. Combustion of Methane
* Equation: $\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$
* Step 1: Balance Carbon (C). There is 1 on the left and 1 on the right. (Balanced)
* Step 2: Balance Hydrogen (H). There are 4 on the left ($\text{CH}_4$). We need 4 on the right, so we put a 2 in front of $\text{H}_2\text{O}$.
* Current: $\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$
* Step 3: Balance Oxygen (O). On the right, we have 2 (from $\text{CO}_2$) + 2 (from $2\text{H}_2\text{O}$) = 4 oxygen atoms. We need 4 on the left, so we put a 2 in front of $\text{O}_2$.
* Result: $1, 2, 1, 2$
2. Formation of Sodium Chloride
* Equation: $\text{Na}^+ + \text{Cl}^- \rightarrow \text{NaCl}$
* Step 1: Check Sodium (Na). 1 on left, 1 on right.
* Step 2: Check Chlorine (Cl). 1 on left, 1 on right.
* Result: It is already balanced. Coefficients are $1, 1, 1$.
3. Oxidation of Aluminum
* Equation: $\text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3$
* Step 1: Balance Oxygen (O). We have 2 on the left and 3 on the right. The least common multiple is 6. Put a 3 in front of $\text{O}_2$ and a 2 in front of $\text{Al}_2\text{O}_3$.
* Current: $\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3$
* Step 2: Balance Aluminum (Al). On the right, we have $2 \times 2 = 4$ Al atoms. Put a 4 in front of Al on the left.
* Result: $4, 3, 2$
4. Synthesis of Ammonia
*(Note: The image shows $\text{N}_2^+$, which is likely a typo for $\text{N}_2$. We will balance it as standard nitrogen gas $\text{N}_2$.)*
* Equation: $\text{N}_2 + \text{H}_2 \rightarrow \text{NH}_3$
* Step 1: Balance Nitrogen (N). 2 on left, 1 on right. Put a 2 in front of $\text{NH}_3$.
* Current: $\text{N}_2 + \text{H}_2 \rightarrow 2\text{NH}_3$
* Step 2: Balance Hydrogen (H). On the right, we have $2 \times 3 = 6$ H atoms. Put a 3 in front of $\text{H}_2$ on the left.
* Result: $1, 3, 2$
5. Fischer-Tropsch Reaction
* Equation: $\text{CO} + \text{H}_2 \rightarrow \text{C}_8\text{H}_{18} + \text{H}_2\text{O}$
* Step 1: Balance Carbon (C). We need 8 carbons on the left. Put an 8 in front of $\text{CO}$.
* Current: $8\text{CO} + \text{H}_2 \rightarrow \text{C}_8\text{H}_{18} + \text{H}_2\text{O}$
* Step 2: Balance Oxygen (O). We have 8 oxygens on the left. Put an 8 in front of $\text{H}_2\text{O}$.
* Current: $8\text{CO} + \text{H}_2 \rightarrow \text{C}_8\text{H}_{18} + 8\text{H}_2\text{O}$
* Step 3: Balance Hydrogen (H). On the right, we have 18 (from $\text{C}_8\text{H}_{18}$) + 16 (from $8\text{H}_2\text{O}$) = 34 hydrogens. Put a 17 in front of $\text{H}_2$ on the left.
* Result: $8, 17, 1, 8$
6. Reduction of Iron Oxide
*(Note: The image writes $\text{FeO}_3$, but standard iron oxide is $\text{Fe}_2\text{O}_3$. Also $\text{Fe(l)}$ should be $\text{Fe(s)}$ or $\text{(l)}$. Assuming standard Hematite reaction $\text{Fe}_2\text{O}_3$.)*
* Equation: $\text{Fe}_2\text{O}_3 + \text{CO} \rightarrow \text{Fe} + \text{CO}_2$
* Step 1: Balance Iron (Fe). 2 on left, 1 on right. Put a 2 in front of $\text{Fe}$.
* Step 2: Balance Oxygen (O). This is tricky because O is in two products. Let's look at the change from CO to $\text{CO}_2$. Each CO takes one oxygen atom from the iron oxide. $\text{Fe}_2\text{O}_3$ has 3 oxygen atoms to give away. So we need 3 CO molecules.
* Current: $\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$
* Check: Left: 2 Fe, 6 O, 3 C. Right: 2 Fe, 6 O, 3 C. Balanced.
* Result: $1, 3, 2, 3$
7. Sulfuric Acid and Lead Hydroxide
* Equation: $\text{H}_2\text{SO}_4 + \text{Pb(OH)}_4 \rightarrow \text{Pb(SO}_4)_2 + \text{H}_2\text{O}$
* Step 1: Balance Sulfate ($\text{SO}_4$). There are 2 sulfates on the right. Put a 2 in front of $\text{H}_2\text{SO}_4$.
* Current: $2\text{H}_2\text{SO}_4 + \text{Pb(OH)}_4 \rightarrow \text{Pb(SO}_4)_2 + \text{H}_2\text{O}$
* Step 2: Balance Hydrogen (H). On the left, we have $2 \times 2 = 4$ H (from acid) + 4 H (from hydroxide) = 8 H total. Put a 4 in front of $\text{H}_2\text{O}$.
* Check Oxygen (O not in sulfate): Left has 4 (in hydroxide). Right has 4 (in water). Balanced.
* Result: $2, 1, 1, 4$
8. Aluminum and Hydrochloric Acid
* Equation: $\text{Al} + \text{HCl} \rightarrow \text{AlCl}_3 + \text{H}_2$
* Step 1: Balance Chlorine (Cl) and Hydrogen (H). We have 3 Cl on right and 2 H on right. The least common multiple for getting even numbers of Cl and multiples of 3 is 6.
* Step 2: Put a 6 in front of $\text{HCl}$.
* Current: $\text{Al} + 6\text{HCl} \rightarrow \text{AlCl}_3 + \text{H}_2$
* Step 3: Now we have 6 Cl on left, so we need 2 $\text{AlCl}_3$ on right. We have 6 H on left, so we need 3 $\text{H}_2$ on right.
* Current: $\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2$
* Step 4: Balance Aluminum (Al). We have 2 Al on right. Put a 2 in front of Al on left.
* Result: $2, 6, 2, 3$
9. Phosphate Rock Reaction
* Equation: $\text{Ca}_3(\text{PO}_4)_2 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + \text{Ca(H}_2\text{PO}_4)_2$
* Step 1: Balance Calcium (Ca). On the right, we have 1 Ca in $\text{CaSO}_4$ and 1 Ca in $\text{Ca(H}_2\text{PO}_4)_2$, totaling 2 Ca? No, wait. Look at the product side carefully.
* Right Side Total Ca: 1 (from first term) + 1 (from second term) = 2?
* Let's check the Phosphate ($\text{PO}_4$). Left: 2 groups. Right: 2 groups (inside the second term). Phosphorus is balanced with coefficient 1 for the main compounds.
* Let's re-evaluate Calcium. Left: 3 Ca. Right: We need 3 Ca total. One is in $\text{Ca(H}_2\text{PO}_4)_2$. That leaves 2 Ca needed for $\text{CaSO}_4$. So put a 2 in front of $\text{CaSO}_4$.
* Current: $\text{Ca}_3(\text{PO}_4)_2 + \text{H}_2\text{SO}_4 \rightarrow 2\text{CaSO}_4 + \text{Ca(H}_2\text{PO}_4)_2$
* Step 2: Balance Sulfate ($\text{SO}_4$). We have 2 sulfates on the right. Put a 2 in front of $\text{H}_2\text{SO}_4$.
* Step 3: Check Hydrogen (H). Left: $2 \times 2 = 4$ H. Right: Inside $\text{Ca(H}_2\text{PO}_4)_2$, there are $2 \times 2 = 4$ H. Balanced.
* Result: $1, 2, 2, 1$
10. Phosphoric Acid and Hydrochloric Acid
*(Note: This equation as written $\text{H}_3\text{PO}_4 + \text{HCl} \rightarrow \text{PCl}_5 + \text{H}_2\text{O}$ is chemically unusual for a simple mix, usually requiring other reagents like $\text{CCl}_4$ or heat, but we will balance the atoms as presented.)*
* Equation: $\text{H}_3\text{PO}_4 + \text{HCl} \rightarrow \text{PCl}_5 + \text{H}_2\text{O}$
* Step 1: Balance Chlorine (Cl). 5 on right. Put a 5 in front of $\text{HCl}$.
* Current: $\text{H}_3\text{PO}_4 + 5\text{HCl} \rightarrow \text{PCl}_5 + \text{H}_2\text{O}$
* Step 2: Balance Hydrogen (H). Left: 3 (from acid) + 5 (from HCl) = 8 H. Right: Need 8 H, so put a 4 in front of $\text{H}_2\text{O}$.
* Step 3: Check Oxygen (O). Left: 4 O. Right: 4 O (in $4\text{H}_2\text{O}$). Balanced.
* Step 4: Check Phosphorus (P). 1 on left, 1 on right. Balanced.
* Result: $1, 5, 1, 4$
Final Answer:
1. 1 $\text{CH}_4$ + 2 $\text{O}_2$ $\rightarrow$ 1 $\text{CO}_2$ + 2 $\text{H}_2\text{O}$
2. 1 $\text{Na}^+$ + 1 $\text{Cl}^-$ $\rightarrow$ 1 $\text{NaCl}$
3. 4 $\text{Al}$ + 3 $\text{O}_2$ $\rightarrow$ 2 $\text{Al}_2\text{O}_3$
4. 1 $\text{N}_2$ + 3 $\text{H}_2$ $\rightarrow$ 2 $\text{NH}_3$
5. 8 $\text{CO}$ + 17 $\text{H}_2$ $\rightarrow$ 1 $\text{C}_8\text{H}_{18}$ + 8 $\text{H}_2\text{O}$
6. 1 $\text{Fe}_2\text{O}_3$ + 3 $\text{CO}$ $\rightarrow$ 2 $\text{Fe}$ + 3 $\text{CO}_2$
7. 2 $\text{H}_2\text{SO}_4$ + 1 $\text{Pb(OH)}_4$ $\rightarrow$ 1 $\text{Pb(SO}_4)_2$ + 4 $\text{H}_2\text{O}$
8. 2 $\text{Al}$ + 6 $\text{HCl}$ $\rightarrow$ 2 $\text{AlCl}_3$ + 3 $\text{H}_2$
9. 1 $\text{Ca}_3(\text{PO}_4)_2$ + 2 $\text{H}_2\text{SO}_4$ $\rightarrow$ 2 $\text{CaSO}_4$ + 1 $\text{Ca(H}_2\text{PO}_4)_2$
10. 1 $\text{H}_3\text{PO}_4$ + 5 $\text{HCl}$ $\rightarrow$ 1 $\text{PCl}_5$ + 4 $\text{H}_2\text{O}$
1. Combustion of Methane
* Equation: $\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$
* Step 1: Balance Carbon (C). There is 1 on the left and 1 on the right. (Balanced)
* Step 2: Balance Hydrogen (H). There are 4 on the left ($\text{CH}_4$). We need 4 on the right, so we put a 2 in front of $\text{H}_2\text{O}$.
* Current: $\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$
* Step 3: Balance Oxygen (O). On the right, we have 2 (from $\text{CO}_2$) + 2 (from $2\text{H}_2\text{O}$) = 4 oxygen atoms. We need 4 on the left, so we put a 2 in front of $\text{O}_2$.
* Result: $1, 2, 1, 2$
2. Formation of Sodium Chloride
* Equation: $\text{Na}^+ + \text{Cl}^- \rightarrow \text{NaCl}$
* Step 1: Check Sodium (Na). 1 on left, 1 on right.
* Step 2: Check Chlorine (Cl). 1 on left, 1 on right.
* Result: It is already balanced. Coefficients are $1, 1, 1$.
3. Oxidation of Aluminum
* Equation: $\text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3$
* Step 1: Balance Oxygen (O). We have 2 on the left and 3 on the right. The least common multiple is 6. Put a 3 in front of $\text{O}_2$ and a 2 in front of $\text{Al}_2\text{O}_3$.
* Current: $\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3$
* Step 2: Balance Aluminum (Al). On the right, we have $2 \times 2 = 4$ Al atoms. Put a 4 in front of Al on the left.
* Result: $4, 3, 2$
4. Synthesis of Ammonia
*(Note: The image shows $\text{N}_2^+$, which is likely a typo for $\text{N}_2$. We will balance it as standard nitrogen gas $\text{N}_2$.)*
* Equation: $\text{N}_2 + \text{H}_2 \rightarrow \text{NH}_3$
* Step 1: Balance Nitrogen (N). 2 on left, 1 on right. Put a 2 in front of $\text{NH}_3$.
* Current: $\text{N}_2 + \text{H}_2 \rightarrow 2\text{NH}_3$
* Step 2: Balance Hydrogen (H). On the right, we have $2 \times 3 = 6$ H atoms. Put a 3 in front of $\text{H}_2$ on the left.
* Result: $1, 3, 2$
5. Fischer-Tropsch Reaction
* Equation: $\text{CO} + \text{H}_2 \rightarrow \text{C}_8\text{H}_{18} + \text{H}_2\text{O}$
* Step 1: Balance Carbon (C). We need 8 carbons on the left. Put an 8 in front of $\text{CO}$.
* Current: $8\text{CO} + \text{H}_2 \rightarrow \text{C}_8\text{H}_{18} + \text{H}_2\text{O}$
* Step 2: Balance Oxygen (O). We have 8 oxygens on the left. Put an 8 in front of $\text{H}_2\text{O}$.
* Current: $8\text{CO} + \text{H}_2 \rightarrow \text{C}_8\text{H}_{18} + 8\text{H}_2\text{O}$
* Step 3: Balance Hydrogen (H). On the right, we have 18 (from $\text{C}_8\text{H}_{18}$) + 16 (from $8\text{H}_2\text{O}$) = 34 hydrogens. Put a 17 in front of $\text{H}_2$ on the left.
* Result: $8, 17, 1, 8$
6. Reduction of Iron Oxide
*(Note: The image writes $\text{FeO}_3$, but standard iron oxide is $\text{Fe}_2\text{O}_3$. Also $\text{Fe(l)}$ should be $\text{Fe(s)}$ or $\text{(l)}$. Assuming standard Hematite reaction $\text{Fe}_2\text{O}_3$.)*
* Equation: $\text{Fe}_2\text{O}_3 + \text{CO} \rightarrow \text{Fe} + \text{CO}_2$
* Step 1: Balance Iron (Fe). 2 on left, 1 on right. Put a 2 in front of $\text{Fe}$.
* Step 2: Balance Oxygen (O). This is tricky because O is in two products. Let's look at the change from CO to $\text{CO}_2$. Each CO takes one oxygen atom from the iron oxide. $\text{Fe}_2\text{O}_3$ has 3 oxygen atoms to give away. So we need 3 CO molecules.
* Current: $\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$
* Check: Left: 2 Fe, 6 O, 3 C. Right: 2 Fe, 6 O, 3 C. Balanced.
* Result: $1, 3, 2, 3$
7. Sulfuric Acid and Lead Hydroxide
* Equation: $\text{H}_2\text{SO}_4 + \text{Pb(OH)}_4 \rightarrow \text{Pb(SO}_4)_2 + \text{H}_2\text{O}$
* Step 1: Balance Sulfate ($\text{SO}_4$). There are 2 sulfates on the right. Put a 2 in front of $\text{H}_2\text{SO}_4$.
* Current: $2\text{H}_2\text{SO}_4 + \text{Pb(OH)}_4 \rightarrow \text{Pb(SO}_4)_2 + \text{H}_2\text{O}$
* Step 2: Balance Hydrogen (H). On the left, we have $2 \times 2 = 4$ H (from acid) + 4 H (from hydroxide) = 8 H total. Put a 4 in front of $\text{H}_2\text{O}$.
* Check Oxygen (O not in sulfate): Left has 4 (in hydroxide). Right has 4 (in water). Balanced.
* Result: $2, 1, 1, 4$
8. Aluminum and Hydrochloric Acid
* Equation: $\text{Al} + \text{HCl} \rightarrow \text{AlCl}_3 + \text{H}_2$
* Step 1: Balance Chlorine (Cl) and Hydrogen (H). We have 3 Cl on right and 2 H on right. The least common multiple for getting even numbers of Cl and multiples of 3 is 6.
* Step 2: Put a 6 in front of $\text{HCl}$.
* Current: $\text{Al} + 6\text{HCl} \rightarrow \text{AlCl}_3 + \text{H}_2$
* Step 3: Now we have 6 Cl on left, so we need 2 $\text{AlCl}_3$ on right. We have 6 H on left, so we need 3 $\text{H}_2$ on right.
* Current: $\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2$
* Step 4: Balance Aluminum (Al). We have 2 Al on right. Put a 2 in front of Al on left.
* Result: $2, 6, 2, 3$
9. Phosphate Rock Reaction
* Equation: $\text{Ca}_3(\text{PO}_4)_2 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + \text{Ca(H}_2\text{PO}_4)_2$
* Step 1: Balance Calcium (Ca). On the right, we have 1 Ca in $\text{CaSO}_4$ and 1 Ca in $\text{Ca(H}_2\text{PO}_4)_2$, totaling 2 Ca? No, wait. Look at the product side carefully.
* Right Side Total Ca: 1 (from first term) + 1 (from second term) = 2?
* Let's check the Phosphate ($\text{PO}_4$). Left: 2 groups. Right: 2 groups (inside the second term). Phosphorus is balanced with coefficient 1 for the main compounds.
* Let's re-evaluate Calcium. Left: 3 Ca. Right: We need 3 Ca total. One is in $\text{Ca(H}_2\text{PO}_4)_2$. That leaves 2 Ca needed for $\text{CaSO}_4$. So put a 2 in front of $\text{CaSO}_4$.
* Current: $\text{Ca}_3(\text{PO}_4)_2 + \text{H}_2\text{SO}_4 \rightarrow 2\text{CaSO}_4 + \text{Ca(H}_2\text{PO}_4)_2$
* Step 2: Balance Sulfate ($\text{SO}_4$). We have 2 sulfates on the right. Put a 2 in front of $\text{H}_2\text{SO}_4$.
* Step 3: Check Hydrogen (H). Left: $2 \times 2 = 4$ H. Right: Inside $\text{Ca(H}_2\text{PO}_4)_2$, there are $2 \times 2 = 4$ H. Balanced.
* Result: $1, 2, 2, 1$
10. Phosphoric Acid and Hydrochloric Acid
*(Note: This equation as written $\text{H}_3\text{PO}_4 + \text{HCl} \rightarrow \text{PCl}_5 + \text{H}_2\text{O}$ is chemically unusual for a simple mix, usually requiring other reagents like $\text{CCl}_4$ or heat, but we will balance the atoms as presented.)*
* Equation: $\text{H}_3\text{PO}_4 + \text{HCl} \rightarrow \text{PCl}_5 + \text{H}_2\text{O}$
* Step 1: Balance Chlorine (Cl). 5 on right. Put a 5 in front of $\text{HCl}$.
* Current: $\text{H}_3\text{PO}_4 + 5\text{HCl} \rightarrow \text{PCl}_5 + \text{H}_2\text{O}$
* Step 2: Balance Hydrogen (H). Left: 3 (from acid) + 5 (from HCl) = 8 H. Right: Need 8 H, so put a 4 in front of $\text{H}_2\text{O}$.
* Step 3: Check Oxygen (O). Left: 4 O. Right: 4 O (in $4\text{H}_2\text{O}$). Balanced.
* Step 4: Check Phosphorus (P). 1 on left, 1 on right. Balanced.
* Result: $1, 5, 1, 4$
Final Answer:
1. 1 $\text{CH}_4$ + 2 $\text{O}_2$ $\rightarrow$ 1 $\text{CO}_2$ + 2 $\text{H}_2\text{O}$
2. 1 $\text{Na}^+$ + 1 $\text{Cl}^-$ $\rightarrow$ 1 $\text{NaCl}$
3. 4 $\text{Al}$ + 3 $\text{O}_2$ $\rightarrow$ 2 $\text{Al}_2\text{O}_3$
4. 1 $\text{N}_2$ + 3 $\text{H}_2$ $\rightarrow$ 2 $\text{NH}_3$
5. 8 $\text{CO}$ + 17 $\text{H}_2$ $\rightarrow$ 1 $\text{C}_8\text{H}_{18}$ + 8 $\text{H}_2\text{O}$
6. 1 $\text{Fe}_2\text{O}_3$ + 3 $\text{CO}$ $\rightarrow$ 2 $\text{Fe}$ + 3 $\text{CO}_2$
7. 2 $\text{H}_2\text{SO}_4$ + 1 $\text{Pb(OH)}_4$ $\rightarrow$ 1 $\text{Pb(SO}_4)_2$ + 4 $\text{H}_2\text{O}$
8. 2 $\text{Al}$ + 6 $\text{HCl}$ $\rightarrow$ 2 $\text{AlCl}_3$ + 3 $\text{H}_2$
9. 1 $\text{Ca}_3(\text{PO}_4)_2$ + 2 $\text{H}_2\text{SO}_4$ $\rightarrow$ 2 $\text{CaSO}_4$ + 1 $\text{Ca(H}_2\text{PO}_4)_2$
10. 1 $\text{H}_3\text{PO}_4$ + 5 $\text{HCl}$ $\rightarrow$ 1 $\text{PCl}_5$ + 4 $\text{H}_2\text{O}$
Parent Tip: Review the logic above to help your child master the concept of balancing chemical reactions worksheet 2.
![49 Balancing Chemical Equations Worksheets [with Answers]](/images2/balancing-chemical-reactions-worksheet-2/balancing-chemical-reactions-worksheet-2-13.jpg)
