Worksheet for balancing chemical equations with ten problems.
Balancing Equations worksheet with ten chemical equations to balance, including CH4 + O2 → CO2 + H2O, Na+ + Cl- → NaCl, and Al + O2 → Al2O3, with spaces for coefficients and a "Name" and "Date" field at the top.
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Show Answer Key & Explanations
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]
To balance chemical equations, we need to ensure that the number of atoms of each element is the same on both sides of the equation. Let's solve each equation step by step.
---
- Reactants: 1 C, 4 H, 2 O
- Products: 1 C, 2 H, 3 O
Balance:
- Start with carbon (C): Already balanced (1 C on both sides).
- Balance hydrogen (H): There are 4 H in CH₄ and 2 H in H₂O. To balance H, we need 2 H₂O.
- Balance oxygen (O): There are 2 O in O₂ and 3 O total on the product side (2 O in CO₂ and 1 O in 2 H₂O). To balance O, we need 2 O₂.
Balanced equation:
$$ \boxed{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O} $$
---
- Reactants: 1 Na, 1 Cl
- Products: 1 Na, 1 Cl
This equation is already balanced because there is 1 Na and 1 Cl on both sides.
Balanced equation:
$$ \boxed{Na^+ + Cl^- \rightarrow NaCl} $$
---
- Reactants: 1 Al, 2 O
- Products: 2 Al, 3 O
Balance:
- Start with aluminum (Al): There are 2 Al in Al₂O₃. To balance Al, we need 2 Al on the reactant side.
- Balance oxygen (O): There are 3 O in Al₂O₃. Since O₂ provides 2 O per molecule, we need 1.5 O₂. To avoid fractions, multiply everything by 2.
Balanced equation:
$$ \boxed{4Al + 3O_2 \rightarrow 2Al_2O_3} $$
---
- Reactants: 2 N, 2 H
- Products: 1 N, 3 H
Balance:
- Start with nitrogen (N): There are 2 N in N₂⁺. To balance N, we need 2 NH₃.
- Balance hydrogen (H): There are 6 H in 2 NH₃. Since H₂ provides 2 H per molecule, we need 3 H₂.
Balanced equation:
$$ \boxed{N_2^+ + 3H_2 \rightarrow 2NH_3} $$
---
- Reactants: 1 C, 1 O, 2 H
- Products: 8 C, 18 H, 1 O
Balance:
- Start with carbon (C): There are 8 C in C₈H₁₈. To balance C, we need 8 CO.
- Balance hydrogen (H): There are 20 H in C₈H₁₈ and 2 H in H₂O, totaling 22 H. Since H₂ provides 2 H per molecule, we need 11 H₂.
- Balance oxygen (O): There are 8 O in 8 CO and 1 O in H₂O, totaling 9 O. Since only one O atom is needed from H₂O, the equation is balanced.
Balanced equation:
$$ \boxed{8CO(g) + 11H_2(g) \rightarrow C_8H_{18}(l) + 9H_2O} $$
---
- Reactants: 1 Fe, 3 O, 1 C
- Products: 1 Fe, 2 O, 1 C
Balance:
- Start with iron (Fe): Already balanced (1 Fe on both sides).
- Balance carbon (C): Already balanced (1 C on both sides).
- Balance oxygen (O): There are 3 O in FeO₃ and 2 O in CO₂. To balance O, we need 3 CO.
Balanced equation:
$$ \boxed{FeO_3(s) + 3CO(g) \rightarrow Fe(l) + 3CO_2(g)} $$
---
- Reactants: 2 H, 1 S, 4 O, 1 Pb, 4 O
- Products: 1 Pb, 2 S, 8 O, 2 H
Balance:
- Start with lead (Pb): Already balanced (1 Pb on both sides).
- Balance sulfur (S): There are 2 S in Pb(SO₄)₂. To balance S, we need 2 H₂SO₄.
- Balance hydrogen (H): There are 4 H in 2 H₂SO₄ and 2 H in H₂O. To balance H, we need 2 H₂O.
- Balance oxygen (O): There are 8 O in 2 H₂SO₄ and 4 O in Pb(OH)₄, totaling 12 O. Since Pb(SO₄)₂ provides 8 O and H₂O provides 2 O, we need 2 H₂O.
Balanced equation:
$$ \boxed{2H_2SO_4 + Pb(OH)_4 \rightarrow Pb(SO_4)_2 + 4H_2O} $$
---
- Reactants: 1 Al, 1 H, 1 Cl
- Products: 1 Al, 3 Cl, 2 H
Balance:
- Start with aluminum (Al): Already balanced (1 Al on both sides).
- Balance chlorine (Cl): There are 3 Cl in AlCl₃. To balance Cl, we need 3 HCl.
- Balance hydrogen (H): There are 6 H in 3 HCl and 2 H in H₂. To balance H, we need 3 H₂.
Balanced equation:
$$ \boxed{2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2} $$
---
- Reactants: 3 Ca, 2 P, 8 O, 2 H, 1 S
- Products: 1 Ca, 1 P, 5 O, 2 H, 1 S
Balance:
- Start with calcium (Ca): There are 3 Ca in Ca₃(PO₄)₂. To balance Ca, we need 3 CaSO₄.
- Balance phosphorus (P): There are 2 P in Ca₃(PO₄)₂. To balance P, we need 2 Ca(H₂PO₄)₂.
- Balance sulfur (S): Already balanced (1 S on both sides).
- Balance hydrogen (H): There are 4 H in 2 Ca(H₂PO₄)₂. To balance H, we need 2 H₂SO₄.
- Balance oxygen (O): There are 14 O in 3 CaSO₄ and 16 O in 2 Ca(H₂PO₄)₂, totaling 30 O. Since H₂SO₄ provides 8 O, we need 2 H₂SO₄.
Balanced equation:
$$ \boxed{Ca_3(PO_4)_2 + 2H_2SO_4 \rightarrow 3CaSO_4 + 2Ca(H_2PO_4)_2} $$
---
- Reactants: 3 H, 1 P, 4 O, 1 Cl
- Products: 1 P, 5 Cl, 2 H, 1 O
Balance:
- Start with phosphorus (P): Already balanced (1 P on both sides).
- Balance chlorine (Cl): There are 5 Cl in PCl₅. To balance Cl, we need 5 HCl.
- Balance hydrogen (H): There are 8 H in 5 HCl and 6 H in H₃PO₄, totaling 14 H. Since H₂O provides 2 H per molecule, we need 7 H₂O.
- Balance oxygen (O): There are 4 O in H₃PO₄ and 7 O in 7 H₂O, totaling 11 O. Since only 1 O is needed from PCl₅, the equation is balanced.
Balanced equation:
$$ \boxed{H_3PO_4 + 5HCl \rightarrow PCl_5 + 4H_2O} $$
---
$$
\boxed{
\begin{aligned}
1. & \ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \\
2. & \ Na^+ + Cl^- \rightarrow NaCl \\
3. & \ 4Al + 3O_2 \rightarrow 2Al_2O_3 \\
4. & \ N_2^+ + 3H_2 \rightarrow 2NH_3 \\
5. & \ 8CO(g) + 11H_2(g) \rightarrow C_8H_{18}(l) + 9H_2O \\
6. & \ FeO_3(s) + 3CO(g) \rightarrow Fe(l) + 3CO_2(g) \\
7. & \ 2H_2SO_4 + Pb(OH)_4 \rightarrow Pb(SO_4)_2 + 4H_2O \\
8. & \ 2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2 \\
9. & \ Ca_3(PO_4)_2 + 2H_2SO_4 \rightarrow 3CaSO_4 + 2Ca(H_2PO_4)_2 \\
10. & \ H_3PO_4 + 5HCl \rightarrow PCl_5 + 4H_2O \\
\end{aligned}
}
$$
---
1. CH₄ + O₂ → CO₂ + H₂O
- Reactants: 1 C, 4 H, 2 O
- Products: 1 C, 2 H, 3 O
Balance:
- Start with carbon (C): Already balanced (1 C on both sides).
- Balance hydrogen (H): There are 4 H in CH₄ and 2 H in H₂O. To balance H, we need 2 H₂O.
- Balance oxygen (O): There are 2 O in O₂ and 3 O total on the product side (2 O in CO₂ and 1 O in 2 H₂O). To balance O, we need 2 O₂.
Balanced equation:
$$ \boxed{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O} $$
---
2. Na⁺ + Cl⁻ → NaCl
- Reactants: 1 Na, 1 Cl
- Products: 1 Na, 1 Cl
This equation is already balanced because there is 1 Na and 1 Cl on both sides.
Balanced equation:
$$ \boxed{Na^+ + Cl^- \rightarrow NaCl} $$
---
3. Al + O₂ → Al₂O₃
- Reactants: 1 Al, 2 O
- Products: 2 Al, 3 O
Balance:
- Start with aluminum (Al): There are 2 Al in Al₂O₃. To balance Al, we need 2 Al on the reactant side.
- Balance oxygen (O): There are 3 O in Al₂O₃. Since O₂ provides 2 O per molecule, we need 1.5 O₂. To avoid fractions, multiply everything by 2.
Balanced equation:
$$ \boxed{4Al + 3O_2 \rightarrow 2Al_2O_3} $$
---
4. N₂⁺ + H₂ → NH₃
- Reactants: 2 N, 2 H
- Products: 1 N, 3 H
Balance:
- Start with nitrogen (N): There are 2 N in N₂⁺. To balance N, we need 2 NH₃.
- Balance hydrogen (H): There are 6 H in 2 NH₃. Since H₂ provides 2 H per molecule, we need 3 H₂.
Balanced equation:
$$ \boxed{N_2^+ + 3H_2 \rightarrow 2NH_3} $$
---
5. CO(g) + H₂(g) → C₈H₁₈(l) + H₂O
- Reactants: 1 C, 1 O, 2 H
- Products: 8 C, 18 H, 1 O
Balance:
- Start with carbon (C): There are 8 C in C₈H₁₈. To balance C, we need 8 CO.
- Balance hydrogen (H): There are 20 H in C₈H₁₈ and 2 H in H₂O, totaling 22 H. Since H₂ provides 2 H per molecule, we need 11 H₂.
- Balance oxygen (O): There are 8 O in 8 CO and 1 O in H₂O, totaling 9 O. Since only one O atom is needed from H₂O, the equation is balanced.
Balanced equation:
$$ \boxed{8CO(g) + 11H_2(g) \rightarrow C_8H_{18}(l) + 9H_2O} $$
---
6. FeO₃(s) + CO(g) → Fe(l) + CO₂(g)
- Reactants: 1 Fe, 3 O, 1 C
- Products: 1 Fe, 2 O, 1 C
Balance:
- Start with iron (Fe): Already balanced (1 Fe on both sides).
- Balance carbon (C): Already balanced (1 C on both sides).
- Balance oxygen (O): There are 3 O in FeO₃ and 2 O in CO₂. To balance O, we need 3 CO.
Balanced equation:
$$ \boxed{FeO_3(s) + 3CO(g) \rightarrow Fe(l) + 3CO_2(g)} $$
---
7. H₂SO₄ + Pb(OH)₄ → Pb(SO₄)₂ + H₂O
- Reactants: 2 H, 1 S, 4 O, 1 Pb, 4 O
- Products: 1 Pb, 2 S, 8 O, 2 H
Balance:
- Start with lead (Pb): Already balanced (1 Pb on both sides).
- Balance sulfur (S): There are 2 S in Pb(SO₄)₂. To balance S, we need 2 H₂SO₄.
- Balance hydrogen (H): There are 4 H in 2 H₂SO₄ and 2 H in H₂O. To balance H, we need 2 H₂O.
- Balance oxygen (O): There are 8 O in 2 H₂SO₄ and 4 O in Pb(OH)₄, totaling 12 O. Since Pb(SO₄)₂ provides 8 O and H₂O provides 2 O, we need 2 H₂O.
Balanced equation:
$$ \boxed{2H_2SO_4 + Pb(OH)_4 \rightarrow Pb(SO_4)_2 + 4H_2O} $$
---
8. Al + HCl → AlCl₃ + H₂
- Reactants: 1 Al, 1 H, 1 Cl
- Products: 1 Al, 3 Cl, 2 H
Balance:
- Start with aluminum (Al): Already balanced (1 Al on both sides).
- Balance chlorine (Cl): There are 3 Cl in AlCl₃. To balance Cl, we need 3 HCl.
- Balance hydrogen (H): There are 6 H in 3 HCl and 2 H in H₂. To balance H, we need 3 H₂.
Balanced equation:
$$ \boxed{2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2} $$
---
9. Ca₃(PO₄)₂ + H₂SO₄ → CaSO₄ + Ca(H₂PO₄)₂
- Reactants: 3 Ca, 2 P, 8 O, 2 H, 1 S
- Products: 1 Ca, 1 P, 5 O, 2 H, 1 S
Balance:
- Start with calcium (Ca): There are 3 Ca in Ca₃(PO₄)₂. To balance Ca, we need 3 CaSO₄.
- Balance phosphorus (P): There are 2 P in Ca₃(PO₄)₂. To balance P, we need 2 Ca(H₂PO₄)₂.
- Balance sulfur (S): Already balanced (1 S on both sides).
- Balance hydrogen (H): There are 4 H in 2 Ca(H₂PO₄)₂. To balance H, we need 2 H₂SO₄.
- Balance oxygen (O): There are 14 O in 3 CaSO₄ and 16 O in 2 Ca(H₂PO₄)₂, totaling 30 O. Since H₂SO₄ provides 8 O, we need 2 H₂SO₄.
Balanced equation:
$$ \boxed{Ca_3(PO_4)_2 + 2H_2SO_4 \rightarrow 3CaSO_4 + 2Ca(H_2PO_4)_2} $$
---
10. H₃PO₄ + HCl → PCl₅ + H₂O
- Reactants: 3 H, 1 P, 4 O, 1 Cl
- Products: 1 P, 5 Cl, 2 H, 1 O
Balance:
- Start with phosphorus (P): Already balanced (1 P on both sides).
- Balance chlorine (Cl): There are 5 Cl in PCl₅. To balance Cl, we need 5 HCl.
- Balance hydrogen (H): There are 8 H in 5 HCl and 6 H in H₃PO₄, totaling 14 H. Since H₂O provides 2 H per molecule, we need 7 H₂O.
- Balance oxygen (O): There are 4 O in H₃PO₄ and 7 O in 7 H₂O, totaling 11 O. Since only 1 O is needed from PCl₅, the equation is balanced.
Balanced equation:
$$ \boxed{H_3PO_4 + 5HCl \rightarrow PCl_5 + 4H_2O} $$
---
Final Answer:
$$
\boxed{
\begin{aligned}
1. & \ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \\
2. & \ Na^+ + Cl^- \rightarrow NaCl \\
3. & \ 4Al + 3O_2 \rightarrow 2Al_2O_3 \\
4. & \ N_2^+ + 3H_2 \rightarrow 2NH_3 \\
5. & \ 8CO(g) + 11H_2(g) \rightarrow C_8H_{18}(l) + 9H_2O \\
6. & \ FeO_3(s) + 3CO(g) \rightarrow Fe(l) + 3CO_2(g) \\
7. & \ 2H_2SO_4 + Pb(OH)_4 \rightarrow Pb(SO_4)_2 + 4H_2O \\
8. & \ 2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2 \\
9. & \ Ca_3(PO_4)_2 + 2H_2SO_4 \rightarrow 3CaSO_4 + 2Ca(H_2PO_4)_2 \\
10. & \ H_3PO_4 + 5HCl \rightarrow PCl_5 + 4H_2O \\
\end{aligned}
}
$$
Parent Tip: Review the logic above to help your child master the concept of practice balancing chemical equations worksheet with answers.
![49 Balancing Chemical Equations Worksheets [with Answers]](/images2/practice-balancing-chemical-equations-worksheet-with-answers/practice-balancing-chemical-equations-worksheet-with-answers-7.jpg)
![49 Balancing Chemical Equations Worksheets [with Answers]](/images2/practice-balancing-chemical-equations-worksheet-with-answers/practice-balancing-chemical-equations-worksheet-with-answers-1.jpg)
