49 Balancing Chemical Equations Worksheets [with Answers] - Free Printable
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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]
To balance the given chemical equations, we need to ensure that the number of atoms of each element is the same on both sides of the equation. Here are the balanced equations along with explanations:
---
- Initial Equation: Fe + O₂ → Fe₃O₄
- Balancing:
- On the right side, there are 3 Fe atoms and 4 O atoms.
- To balance Fe, multiply Fe by 3: 3Fe + O₂ → Fe₃O₄.
- Now, there are 3 Fe atoms on both sides.
- To balance O, multiply O₂ by 2: 3Fe + 2O₂ → Fe₃O₄.
- Now, there are 4 O atoms on both sides.
- Balanced Equation: 3Fe + 2O₂ → Fe₃O₄
---
- Initial Equation: Sr + O₂ → SrO
- Balancing:
- On the right side, there is 1 Sr atom and 1 O atom.
- To balance Sr, keep Sr as is: Sr + O₂ → SrO.
- To balance O, multiply SrO by 2: Sr + O₂ → 2SrO.
- Now, there are 2 O atoms on both sides.
- Balanced Equation: 2Sr + O₂ → 2SrO
---
- Initial Equation: Sn + NaOH → Na₂SnO₂ + H₂
- Balancing:
- On the right side, there are 2 Na atoms, 1 Sn atom, 2 O atoms, and 2 H atoms.
- To balance Na, multiply NaOH by 2: Sn + 2NaOH → Na₂SnO₂ + H₂.
- Now, there are 2 Na atoms on both sides.
- The equation is already balanced for Sn, O, and H.
- Balanced Equation: Sn + 2NaOH → Na₂SnO₂ + H₂
---
- Initial Equation: K + Br₂ → KBr
- Balancing:
- On the right side, there is 1 K atom and 1 Br atom.
- To balance K, keep K as is: K + Br₂ → KBr.
- To balance Br, multiply KBr by 2: K + Br₂ → 2KBr.
- Now, there are 2 Br atoms on both sides.
- Balanced Equation: 2K + Br₂ → 2KBr
---
- Initial Equation: C₈H₁₈ + O₂ → CO₂ + H₂O
- Balancing:
- Start with carbon (C): There are 8 C atoms in C₈H₁₈, so multiply CO₂ by 8: C₈H₁₈ + O₂ → 8CO₂ + H₂O.
- Next, balance hydrogen (H): There are 18 H atoms in C₈H₁₈, so multiply H₂O by 9: C₈H₁₈ + O₂ → 8CO₂ + 9H₂O.
- Now, balance oxygen (O): There are 25 O atoms on the right side (8 from CO₂ and 9 from H₂O), so multiply O₂ by 25/2 = 12.5. Since coefficients must be whole numbers, multiply the entire equation by 2: 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O.
- Balanced Equation: 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O
---
- Initial Equation: Sb + I₂ → SbI₃
- Balancing:
- On the right side, there is 1 Sb atom and 3 I atoms.
- To balance Sb, keep Sb as is: Sb + I₂ → SbI₃.
- To balance I, multiply I₂ by 3/2 (or 1.5). Since coefficients must be whole numbers, multiply the entire equation by 2: 2Sb + 3I₂ → 2SbI₃.
- Balanced Equation: 2Sb + 3I₂ → 2SbI₃
---
- Initial Equation: COCl₂ + H₂O → HCl + CO₂
- Balancing:
- Start with chlorine (Cl): There are 2 Cl atoms in COCl₂, so multiply HCl by 2: COCl₂ + H₂O → 2HCl + CO₂.
- Next, balance hydrogen (H): There are 2 H atoms in H₂O, so the equation is already balanced for H.
- Finally, balance oxygen (O): There is 1 O atom in H₂O and 2 O atoms in CO₂, so the equation is already balanced for O.
- Balanced Equation: COCl₂ + H₂O → 2HCl + CO₂
---
- Initial Equation: CS₂ + O₂ → CO₂ + SO₂
- Balancing:
- Start with carbon (C): There is 1 C atom in CS₂, so keep CO₂ as is: CS₂ + O₂ → CO₂ + SO₂.
- Next, balance sulfur (S): There are 2 S atoms in CS₂, so multiply SO₂ by 2: CS₂ + O₂ → CO₂ + 2SO₂.
- Now, balance oxygen (O): There are 6 O atoms on the right side (2 from CO₂ and 4 from 2SO₂), so multiply O₂ by 3: CS₂ + 3O₂ → CO₂ + 2SO₂.
- Balanced Equation: CS₂ + 3O₂ → CO₂ + 2SO₂
---
- Initial Equation: H₂SO₄ + NaCN → HCN + Na₂SO₄
- Balancing:
- Start with sodium (Na): There are 2 Na atoms in Na₂SO₄, so multiply NaCN by 2: H₂SO₄ + 2NaCN → HCN + Na₂SO₄.
- Next, balance hydrogen (H): There are 2 H atoms in H₂SO₄, so multiply HCN by 2: H₂SO₄ + 2NaCN → 2HCN + Na₂SO₄.
- The equation is now balanced for all elements.
- Balanced Equation: H₂SO₄ + 2NaCN → 2HCN + Na₂SO₄
---
- Initial Equation: KClO₃ → KCl + O₂
- Balancing:
- Start with potassium (K): There is 1 K atom in KClO₃, so keep KCl as is: KClO₃ → KCl + O₂.
- Next, balance chlorine (Cl): There is 1 Cl atom in KClO₃, so the equation is already balanced for Cl.
- Balance oxygen (O): There are 3 O atoms in KClO₃, so multiply O₂ by 3/2 (or 1.5). Since coefficients must be whole numbers, multiply the entire equation by 2: 2KClO₃ → 2KCl + 3O₂.
- Balanced Equation: 2KClO₃ → 2KCl + 3O₂
---
- Initial Equation: H₂ + F₂ → HF
- Balancing:
- On the right side, there is 1 H atom and 1 F atom.
- To balance H, multiply HF by 2: H₂ + F₂ → 2HF.
- Now, there are 2 H atoms and 2 F atoms on both sides.
- Balanced Equation: H₂ + F₂ → 2HF
---
- Initial Equation: BaCl₂ + KIO₃ → Ba(IO₃)₂ + KCl
- Balancing:
- Start with barium (Ba): There is 1 Ba atom in BaCl₂, so keep Ba(IO₃)₂ as is: BaCl₂ + KIO₃ → Ba(IO₃)₂ + KCl.
- Next, balance iodate (IO₃⁻): There are 2 IO₃⁻ ions in Ba(IO₃)₂, so multiply KIO₃ by 2: BaCl₂ + 2KIO₃ → Ba(IO₃)₂ + KCl.
- Now, balance potassium (K): There are 2 K atoms in 2KIO₃, so multiply KCl by 2: BaCl₂ + 2KIO₃ → Ba(IO₃)₂ + 2KCl.
- The equation is now balanced for all elements.
- Balanced Equation: BaCl₂ + 2KIO₃ → Ba(IO₃)₂ + 2KCl
---
- Initial Equation: Mg + HCl → MgCl₂ + H₂
- Balancing:
- Start with magnesium (Mg): There is 1 Mg atom in Mg, so keep MgCl₂ as is: Mg + HCl → MgCl₂ + H₂.
- Next, balance chlorine (Cl): There are 2 Cl atoms in MgCl₂, so multiply HCl by 2: Mg + 2HCl → MgCl₂ + H₂.
- Now, there are 2 H atoms on both sides.
- Balanced Equation: Mg + 2HCl → MgCl₂ + H₂
---
\[
\boxed{
\begin{aligned}
1. & \quad 3Fe + 2O_2 \rightarrow Fe_3O_4 \\
2. & \quad 2Sr + O_2 \rightarrow 2SrO \\
3. & \quad Sn + 2NaOH \rightarrow Na_2SnO_2 + H_2 \\
4. & \quad 2K + Br_2 \rightarrow 2KBr \\
5. & \quad 2C_8H_{18} + 25O_2 \rightarrow 16CO_2 + 18H_2O \\
6. & \quad 2Sb + 3I_2 \rightarrow 2SbI_3 \\
7. & \quad COCl_2 + H_2O \rightarrow 2HCl + CO_2 \\
8. & \quad CS_2 + 3O_2 \rightarrow CO_2 + 2SO_2 \\
9. & \quad H_2SO_4 + 2NaCN \rightarrow 2HCN + Na_2SO_4 \\
10. & \quad 2KClO_3 \rightarrow 2KCl + 3O_2 \\
11. & \quad H_2 + F_2 \rightarrow 2HF \\
12. & \quad BaCl_2 + 2KIO_3 \rightarrow Ba(IO_3)_2 + 2KCl \\
13. & \quad Mg + 2HCl \rightarrow MgCl_2 + H_2 \\
\end{aligned}
}
\]
---
1. Fe + O₂ → Fe₃O₄
- Initial Equation: Fe + O₂ → Fe₃O₄
- Balancing:
- On the right side, there are 3 Fe atoms and 4 O atoms.
- To balance Fe, multiply Fe by 3: 3Fe + O₂ → Fe₃O₄.
- Now, there are 3 Fe atoms on both sides.
- To balance O, multiply O₂ by 2: 3Fe + 2O₂ → Fe₃O₄.
- Now, there are 4 O atoms on both sides.
- Balanced Equation: 3Fe + 2O₂ → Fe₃O₄
---
2. Sr + O₂ → SrO
- Initial Equation: Sr + O₂ → SrO
- Balancing:
- On the right side, there is 1 Sr atom and 1 O atom.
- To balance Sr, keep Sr as is: Sr + O₂ → SrO.
- To balance O, multiply SrO by 2: Sr + O₂ → 2SrO.
- Now, there are 2 O atoms on both sides.
- Balanced Equation: 2Sr + O₂ → 2SrO
---
3. Sn + NaOH → Na₂SnO₂ + H₂
- Initial Equation: Sn + NaOH → Na₂SnO₂ + H₂
- Balancing:
- On the right side, there are 2 Na atoms, 1 Sn atom, 2 O atoms, and 2 H atoms.
- To balance Na, multiply NaOH by 2: Sn + 2NaOH → Na₂SnO₂ + H₂.
- Now, there are 2 Na atoms on both sides.
- The equation is already balanced for Sn, O, and H.
- Balanced Equation: Sn + 2NaOH → Na₂SnO₂ + H₂
---
4. K + Br₂ → KBr
- Initial Equation: K + Br₂ → KBr
- Balancing:
- On the right side, there is 1 K atom and 1 Br atom.
- To balance K, keep K as is: K + Br₂ → KBr.
- To balance Br, multiply KBr by 2: K + Br₂ → 2KBr.
- Now, there are 2 Br atoms on both sides.
- Balanced Equation: 2K + Br₂ → 2KBr
---
5. C₈H₁₈ + O₂ → CO₂ + H₂O
- Initial Equation: C₈H₁₈ + O₂ → CO₂ + H₂O
- Balancing:
- Start with carbon (C): There are 8 C atoms in C₈H₁₈, so multiply CO₂ by 8: C₈H₁₈ + O₂ → 8CO₂ + H₂O.
- Next, balance hydrogen (H): There are 18 H atoms in C₈H₁₈, so multiply H₂O by 9: C₈H₁₈ + O₂ → 8CO₂ + 9H₂O.
- Now, balance oxygen (O): There are 25 O atoms on the right side (8 from CO₂ and 9 from H₂O), so multiply O₂ by 25/2 = 12.5. Since coefficients must be whole numbers, multiply the entire equation by 2: 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O.
- Balanced Equation: 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O
---
6. Sb + I₂ → SbI₃
- Initial Equation: Sb + I₂ → SbI₃
- Balancing:
- On the right side, there is 1 Sb atom and 3 I atoms.
- To balance Sb, keep Sb as is: Sb + I₂ → SbI₃.
- To balance I, multiply I₂ by 3/2 (or 1.5). Since coefficients must be whole numbers, multiply the entire equation by 2: 2Sb + 3I₂ → 2SbI₃.
- Balanced Equation: 2Sb + 3I₂ → 2SbI₃
---
7. COCl₂ + H₂O → HCl + CO₂
- Initial Equation: COCl₂ + H₂O → HCl + CO₂
- Balancing:
- Start with chlorine (Cl): There are 2 Cl atoms in COCl₂, so multiply HCl by 2: COCl₂ + H₂O → 2HCl + CO₂.
- Next, balance hydrogen (H): There are 2 H atoms in H₂O, so the equation is already balanced for H.
- Finally, balance oxygen (O): There is 1 O atom in H₂O and 2 O atoms in CO₂, so the equation is already balanced for O.
- Balanced Equation: COCl₂ + H₂O → 2HCl + CO₂
---
8. CS₂ + O₂ → CO₂ + SO₂
- Initial Equation: CS₂ + O₂ → CO₂ + SO₂
- Balancing:
- Start with carbon (C): There is 1 C atom in CS₂, so keep CO₂ as is: CS₂ + O₂ → CO₂ + SO₂.
- Next, balance sulfur (S): There are 2 S atoms in CS₂, so multiply SO₂ by 2: CS₂ + O₂ → CO₂ + 2SO₂.
- Now, balance oxygen (O): There are 6 O atoms on the right side (2 from CO₂ and 4 from 2SO₂), so multiply O₂ by 3: CS₂ + 3O₂ → CO₂ + 2SO₂.
- Balanced Equation: CS₂ + 3O₂ → CO₂ + 2SO₂
---
9. H₂SO₄ + NaCN → HCN + Na₂SO₄
- Initial Equation: H₂SO₄ + NaCN → HCN + Na₂SO₄
- Balancing:
- Start with sodium (Na): There are 2 Na atoms in Na₂SO₄, so multiply NaCN by 2: H₂SO₄ + 2NaCN → HCN + Na₂SO₄.
- Next, balance hydrogen (H): There are 2 H atoms in H₂SO₄, so multiply HCN by 2: H₂SO₄ + 2NaCN → 2HCN + Na₂SO₄.
- The equation is now balanced for all elements.
- Balanced Equation: H₂SO₄ + 2NaCN → 2HCN + Na₂SO₄
---
10. KClO₃ → KCl + O₂
- Initial Equation: KClO₃ → KCl + O₂
- Balancing:
- Start with potassium (K): There is 1 K atom in KClO₃, so keep KCl as is: KClO₃ → KCl + O₂.
- Next, balance chlorine (Cl): There is 1 Cl atom in KClO₃, so the equation is already balanced for Cl.
- Balance oxygen (O): There are 3 O atoms in KClO₃, so multiply O₂ by 3/2 (or 1.5). Since coefficients must be whole numbers, multiply the entire equation by 2: 2KClO₃ → 2KCl + 3O₂.
- Balanced Equation: 2KClO₃ → 2KCl + 3O₂
---
11. H₂ + F₂ → HF
- Initial Equation: H₂ + F₂ → HF
- Balancing:
- On the right side, there is 1 H atom and 1 F atom.
- To balance H, multiply HF by 2: H₂ + F₂ → 2HF.
- Now, there are 2 H atoms and 2 F atoms on both sides.
- Balanced Equation: H₂ + F₂ → 2HF
---
12. BaCl₂ + KIO₃ → Ba(IO₃)₂ + KCl
- Initial Equation: BaCl₂ + KIO₃ → Ba(IO₃)₂ + KCl
- Balancing:
- Start with barium (Ba): There is 1 Ba atom in BaCl₂, so keep Ba(IO₃)₂ as is: BaCl₂ + KIO₃ → Ba(IO₃)₂ + KCl.
- Next, balance iodate (IO₃⁻): There are 2 IO₃⁻ ions in Ba(IO₃)₂, so multiply KIO₃ by 2: BaCl₂ + 2KIO₃ → Ba(IO₃)₂ + KCl.
- Now, balance potassium (K): There are 2 K atoms in 2KIO₃, so multiply KCl by 2: BaCl₂ + 2KIO₃ → Ba(IO₃)₂ + 2KCl.
- The equation is now balanced for all elements.
- Balanced Equation: BaCl₂ + 2KIO₃ → Ba(IO₃)₂ + 2KCl
---
13. Mg + HCl → MgCl₂ + H₂
- Initial Equation: Mg + HCl → MgCl₂ + H₂
- Balancing:
- Start with magnesium (Mg): There is 1 Mg atom in Mg, so keep MgCl₂ as is: Mg + HCl → MgCl₂ + H₂.
- Next, balance chlorine (Cl): There are 2 Cl atoms in MgCl₂, so multiply HCl by 2: Mg + 2HCl → MgCl₂ + H₂.
- Now, there are 2 H atoms on both sides.
- Balanced Equation: Mg + 2HCl → MgCl₂ + H₂
---
Final Answer
\[
\boxed{
\begin{aligned}
1. & \quad 3Fe + 2O_2 \rightarrow Fe_3O_4 \\
2. & \quad 2Sr + O_2 \rightarrow 2SrO \\
3. & \quad Sn + 2NaOH \rightarrow Na_2SnO_2 + H_2 \\
4. & \quad 2K + Br_2 \rightarrow 2KBr \\
5. & \quad 2C_8H_{18} + 25O_2 \rightarrow 16CO_2 + 18H_2O \\
6. & \quad 2Sb + 3I_2 \rightarrow 2SbI_3 \\
7. & \quad COCl_2 + H_2O \rightarrow 2HCl + CO_2 \\
8. & \quad CS_2 + 3O_2 \rightarrow CO_2 + 2SO_2 \\
9. & \quad H_2SO_4 + 2NaCN \rightarrow 2HCN + Na_2SO_4 \\
10. & \quad 2KClO_3 \rightarrow 2KCl + 3O_2 \\
11. & \quad H_2 + F_2 \rightarrow 2HF \\
12. & \quad BaCl_2 + 2KIO_3 \rightarrow Ba(IO_3)_2 + 2KCl \\
13. & \quad Mg + 2HCl \rightarrow MgCl_2 + H_2 \\
\end{aligned}
}
\]
Parent Tip: Review the logic above to help your child master the concept of balancing chemical equations practice worksheets.
