Free Printable Balancing and Classifying Chemical Equations Worksheets - Free Printable

To balance the given acid-base reactions, 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 reaction step by step.

---

i. HBr + NaOH → NaBr + H₂O

- Reactants: HBr, NaOH
- Products: NaBr, H₂O

1. Start with the unbalanced equation:
\[
\text{HBr} + \text{NaOH} \rightarrow \text{NaBr} + \text{H}_2\text{O}
\]

2. Balance the equation:
- There is 1 H atom in HBr and 1 H atom in NaOH, which totals 2 H atoms.
- There is 1 O atom in NaOH.
- The products already have 2 H atoms (in H₂O) and 1 O atom.
- All other elements (Na, Br) are balanced with 1 atom each.

The balanced equation is:
\[
\boxed{1 \text{ HBr} + 1 \text{ NaOH} \rightarrow 1 \text{ NaBr} + 1 \text{ H}_2\text{O}}
\]

---

ii. H₂SO₄ + KOH → K₂SO₄ + H₂O

- Reactants: H₂SO₄, KOH
- Products: K₂SO₄, H₂O

1. Start with the unbalanced equation:
\[
\text{H}_2\text{SO}_4 + \text{KOH} \rightarrow \text{K}_2\text{SO}_4 + \text{H}_2\text{O}
\]

2. Balance the equation:
- There are 2 K atoms in K₂SO₄, so we need 2 KOH molecules.
- With 2 KOH molecules, there are 2 H atoms and 2 O atoms from KOH.
- The H₂SO₄ molecule provides 2 H atoms and 4 O atoms.
- The products now have 2 K atoms, 1 S atom, 4 O atoms, and 2 H atoms.

The balanced equation is:
\[
\boxed{1 \text{ H}_2\text{SO}_4 + 2 \text{ KOH} \rightarrow 1 \text{ K}_2\text{SO}_4 + 2 \text{ H}_2\text{O}}
\]

---

iii. HCl + Ca(OH)₂ → CaCl₂ + H₂O

- Reactants: HCl, Ca(OH)₂
- Products: CaCl₂, H₂O

1. Start with the unbalanced equation:
\[
\text{HCl} + \text{Ca(OH)}_2 \rightarrow \text{CaCl}_2 + \text{H}_2\text{O}
\]

2. Balance the equation:
- There are 2 OH⁻ ions in Ca(OH)₂, so we need 2 HCl molecules.
- With 2 HCl molecules, there are 2 H atoms and 2 Cl atoms.
- The products now have 1 Ca atom, 2 Cl atoms, 2 H atoms, and 2 O atoms.

The balanced equation is:
\[
\boxed{2 \text{ HCl} + 1 \text{ Ca(OH)}_2 \rightarrow 1 \text{ CaCl}_2 + 2 \text{ H}_2\text{O}}
\]

---

iv. Fe(OH)₃ + H₂SO₄ → Fe₂(SO₄)₃ + H₂O

- Reactants: Fe(OH)₃, H₂SO₄
- Products: Fe₂(SO₄)₃, H₂O

1. Start with the unbalanced equation:
\[
\text{Fe(OH)}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + \text{H}_2\text{O}
\]

2. Balance the equation:
- There are 2 Fe atoms in Fe₂(SO₄)₃, so we need 2 Fe(OH)₃ molecules.
- With 2 Fe(OH)₃ molecules, there are 6 OH⁻ ions, requiring 3 H₂SO₄ molecules.
- With 3 H₂SO₄ molecules, there are 6 H atoms and 3 SO₄²⁻ ions.
- The products now have 2 Fe atoms, 3 SO₄²⁻ ions, 6 H atoms, and 6 O atoms.

The balanced equation is:
\[
\boxed{2 \text{ Fe(OH)}_3 + 3 \text{ H}_2\text{SO}_4 \rightarrow 1 \text{ Fe}_2(\text{SO}_4)_3 + 6 \text{ H}_2\text{O}}
\]

---

v. H₂SO₄ + B(OH)₃ → B₂(SO₄)₃ + H₂O

- Reactants: H₂SO₄, B(OH)₃
- Products: B₂(SO₄)₃, H₂O

1. Start with the unbalanced equation:
\[
\text{H}_2\text{SO}_4 + \text{B(OH)}_3 \rightarrow \text{B}_2(\text{SO}_4)_3 + \text{H}_2\text{O}
\]

2. Balance the equation:
- There are 2 B atoms in B₂(SO₄)₃, so we need 2 B(OH)₃ molecules.
- With 2 B(OH)₃ molecules, there are 6 OH⁻ ions, requiring 3 H₂SO₄ molecules.
- With 3 H₂SO₄ molecules, there are 6 H atoms and 3 SO₄²⁻ ions.
- The products now have 2 B atoms, 3 SO₄²⁻ ions, 6 H atoms, and 6 O atoms.

The balanced equation is:
\[
\boxed{3 \text{ H}_2\text{SO}_4 + 2 \text{ B(OH)}_3 \rightarrow 1 \text{ B}_2(\text{SO}_4)_3 + 6 \text{ H}_2\text{O}}
\]

---

vi. Pb(OH)₂ + HCl → PbCl₂ + H₂O

- Reactants: Pb(OH)₂, HCl
- Products: PbCl₂, H₂O

1. Start with the unbalanced equation:
\[
\text{Pb(OH)}_2 + \text{HCl} \rightarrow \text{PbCl}_2 + \text{H}_2\text{O}
\]

2. Balance the equation:
- There are 2 OH⁻ ions in Pb(OH)₂, so we need 2 HCl molecules.
- With 2 HCl molecules, there are 2 H atoms and 2 Cl atoms.
- The products now have 1 Pb atom, 2 Cl atoms, 2 H atoms, and 2 O atoms.

The balanced equation is:
\[
\boxed{1 \text{ Pb(OH)}_2 + 2 \text{ HCl} \rightarrow 1 \text{ PbCl}_2 + 2 \text{ H}_2\text{O}}
\]

---

vii. H₂SO₄ + NH₄OH → (NH₄)₂SO₄ + H₂O

- Reactants: H₂SO₄, NH₄OH
- Products: (NH₄)₂SO₄, H₂O

1. Start with the unbalanced equation:
\[
\text{H}_2\text{SO}_4 + \text{NH}_4\text{OH} \rightarrow \text{(NH}_4\text{)}_2\text{SO}_4 + \text{H}_2\text{O}
\]

2. Balance the equation:
- There are 2 NH₄⁺ ions in (NH₄)₂SO₄, so we need 2 NH₄OH molecules.
- With 2 NH₄OH molecules, there are 2 H atoms and 2 OH⁻ ions.
- The products now have 2 N atoms, 8 H atoms, 1 S atom, 4 O atoms.

The balanced equation is:
\[
\boxed{1 \text{ H}_2\text{SO}_4 + 2 \text{ NH}_4\text{OH} \rightarrow 1 \text{ (NH}_4\text{)}_2\text{SO}_4 + 2 \text{ H}_2\text{O}}
\]

---

viii. H₂CO₃ + CsOH → Cs₂CO₃ + H₂O

- Reactants: H₂CO₃, CsOH
- Products: Cs₂CO₃, H₂O

1. Start with the unbalanced equation:
\[
\text{H}_2\text{CO}_3 + \text{CsOH} \rightarrow \text{Cs}_2\text{CO}_3 + \text{H}_2\text{O}
\]

2. Balance the equation:
- There are 2 Cs atoms in Cs₂CO₃, so we need 2 CsOH molecules.
- With 2 CsOH molecules, there are 2 H atoms and 2 O atoms.
- The products now have 2 Cs atoms, 1 C atom, 3 O atoms, and 2 H atoms.

The balanced equation is:
\[
\boxed{1 \text{ H}_2\text{CO}_3 + 2 \text{ CsOH} \rightarrow 1 \text{ Cs}_2\text{CO}_3 + 2 \text{ H}_2\text{O}}
\]

---

ix. HF + Mg(OH)₂ → MgF₂ + H₂O

- Reactants: HF, Mg(OH)₂
- Products: MgF₂, H₂O

1. Start with the unbalanced equation:
\[
\text{HF} + \text{Mg(OH)}_2 \rightarrow \text{MgF}_2 + \text{H}_2\text{O}
\]

2. Balance the equation:
- There are 2 OH⁻ ions in Mg(OH)₂, so we need 2 HF molecules.
- With 2 HF molecules, there are 2 H atoms and 2 F atoms.
- The products now have 1 Mg atom, 2 F atoms, 2 H atoms, and 2 O atoms.

The balanced equation is:
\[
\boxed{2 \text{ HF} + 1 \text{ Mg(OH)}_2 \rightarrow 1 \text{ MgF}_2 + 2 \text{ H}_2\text{O}}
\]

---

x. HNO₃ + Al(OH)₃ → Al(NO₃)₃ + H₂O

- Reactants: HNO₃, Al(OH)₃
- Products: Al(NO₃)₃, H₂O

1. Start with the unbalanced equation:
\[
\text{HNO}_3 + \text{Al(OH)}_3 \rightarrow \text{Al(NO}_3\text{)}_3 + \text{H}_2\text{O}
\]

2. Balance the equation:
- There are 3 NO₃⁻ ions in Al(NO₃)₃, so we need 3 HNO₃ molecules.
- With 3 HNO₃ molecules, there are 3 H atoms.
- The products now have 1 Al atom, 3 N atoms, 9 O atoms, and 3 H atoms.

The balanced equation is:
\[
\boxed{3 \text{ HNO}_3 + 1 \text{ Al(OH)}_3 \rightarrow 1 \text{ Al(NO}_3\text{)}_3 + 3 \text{ H}_2\text{O}}
\]

---

xi. HNO₃ + Zn(OH)₂ → Zn(NO₃)₂ + H₂O

- Reactants: HNO₃, Zn(OH)₂
- Products: Zn(NO₃)₂, H₂O

1. Start with the unbalanced equation:
\[
\text{HNO}_3 + \text{Zn(OH)}_2 \rightarrow \text{Zn(NO}_3\text{)}_2 + \text{H}_2\text{O}
\]

2. Balance the equation:
- There are 2 NO₃⁻ ions in Zn(NO₃)₂, so we need 2 HNO₃ molecules.
- With 2 HNO₃ molecules, there are 2 H atoms.
- The products now have 1 Zn atom, 2 N atoms, 6 O atoms, and 2 H atoms.

The balanced equation is:
\[
\boxed{2 \text{ HNO}_3 + 1 \text{ Zn(OH)}_2 \rightarrow 1 \text{ Zn(NO}_3\text{)}_2 + 2 \text{ H}_2\text{O}}
\]

---

xii. H₃PO₄ + Ca(OH)₂ → Ca₃(PO₄)₂ + H₂O

- Reactants: H₃PO₄, Ca(OH)₂
- Products: Ca₃(PO₄)₂, H₂O

1. Start with the unbalanced equation:
\[
\text{H}_3\text{PO}_4 + \text{Ca(OH)}_2 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + \text{H}_2\text{O}
\]

2. Balance the equation:
- There are 3 Ca atoms in Ca₃(PO₄)₂, so we need 3 Ca(OH)₂ molecules.
- With 3 Ca(OH)₂ molecules, there are 6 OH⁻ ions, requiring 2 H₃PO₄ molecules.
- With 2 H₃PO₄ molecules, there are 6 H atoms and 2 PO₄³⁻ ions.
- The products now have 3 Ca atoms, 2 P atoms, 8 O atoms, and 6 H atoms.

The balanced equation is:
\[
\boxed{2 \text{ H}_3\text{PO}_4 + 3 \text{ Ca(OH)}_2 \rightarrow 1 \text{ Ca}_3(\text{PO}_4)_2 + 6 \text{ H}_2\text{O}}
\]

---

xiii. HClO₃ + Al(OH)₃ → Al(ClO₃)₃ + H₂O

- Reactants: HClO₃, Al(OH)₃
- Products: Al(ClO₃)₃, H₂O

1. Start with the unbalanced equation:
\[
\text{HClO}_3 + \text{Al(OH)}_3 \rightarrow \text{Al(ClO}_3\text{)}_3 + \text{H}_2\text{O}
\]

2. Balance the equation:
- There are 3 ClO₃⁻ ions in Al(ClO₃)₃, so we need 3 HClO₃ molecules.
- With 3 HClO₃ molecules, there are 3 H atoms.
- The products now have 1 Al atom, 3 Cl atoms, 9 O atoms, and 3 H atoms.

The balanced equation is:
\[
\boxed{3 \text{ HClO}_3 + 1 \text{ Al(OH)}_3 \rightarrow 1 \text{ Al(ClO}_3\text{)}_3 + 3 \text{ H}_2\text{O}}
\]

---

xiv. HF + Ba(OH)₂ → BaF₂ + H₂O

- Reactants: HF, Ba(OH)₂
- Products: BaF₂, H₂O

1. Start with the unbalanced equation:
\[
\text{HF} + \text{Ba(OH)}_2 \rightarrow \text{BaF}_2 + \text{H}_2\text{O}
\]

2. Balance the equation:
- There are 2 OH⁻ ions in Ba(OH)₂, so we need 2 HF molecules.
- With 2 HF molecules, there are 2 H atoms and 2 F atoms.
- The products now have 1 Ba atom, 2 F atoms, 2 H atoms, and 2 O atoms.

The balanced equation is:
\[
\boxed{2 \text{ HF} + 1 \text{ Ba(OH)}_2 \rightarrow 1 \text{ BaF}_2 + 2 \text{ H}_2\text{O}}
\]

---

xv. HCl + Al(OH)₃ → AlCl₃ + H₂O

- Reactants: HCl, Al(OH)₃
- Products: AlCl₃, H₂O

1. Start with the unbalanced equation:
\[
\text{HCl} + \text{Al(OH)}_3 \rightarrow \text{AlCl}_3 + \text{H}_2\text{O}
\]

2. Balance the equation:
- There are 3 Cl atoms in AlCl₃, so we need 3 HCl molecules.
- With 3 HCl molecules, there are 3 H atoms.
- The products now have 1 Al atom, 3 Cl atoms, 3 H atoms, and 3 O atoms.

The balanced equation is:
\[
\boxed{3 \text{ HCl} + 1 \text{ Al(OH)}_3 \rightarrow 1 \text{ AlCl}_3 + 3 \text{ H}_2\text{O}}
\]

---

Final Answer:

\[
\boxed{
\begin{aligned}
&\text{i. } 1 \text{ HBr} + 1 \text{ NaOH} \rightarrow 1 \text{ NaBr} + 1 \text{ H}_2\text{O} \\
&\text{ii. } 1 \text{ H}_2\text{SO}_4 + 2 \text{ KOH} \rightarrow 1 \text{ K}_2\text{SO}_4 + 2 \text{ H}_2\text{O} \\
&\text{iii. } 2 \text{ HCl} + 1 \text{ Ca(OH)}_2 \rightarrow 1 \text{ CaCl}_2 + 2 \text{ H}_2\text{O} \\
&\text{iv. } 2 \text{ Fe(OH)}_3 + 3 \text{ H}_2\text{SO}_4 \rightarrow 1 \text{ Fe}_2(\text{SO}_4)_3 + 6 \text{ H}_2\text{O} \\
&\text{v. } 3 \text{ H}_2\text{SO}_4 + 2 \text{ B(OH)}_3 \rightarrow 1 \text{ B}_2(\text{SO}_4)_3 + 6 \text{ H}_2\text{O} \\
&\text{vi. } 1 \text{ Pb(OH)}_2 + 2 \text{ HCl} \rightarrow 1 \text{ PbCl}_2 + 2 \text{ H}_2\text{O} \\
&\text{vii. } 1 \text{ H}_2\text{SO}_4 + 2 \text{ NH}_4\text{OH} \rightarrow 1 \text{ (NH}_4\text{)}_2\text{SO}_4 + 2 \text{ H}_2\text{O} \\
&\text{viii. } 1 \text{ H}_2\text{CO}_3 + 2 \text{ CsOH} \rightarrow 1 \text{ Cs}_2\text{CO}_3 + 2 \text{ H}_2\text{O} \\
&\text{ix. } 2 \text{ HF} + 1 \text{ Mg(OH)}_2 \rightarrow 1 \text{ MgF}_2 + 2 \text{ H}_2\text{O} \\
&\text{x. } 3 \text{ HNO}_3 + 1 \text{ Al(OH)}_3 \rightarrow 1 \text{ Al(NO}_3\text{)}_3 + 3 \text{ H}_2\text{O} \\
&\text{xi. } 2 \text{ HNO}_3 + 1 \text{ Zn(OH)}_2 \rightarrow 1 \text{ Zn(NO}_3\text{)}_2 + 2 \text{ H}_2\text{O} \\
&\text{xii. } 2 \text{ H}_3\text{PO}_4 + 3 \text{ Ca(OH)}_2 \rightarrow 1 \text{ Ca}_3(\text{PO}_4)_2 + 6 \text{ H}_2\text{O} \\
&\text{xiii. } 3 \text{ HClO}_3 + 1 \text{ Al(OH)}_3 \rightarrow 1 \text{ Al(ClO}_3\text{)}_3 + 3 \text{ H}_2\text{O} \\
&\text{xiv. } 2 \text{ HF} + 1 \text{ Ba(OH)}_2 \rightarrow 1 \text{ BaF}_2 + 2 \text{ H}_2\text{O} \\
&\text{xv. } 3 \text{ HCl} + 1 \text{ Al(OH)}_3 \rightarrow 1 \text{ AlCl}_3 + 3 \text{ H}_2\text{O}
\end{aligned}
}
\]