Balancing Equations Practice Worksheet | StudyX - Free Printable

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. Here are the balanced equations for each problem, along with explanations:

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1. \( \_\_ \text{NaNO}_3 + \_\_ \text{PbO} \rightarrow \_\_ \text{Pb(NO}_3\text{)}_2 + \_\_ \text{Na}_2\text{O} \)

- Step 1: Start by balancing the lead (Pb) atoms.
- There is 1 Pb atom in \(\text{PbO}\) and 1 Pb atom in \(\text{Pb(NO}_3\text{)}_2\). So, place a coefficient of 1 in front of \(\text{PbO}\) and \(\text{Pb(NO}_3\text{)}_2\).

\[
\_\_ \text{NaNO}_3 + 1 \text{PbO} \rightarrow 1 \text{Pb(NO}_3\text{)}_2 + \_\_ \text{Na}_2\text{O}
\]

- Step 2: Balance the nitrogen (N) atoms.
- There are 2 N atoms in \(\text{Pb(NO}_3\text{)}_2\), so there must be 2 N atoms from \(\text{NaNO}_3\). Place a coefficient of 2 in front of \(\text{NaNO}_3\).

\[
2 \text{NaNO}_3 + 1 \text{PbO} \rightarrow 1 \text{Pb(NO}_3\text{)}_2 + \_\_ \text{Na}_2\text{O}
\]

- Step 3: Balance the sodium (Na) atoms.
- There are 2 Na atoms in \(\text{Na}_2\text{O}\) and 2 Na atoms in \(2 \text{NaNO}_3\). So, place a coefficient of 1 in front of \(\text{Na}_2\text{O}\).

\[
2 \text{NaNO}_3 + 1 \text{PbO} \rightarrow 1 \text{Pb(NO}_3\text{)}_2 + 1 \text{Na}_2\text{O}
\]

- Step 4: Verify all atoms.
- Na: 2 on both sides
- N: 2 on both sides
- O: \(6\) (from \(2 \times 3\) in \(\text{NaNO}_3\) + 1 from \(\text{PbO}\)) = \(6\) (from \(2 \times 3\) in \(\text{Pb(NO}_3\text{)}_2\) + 1 in \(\text{Na}_2\text{O}\))
- Pb: 1 on both sides

The balanced equation is:
\[
\boxed{2 \text{NaNO}_3 + 1 \text{PbO} \rightarrow 1 \text{Pb(NO}_3\text{)}_2 + 1 \text{Na}_2\text{O}}
\]

---

2. \( \_\_ \text{AgI} + \_\_ \text{Fe}_2\text{(CO}_3\text{)}_3 \rightarrow \_\_ \text{FeI}_3 + \_\_ \text{Ag}_2\text{CO}_3 \)

- Step 1: Balance the iron (Fe) atoms.
- There are 2 Fe atoms in \(\text{Fe}_2\text{(CO}_3\text{)}_3\) and 3 Fe atoms in \(\text{FeI}_3\). To balance Fe, place a coefficient of 2 in front of \(\text{FeI}_3\).

\[
\_\_ \text{AgI} + 1 \text{Fe}_2\text{(CO}_3\text{)}_3 \rightarrow 2 \text{FeI}_3 + \_\_ \text{Ag}_2\text{CO}_3
\]

- Step 2: Balance the iodine (I) atoms.
- There are 6 I atoms in \(2 \text{FeI}_3\) (since \(2 \times 3 = 6\)). Place a coefficient of 6 in front of \(\text{AgI}\).

\[
6 \text{AgI} + 1 \text{Fe}_2\text{(CO}_3\text{)}_3 \rightarrow 2 \text{FeI}_3 + \_\_ \text{Ag}_2\text{CO}_3
\]

- Step 3: Balance the silver (Ag) atoms.
- There are 6 Ag atoms in \(6 \text{AgI}\). Since each \(\text{Ag}_2\text{CO}_3\) contains 2 Ag atoms, place a coefficient of 3 in front of \(\text{Ag}_2\text{CO}_3\).

\[
6 \text{AgI} + 1 \text{Fe}_2\text{(CO}_3\text{)}_3 \rightarrow 2 \text{FeI}_3 + 3 \text{Ag}_2\text{CO}_3
\]

- Step 4: Verify all atoms.
- Ag: 6 on both sides
- I: 6 on both sides
- Fe: 2 on both sides
- C: 3 on both sides
- O: \(9\) (from \(3 \times 3\) in \(\text{Fe}_2\text{(CO}_3\text{)}_3\)) = \(9\) (from \(3 \times 3\) in \(\text{Ag}_2\text{CO}_3\))

The balanced equation is:
\[
\boxed{6 \text{AgI} + 1 \text{Fe}_2\text{(CO}_3\text{)}_3 \rightarrow 2 \text{FeI}_3 + 3 \text{Ag}_2\text{CO}_3}
\]

---

3. \( \_\_ \text{C}_2\text{H}_5\text{O}_2 + \_\_ \text{O}_2 \rightarrow \_\_ \text{CO}_2 + \_\_ \text{H}_2\text{O} \)

- Step 1: Balance the carbon (C) atoms.
- There are 2 C atoms in \(\text{C}_2\text{H}_5\text{O}_2\) and 1 C atom in \(\text{CO}_2\). Place a coefficient of 2 in front of \(\text{CO}_2\).

\[
1 \text{C}_2\text{H}_5\text{O}_2 + \_\_ \text{O}_2 \rightarrow 2 \text{CO}_2 + \_\_ \text{H}_2\text{O}
\]

- Step 2: Balance the hydrogen (H) atoms.
- There are 5 H atoms in \(\text{C}_2\text{H}_5\text{O}_2\) and 2 H atoms in \(\text{H}_2\text{O}\). Place a coefficient of \( \frac{5}{2} \) in front of \(\text{H}_2\text{O}\). To avoid fractions, multiply all coefficients by 2.

\[
2 \text{C}_2\text{H}_5\text{O}_2 + \_\_ \text{O}_2 \rightarrow 4 \text{CO}_2 + 5 \text{H}_2\text{O}
\]

- Step 3: Balance the oxygen (O) atoms.
- There are 7 O atoms on the left side (\(2 \times 2\) from \(\text{C}_2\text{H}_5\text{O}_2\)) and 13 O atoms on the right side (\(4 \times 2\) from \(\text{CO}_2\) + \(5 \times 1\) from \(\text{H}_2\text{O}\)). Place a coefficient of 6 in front of \(\text{O}_2\).

\[
2 \text{C}_2\text{H}_5\text{O}_2 + 6 \text{O}_2 \rightarrow 4 \text{CO}_2 + 5 \text{H}_2\text{O}
\]

- Step 4: Verify all atoms.
- C: 4 on both sides
- H: 10 on both sides
- O: 14 on both sides

The balanced equation is:
\[
\boxed{2 \text{C}_2\text{H}_5\text{O}_2 + 6 \text{O}_2 \rightarrow 4 \text{CO}_2 + 5 \text{H}_2\text{O}}
\]

---

4. \( \_\_ \text{ZnSO}_4 + \_\_ \text{Li}_2\text{CO}_3 \rightarrow \_\_ \text{ZnCO}_3 + \_\_ \text{Li}_2\text{SO}_4 \)

- Step 1: Balance the zinc (Zn) atoms.
- There is 1 Zn atom in \(\text{ZnSO}_4\) and 1 Zn atom in \(\text{ZnCO}_3\). Place a coefficient of 1 in front of both \(\text{ZnSO}_4\) and \(\text{ZnCO}_3\).

\[
1 \text{ZnSO}_4 + \_\_ \text{Li}_2\text{CO}_3 \rightarrow 1 \text{ZnCO}_3 + \_\_ \text{Li}_2\text{SO}_4
\]

- Step 2: Balance the lithium (Li) atoms.
- There are 2 Li atoms in \(\text{Li}_2\text{CO}_3\) and 2 Li atoms in \(\text{Li}_2\text{SO}_4\). Place a coefficient of 1 in front of both \(\text{Li}_2\text{CO}_3\) and \(\text{Li}_2\text{SO}_4\).

\[
1 \text{ZnSO}_4 + 1 \text{Li}_2\text{CO}_3 \rightarrow 1 \text{ZnCO}_3 + 1 \text{Li}_2\text{SO}_4
\]

- Step 3: Verify all atoms.
- Zn: 1 on both sides
- S: 1 on both sides
- O: 7 on both sides
- Li: 2 on both sides
- C: 1 on both sides

The balanced equation is:
\[
\boxed{1 \text{ZnSO}_4 + 1 \text{Li}_2\text{CO}_3 \rightarrow 1 \text{ZnCO}_3 + 1 \text{Li}_2\text{SO}_4}
\]

---

5. \( \_\_ \text{V}_2\text{O}_5 + \_\_ \text{CaS} \rightarrow \_\_ \text{CaO} + \_\_ \text{V}_2\text{S}_3 \)

- Step 1: Balance the vanadium (V) atoms.
- There are 2 V atoms in \(\text{V}_2\text{O}_5\) and 2 V atoms in \(\text{V}_2\text{S}_3\). Place a coefficient of 1 in front of both \(\text{V}_2\text{O}_5\) and \(\text{V}_2\text{S}_3\).

\[
1 \text{V}_2\text{O}_5 + \_\_ \text{CaS} \rightarrow \_\_ \text{CaO} + 1 \text{V}_2\text{S}_3
\]

- Step 2: Balance the sulfur (S) atoms.
- There are 3 S atoms in \(\text{V}_2\text{S}_3\) and 1 S atom in \(\text{CaS}\). Place a coefficient of 3 in front of \(\text{CaS}\).

\[
1 \text{V}_2\text{O}_5 + 3 \text{CaS} \rightarrow \_\_ \text{CaO} + 1 \text{V}_2\text{S}_3
\]

- Step 3: Balance the calcium (Ca) atoms.
- There are 3 Ca atoms in \(3 \text{CaS}\) and 1 Ca atom in \(\text{CaO}\). Place a coefficient of 3 in front of \(\text{CaO}\).

\[
1 \text{V}_2\text{O}_5 + 3 \text{CaS} \rightarrow 3 \text{CaO} + 1 \text{V}_2\text{S}_3
\]

- Step 4: Verify all atoms.
- V: 2 on both sides
- O: 5 on both sides
- Ca: 3 on both sides
- S: 3 on both sides

The balanced equation is:
\[
\boxed{1 \text{V}_2\text{O}_5 + 3 \text{CaS} \rightarrow 3 \text{CaO} + 1 \text{V}_2\text{S}_3}
\]

---

6. \( \_\_ \text{Mn(NO}_3\text{)}_2 + \_\_ \text{BeCl}_2 \rightarrow \_\_ \text{Be(NO}_3\text{)}_2 + \_\_ \text{MnCl}_2 \)

- Step 1: Balance the manganese (Mn) atoms.
- There is 1 Mn atom in \(\text{Mn(NO}_3\text{)}_2\) and 1 Mn atom in \(\text{MnCl}_2\). Place a coefficient of 1 in front of both \(\text{Mn(NO}_3\text{)}_2\) and \(\text{MnCl}_2\).

\[
1 \text{Mn(NO}_3\text{)}_2 + \_\_ \text{BeCl}_2 \rightarrow \_\_ \text{Be(NO}_3\text{)}_2 + 1 \text{MnCl}_2
\]

- Step 2: Balance the beryllium (Be) atoms.
- There is 1 Be atom in \(\text{BeCl}_2\) and 1 Be atom in \(\text{Be(NO}_3\text{)}_2\). Place a coefficient of 1 in front of both \(\text{BeCl}_2\) and \(\text{Be(NO}_3\text{)}_2\).

\[
1 \text{Mn(NO}_3\text{)}_2 + 1 \text{BeCl}_2 \rightarrow 1 \text{Be(NO}_3\text{)}_2 + 1 \text{MnCl}_2
\]

- Step 3: Verify all atoms.
- Mn: 1 on both sides
- Be: 1 on both sides
- N: 2 on both sides
- O: 6 on both sides
- Cl: 2 on both sides

The balanced equation is:
\[
\boxed{1 \text{Mn(NO}_3\text{)}_2 + 1 \text{BeCl}_2 \rightarrow 1 \text{Be(NO}_3\text{)}_2 + 1 \text{MnCl}_2}
\]

---

7. \( \_\_ \text{AgBr} + \_\_ \text{GaPO}_4 \rightarrow \_\_ \text{Ag}_3\text{PO}_4 + \_\_ \text{GaBr}_3 \)

- Step 1: Balance the gallium (Ga) atoms.
- There is 1 Ga atom in \(\text{GaPO}_4\) and 1 Ga atom in \(\text{GaBr}_3\). Place a coefficient of 1 in front of both \(\text{GaPO}_4\) and \(\text{GaBr}_3\).

\[
\_\_ \text{AgBr} + 1 \text{GaPO}_4 \rightarrow \_\_ \text{Ag}_3\text{PO}_4 + 1 \text{GaBr}_3
\]

- Step 2: Balance the bromine (Br) atoms.
- There are 3 Br atoms in \(\text{GaBr}_3\) and 1 Br atom in \(\text{AgBr}\). Place a coefficient of 3 in front of \(\text{AgBr}\).

\[
3 \text{AgBr} + 1 \text{GaPO}_4 \rightarrow \_\_ \text{Ag}_3\text{PO}_4 + 1 \text{GaBr}_3
\]

- Step 3: Balance the silver (Ag) atoms.
- There are 3 Ag atoms in \(3 \text{AgBr}\) and 3 Ag atoms in \(\text{Ag}_3\text{PO}_4\). Place a coefficient of 1 in front of \(\text{Ag}_3\text{PO}_4\).

\[
3 \text{AgBr} + 1 \text{GaPO}_4 \rightarrow 1 \text{Ag}_3\text{PO}_4 + 1 \text{GaBr}_3
\]

- Step 4: Verify all atoms.
- Ag: 3 on both sides
- Br: 3 on both sides
- Ga: 1 on both sides
- P: 1 on both sides
- O: 4 on both sides

The balanced equation is:
\[
\boxed{3 \text{AgBr} + 1 \text{GaPO}_4 \rightarrow 1 \text{Ag}_3\text{PO}_4 + 1 \text{GaBr}_3}
\]

---

8. \( \_\_ \text{H}_2\text{SO}_4 + \_\_ \text{B(OH)}_3 \rightarrow \_\_ \text{B}_2\text{(SO}_4\text{)}_3 + \_\_ \text{H}_2\text{O} \)

- Step 1: Balance the boron (B) atoms.
- There is 1 B atom in \(\text{B(OH)}_3\) and 2 B atoms in \(\text{B}_2\text{(SO}_4\text{)}_3\). Place a coefficient of 2 in front of \(\text{B(OH)}_3\).

\[
\_\_ \text{H}_2\text{SO}_4 + 2 \text{B(OH)}_3 \rightarrow 1 \text{B}_2\text{(SO}_4\text{)}_3 + \_\_ \text{H}_2\text{O}
\]

- Step 2: Balance the sulfur (S) atoms.
- There is 1 S atom in \(\text{H}_2\text{SO}_4\) and 3 S atoms in \(\text{B}_2\text{(SO}_4\text{)}_3\). Place a coefficient of 3 in front of \(\text{H}_2\text{SO}_4\).

\[
3 \text{H}_2\text{SO}_4 + 2 \text{B(OH)}_3 \rightarrow 1 \text{B}_2\text{(SO}_4\text{)}_3 + \_\_ \text{H}_2\text{O}
\]

- Step 3: Balance the hydrogen (H) atoms.
- There are 12 H atoms on the left side (\(3 \times 2\) from \(\text{H}_2\text{SO}_4\) + \(2 \times 3\) from \(\text{B(OH)}_3\)) and 2 H atoms in \(\text{H}_2\text{O}\). Place a coefficient of 6 in front of \(\text{H}_2\text{O}\).

\[
3 \text{H}_2\text{SO}_4 + 2 \text{B(OH)}_3 \rightarrow 1 \text{B}_2\text{(SO}_4\text{)}_3 + 6 \text{H}_2\text{O}
\]

- Step 4: Verify all atoms.
- H: 12 on both sides
- S: 3 on both sides
- O: 21 on both sides
- B: 2 on both sides

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

---

9. \( \_\_ \text{S}_8 + \_\_ \text{O}_2 \rightarrow \_\_ \text{SO}_2 \)

- Step 1: Balance the sulfur (S) atoms.
- There are 8 S atoms in \(\text{S}_8\) and 1 S atom in \(\text{SO}_2\). Place a coefficient of 8 in front of \(\text{SO}_2\).

\[
1 \text{S}_8 + \_\_ \text{O}_2 \rightarrow 8 \text{SO}_2
\]

- Step 2: Balance the oxygen (O) atoms.
- There are 16 O atoms in \(8 \text{SO}_2\) and 2 O atoms in \(\text{O}_2\). Place a coefficient of 8 in front of \(\text{O}_2\).

\[
1 \text{S}_8 + 8 \text{O}_2 \rightarrow 8 \text{SO}_2
\]

- Step 3: Verify all atoms.
- S: 8 on both sides
- O: 16 on both sides

The balanced equation is:
\[
\boxed{1 \text{S}_8 + 8 \text{O}_2 \rightarrow 8 \text{SO}_2}
\]

---

10. \( \_\_ \text{Fe} + \_\_ \text{AgNO}_3 \rightarrow \_\_ \text{Fe(NO}_3\text{)}_2 + \_\_ \text{Ag} \)

- Step 1: Balance the iron (Fe) atoms.
- There is 1 Fe atom in \(\text{Fe}\) and 1 Fe atom in \(\text{Fe(NO}_3\text{)}_2\). Place a coefficient of 1 in front of both \(\text{Fe}\) and \(\text{Fe(NO}_3\text{)}_2\).

\[
1 \text{Fe} + \_\_ \text{AgNO}_3 \rightarrow 1 \text{Fe(NO}_3\text{)}_2 + \_\_ \text{Ag}
\]

- Step 2: Balance the silver (Ag) atoms.
- There is 1 Ag atom in \(\text{AgNO}_3\) and 1 Ag atom in \(\text{Ag}\). Since there are 2 nitrate ions in \(\text{Fe(NO}_3\text{)}_2\), place a coefficient of 2 in front of \(\text{AgNO}_3\) and \(\text{Ag}\).

\[
1 \text{Fe} + 2 \text{AgNO}_3 \rightarrow 1 \text{Fe(NO}_3\text{)}_2 + 2 \text{Ag}
\]

- Step 3: Verify all atoms.
- Fe: 1 on both sides
- Ag: 2 on both sides
- N: 2 on both sides
- O: 6 on both sides

The balanced equation is:
\[
\boxed{1 \text{Fe} + 2 \text{AgNO}_3 \rightarrow 1 \text{Fe(NO}_3\text{)}_2 + 2 \text{Ag}}
\]

---

Final Answer:

\[
\boxed{
\begin{aligned}
1. & \quad 2 \text{NaNO}_3 + 1 \text{PbO} \rightarrow 1 \text{Pb(NO}_3\text{)}_2 + 1 \text{Na}_2\text{O} \\
2. & \quad 6 \text{AgI} + 1 \text{Fe}_2\text{(CO}_3\text{)}_3 \rightarrow 2 \text{FeI}_3 + 3 \text{Ag}_2\text{CO}_3 \\
3. & \quad 2 \text{C}_2\text{H}_5\text{O}_2 + 6 \text{O}_2 \rightarrow 4 \text{CO}_2 + 5 \text{H}_2\text{O} \\
4. & \quad 1 \text{ZnSO}_4 + 1 \text{Li}_2\text{CO}_3 \rightarrow 1 \text{ZnCO}_3 + 1 \text{Li}_2\text{SO}_4 \\
5. & \quad 1 \text{V}_2\text{O}_5 + 3 \text{CaS} \rightarrow 3 \text{CaO} + 1 \text{V}_2\text{S}_3 \\
6. & \quad 1 \text{Mn(NO}_3\text{)}_2 + 1 \text{BeCl}_2 \rightarrow 1 \text{Be(NO}_3\text{)}_2 + 1 \text{MnCl}_2 \\
7. & \quad 3 \text{AgBr} + 1 \text{GaPO}_4 \rightarrow 1 \text{Ag}_3\text{PO}_4 + 1 \text{GaBr}_3 \\
8. & \quad 3 \text{H}_2\text{SO}_4 + 2 \text{B(OH)}_3 \rightarrow 1 \text{B}_2\text{(SO}_4\text{)}_3 + 6 \text{H}_2\text{O} \\
9. & \quad 1 \text{S}_8 + 8 \text{O}_2 \rightarrow 8 \text{SO}_2 \\
10. & \quad 1 \text{Fe} + 2 \text{AgNO}_3 \rightarrow 1 \text{Fe(NO}_3\text{)}_2 + 2 \text{Ag}
\end{aligned}
}
\]