Balancing Equations Worksheet featuring 14 chemical equations to practice balancing.

To solve the problem of balancing 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:

---

1) \( \_\_ \text{Na}_3\text{PO}_4 + \_\_ \text{KOH} \rightarrow \_\_ \text{NaOH} + \_\_ \text{K}_3\text{PO}_4 \)

- Start by balancing sodium (Na): There are 3 Na atoms in \(\text{Na}_3\text{PO}_4\) and 1 Na atom in \(\text{NaOH}\). To balance Na, we need 3 \(\text{NaOH}\).
- Next, balance potassium (K): There is 1 K atom in \(\text{KOH}\) and 3 K atoms in \(\text{K}_3\text{PO}_4\). To balance K, we need 3 \(\text{KOH}\).
- Finally, check phosphorus (P) and oxygen (O): Both are already balanced.

Balanced Equation:
\[ \boxed{1 \text{Na}_3\text{PO}_4 + 3 \text{KOH} \rightarrow 3 \text{NaOH} + 1 \text{K}_3\text{PO}_4} \]

---

2) \( \_\_ \text{MgF}_2 + \_\_ \text{Li}_2\text{CO}_3 \rightarrow \_\_ \text{MgCO}_3 + \_\_ \text{LiF} \)

- Balance magnesium (Mg): There is 1 Mg atom in \(\text{MgF}_2\) and 1 Mg atom in \(\text{MgCO}_3\). So, 1 \(\text{MgF}_2\) is needed.
- Balance lithium (Li): There are 2 Li atoms in \(\text{Li}_2\text{CO}_3\) and 1 Li atom in \(\text{LiF}\). To balance Li, we need 2 \(\text{LiF}\).
- Balance fluorine (F): There are 2 F atoms in \(\text{MgF}_2\) and 1 F atom in \(\text{LiF}\). Since we have 2 \(\text{LiF}\), F is balanced.
- Check carbon (C) and oxygen (O): Both are already balanced.

Balanced Equation:
\[ \boxed{1 \text{MgF}_2 + 1 \text{Li}_2\text{CO}_3 \rightarrow 1 \text{MgCO}_3 + 2 \text{LiF}} \]

---

3) \( \_\_ \text{P}_4 + \_\_ \text{O}_2 \rightarrow \_\_ \text{P}_2\text{O}_3 \)

- Balance phosphorus (P): There are 4 P atoms in \(\text{P}_4\) and 2 P atoms in \(\text{P}_2\text{O}_3\). To balance P, we need 2 \(\text{P}_2\text{O}_3\).
- Balance oxygen (O): There are 3 O atoms in \(\text{P}_2\text{O}_3\) and 2 O atoms in \(\text{O}_2\). Since we have 2 \(\text{P}_2\text{O}_3\), there are \(2 \times 3 = 6\) O atoms total. To balance O, we need 3 \(\text{O}_2\).

Balanced Equation:
\[ \boxed{1 \text{P}_4 + 3 \text{O}_2 \rightarrow 2 \text{P}_2\text{O}_3} \]

---

4) \( \_\_ \text{RbNO}_3 + \_\_ \text{BeF}_2 \rightarrow \_\_ \text{Be(NO}_3)_2 + \_\_ \text{RbF} \)

- Balance beryllium (Be): There is 1 Be atom in \(\text{BeF}_2\) and 1 Be atom in \(\text{Be(NO}_3)_2\). So, 1 \(\text{BeF}_2\) is needed.
- Balance rubidium (Rb): There is 1 Rb atom in \(\text{RbNO}_3\) and 1 Rb atom in \(\text{RbF}\). Since we have 2 \(\text{RbF}\), we need 2 \(\text{RbNO}_3\).
- Balance nitrogen (N) and oxygen (O): There are 2 N atoms and 6 O atoms in \(\text{Be(NO}_3)_2\). Since we have 2 \(\text{RbNO}_3\), N and O are balanced.
- Check fluorine (F): There are 2 F atoms in \(\text{BeF}_2\) and 1 F atom in \(\text{RbF}\). Since we have 2 \(\text{RbF}\), F is balanced.

Balanced Equation:
\[ \boxed{2 \text{RbNO}_3 + 1 \text{BeF}_2 \rightarrow 1 \text{Be(NO}_3)_2 + 2 \text{RbF}} \]

---

5) \( \_\_ \text{AgNO}_3 + \_\_ \text{Cu} \rightarrow \_\_ \text{Cu(NO}_3)_2 + \_\_ \text{Ag} \)

- Balance copper (Cu): There is 1 Cu atom in \(\text{Cu}\) and 1 Cu atom in \(\text{Cu(NO}_3)_2\). So, 1 \(\text{Cu}\) is needed.
- Balance silver (Ag): There is 1 Ag atom in \(\text{AgNO}_3\) and 1 Ag atom in \(\text{Ag}\). Since we have 2 \(\text{Ag}\), we need 2 \(\text{AgNO}_3\).
- Balance nitrogen (N) and oxygen (O): There are 2 N atoms and 6 O atoms in \(\text{Cu(NO}_3)_2\). Since we have 2 \(\text{AgNO}_3\), N and O are balanced.

Balanced Equation:
\[ \boxed{2 \text{AgNO}_3 + 1 \text{Cu} \rightarrow 1 \text{Cu(NO}_3)_2 + 2 \text{Ag}} \]

---

6) \( \_\_ \text{CF}_4 + \_\_ \text{Br}_2 \rightarrow \_\_ \text{CBr}_4 + \_\_ \text{F}_2 \)

- Balance carbon (C): There is 1 C atom in \(\text{CF}_4\) and 1 C atom in \(\text{CBr}_4\). So, 1 \(\text{CF}_4\) is needed.
- Balance fluorine (F): There are 4 F atoms in \(\text{CF}_4\) and 2 F atoms in \(\text{F}_2\). To balance F, we need 2 \(\text{F}_2\).
- Balance bromine (Br): There are 2 Br atoms in \(\text{Br}_2\) and 4 Br atoms in \(\text{CBr}_4\). To balance Br, we need 2 \(\text{Br}_2\).

Balanced Equation:
\[ \boxed{1 \text{CF}_4 + 2 \text{Br}_2 \rightarrow 1 \text{CBr}_4 + 2 \text{F}_2} \]

---

7) \( \_\_ \text{HCN} + \_\_ \text{CuSO}_4 \rightarrow \_\_ \text{H}_2\text{SO}_4 + \_\_ \text{Cu(CN)}_2 \)

- Balance hydrogen (H): There is 1 H atom in \(\text{HCN}\) and 2 H atoms in \(\text{H}_2\text{SO}_4\). To balance H, we need 2 \(\text{HCN}\).
- Balance sulfur (S): There is 1 S atom in \(\text{CuSO}_4\) and 1 S atom in \(\text{H}_2\text{SO}_4\). So, 1 \(\text{CuSO}_4\) is needed.
- Balance copper (Cu): There is 1 Cu atom in \(\text{CuSO}_4\) and 1 Cu atom in \(\text{Cu(CN)}_2\). So, 1 \(\text{Cu(CN)}_2\) is needed.
- Balance nitrogen (N): There is 1 N atom in \(\text{HCN}\) and 2 N atoms in \(\text{Cu(CN)}_2\). Since we have 2 \(\text{HCN}\), N is balanced.
- Check oxygen (O): There are 4 O atoms in \(\text{CuSO}_4\) and 4 O atoms in \(\text{H}_2\text{SO}_4\). O is balanced.

Balanced Equation:
\[ \boxed{2 \text{HCN} + 1 \text{CuSO}_4 \rightarrow 1 \text{H}_2\text{SO}_4 + 1 \text{Cu(CN)}_2} \]

---

8) \( \_\_ \text{GaF}_3 + \_\_ \text{Cs} \rightarrow \_\_ \text{CsF} + \_\_ \text{Ga} \)

- Balance gallium (Ga): There is 1 Ga atom in \(\text{GaF}_3\) and 1 Ga atom in \(\text{Ga}\). So, 1 \(\text{GaF}_3\) is needed.
- Balance cesium (Cs): There is 1 Cs atom in \(\text{Cs}\) and 1 Cs atom in \(\text{CsF}\). Since we have 3 \(\text{CsF}\), we need 3 \(\text{Cs}\).
- Balance fluorine (F): There are 3 F atoms in \(\text{GaF}_3\) and 1 F atom in \(\text{CsF}\). Since we have 3 \(\text{CsF}\), F is balanced.

Balanced Equation:
\[ \boxed{1 \text{GaF}_3 + 3 \text{Cs} \rightarrow 3 \text{CsF} + 1 \text{Ga}} \]

---

9) \( \_\_ \text{BaS} + \_\_ \text{PtF}_2 \rightarrow \_\_ \text{BaF}_2 + \_\_ \text{PtS} \)

- Balance barium (Ba): There is 1 Ba atom in \(\text{BaS}\) and 1 Ba atom in \(\text{BaF}_2\). So, 1 \(\text{BaS}\) is needed.
- Balance platinum (Pt): There is 1 Pt atom in \(\text{PtF}_2\) and 1 Pt atom in \(\text{PtS}\). So, 1 \(\text{PtF}_2\) is needed.
- Balance sulfur (S): There is 1 S atom in \(\text{BaS}\) and 1 S atom in \(\text{PtS}\). S is balanced.
- Balance fluorine (F): There are 2 F atoms in \(\text{PtF}_2\) and 2 F atoms in \(\text{BaF}_2\). F is balanced.

Balanced Equation:
\[ \boxed{1 \text{BaS} + 1 \text{PtF}_2 \rightarrow 1 \text{BaF}_2 + 1 \text{PtS}} \]

---

10) \( \_\_ \text{N}_2 + \_\_ \text{H}_2 \rightarrow \_\_ \text{NH}_3 \)

- Balance nitrogen (N): There are 2 N atoms in \(\text{N}_2\) and 1 N atom in \(\text{NH}_3\). To balance N, we need 2 \(\text{NH}_3\).
- Balance hydrogen (H): There are 2 H atoms in \(\text{H}_2\) and 3 H atoms in \(\text{NH}_3\). Since we have 2 \(\text{NH}_3\), there are \(2 \times 3 = 6\) H atoms total. To balance H, we need 3 \(\text{H}_2\).

Balanced Equation:
\[ \boxed{1 \text{N}_2 + 3 \text{H}_2 \rightarrow 2 \text{NH}_3} \]

---

11) \( \_\_ \text{NaF} + \_\_ \text{Br}_2 \rightarrow \_\_ \text{NaBr} + \_\_ \text{F}_2 \)

- Balance sodium (Na): There is 1 Na atom in \(\text{NaF}\) and 1 Na atom in \(\text{NaBr}\). Since we have 2 \(\text{NaBr}\), we need 2 \(\text{NaF}\).
- Balance bromine (Br): There are 2 Br atoms in \(\text{Br}_2\) and 1 Br atom in \(\text{NaBr}\). Since we have 2 \(\text{NaBr}\), Br is balanced.
- Balance fluorine (F): There is 1 F atom in \(\text{NaF}\) and 2 F atoms in \(\text{F}_2\). Since we have 2 \(\text{NaF}\), F is balanced.

Balanced Equation:
\[ \boxed{2 \text{NaF} + 1 \text{Br}_2 \rightarrow 2 \text{NaBr} + 1 \text{F}_2} \]

---

12) \( \_\_ \text{Pb(OH)}_2 + \_\_ \text{HCl} \rightarrow \_\_ \text{H}_2\text{O} + \_\_ \text{PbCl}_2 \)

- Balance lead (Pb): There is 1 Pb atom in \(\text{Pb(OH)}_2\) and 1 Pb atom in \(\text{PbCl}_2\). So, 1 \(\text{Pb(OH)}_2\) is needed.
- Balance chlorine (Cl): There is 1 Cl atom in \(\text{HCl}\) and 2 Cl atoms in \(\text{PbCl}_2\). To balance Cl, we need 2 \(\text{HCl}\).
- Balance hydrogen (H): There are 2 H atoms in \(\text{Pb(OH)}_2\) and 1 H atom in \(\text{HCl}\). Since we have 2 \(\text{HCl}\), there are \(2 + 2 = 4\) H atoms total. To balance H, we need 2 \(\text{H}_2\text{O}\).
- Balance oxygen (O): There are 2 O atoms in \(\text{Pb(OH)}_2\) and 1 O atom in \(\text{H}_2\text{O}\). Since we have 2 \(\text{H}_2\text{O}\), O is balanced.

Balanced Equation:
\[ \boxed{1 \text{Pb(OH)}_2 + 2 \text{HCl} \rightarrow 2 \text{H}_2\text{O} + 1 \text{PbCl}_2} \]

---

13) \( \_\_ \text{AlBr}_3 + \_\_ \text{K}_2\text{SO}_4 \rightarrow \_\_ \text{KBr} + \_\_ \text{Al}_2(\text{SO}_4)_3 \)

- Balance aluminum (Al): There is 1 Al atom in \(\text{AlBr}_3\) and 2 Al atoms in \(\text{Al}_2(\text{SO}_4)_3\). To balance Al, we need 2 \(\text{AlBr}_3\).
- Balance potassium (K): There are 2 K atoms in \(\text{K}_2\text{SO}_4\) and 1 K atom in \(\text{KBr}\). Since we have 6 \(\text{KBr}\), we need 3 \(\text{K}_2\text{SO}_4\).
- Balance bromine (Br): There are 3 Br atoms in \(\text{AlBr}_3\) and 1 Br atom in \(\text{KBr}\). Since we have 6 \(\text{KBr}\), we need 2 \(\text{AlBr}_3\).
- Balance sulfur (S) and oxygen (O): There are 3 S atoms and 12 O atoms in \(\text{Al}_2(\text{SO}_4)_3\). Since we have 3 \(\text{K}_2\text{SO}_4\), S and O are balanced.

Balanced Equation:
\[ \boxed{2 \text{AlBr}_3 + 3 \text{K}_2\text{SO}_4 \rightarrow 6 \text{KBr} + 1 \text{Al}_2(\text{SO}_4)_3} \]

---

14) \( \_\_ \text{CH}_4 + \_\_ \text{O}_2 \rightarrow \_\_ \text{CO}_2 + \_\_ \text{H}_2\text{O} \)

- Balance carbon (C): There is 1 C atom in \(\text{CH}_4\) and 1 C atom in \(\text{CO}_2\). So, 1 \(\text{CH}_4\) is needed.
- Balance hydrogen (H): There are 4 H atoms in \(\text{CH}_4\) and 2 H atoms in \(\text{H}_2\text{O}\). To balance H, we need 2 \(\text{H}_2\text{O}\).
- Balance oxygen (O): There are 2 O atoms in \(\text{O}_2\) and 2 O atoms in \(\text{CO}_2\) plus 1 O atom in \(\text{H}_2\text{O}\). Since we have 2 \(\text{H}_2\text{O}\), there are \(2 + 2 = 4\) O atoms total. To balance O, we need 2 \(\text{O}_2\).

Balanced Equation:
\[ \boxed{1 \text{CH}_4 + 2 \text{O}_2 \rightarrow 1 \text{CO}_2 + 2 \text{H}_2\text{O}} \]

---

Final Answer:

\[
\boxed{
\begin{aligned}
1. & \ 1 \text{Na}_3\text{PO}_4 + 3 \text{KOH} \rightarrow 3 \text{NaOH} + 1 \text{K}_3\text{PO}_4 \\
2. & \ 1 \text{MgF}_2 + 1 \text{Li}_2\text{CO}_3 \rightarrow 1 \text{MgCO}_3 + 2 \text{LiF} \\
3. & \ 1 \text{P}_4 + 3 \text{O}_2 \rightarrow 2 \text{P}_2\text{O}_3 \\
4. & \ 2 \text{RbNO}_3 + 1 \text{BeF}_2 \rightarrow 1 \text{Be(NO}_3)_2 + 2 \text{RbF} \\
5. & \ 2 \text{AgNO}_3 + 1 \text{Cu} \rightarrow 1 \text{Cu(NO}_3)_2 + 2 \text{Ag} \\
6. & \ 1 \text{CF}_4 + 2 \text{Br}_2 \rightarrow 1 \text{CBr}_4 + 2 \text{F}_2 \\
7. & \ 2 \text{HCN} + 1 \text{CuSO}_4 \rightarrow 1 \text{H}_2\text{SO}_4 + 1 \text{Cu(CN)}_2 \\
8. & \ 1 \text{GaF}_3 + 3 \text{Cs} \rightarrow 3 \text{CsF} + 1 \text{Ga} \\
9. & \ 1 \text{BaS} + 1 \text{PtF}_2 \rightarrow 1 \text{BaF}_2 + 1 \text{PtS} \\
10. & \ 1 \text{N}_2 + 3 \text{H}_2 \rightarrow 2 \text{NH}_3 \\
11. & \ 2 \text{NaF} + 1 \text{Br}_2 \rightarrow 2 \text{NaBr} + 1 \text{F}_2 \\
12. & \ 1 \text{Pb(OH)}_2 + 2 \text{HCl} \rightarrow 2 \text{H}_2\text{O} + 1 \text{PbCl}_2 \\
13. & \ 2 \text{AlBr}_3 + 3 \text{K}_2\text{SO}_4 \rightarrow 6 \text{KBr} + 1 \text{Al}_2(\text{SO}_4)_3 \\
14. & \ 1 \text{CH}_4 + 2 \text{O}_2 \rightarrow 1 \text{CO}_2 + 2 \text{H}_2\text{O}
\end{aligned}
}
\]