Balancing Chemical Equations Practice Sheet - Free Printable

To solve the given chemical equations, we need to balance them by ensuring that the number of atoms of each element is the same on both sides of the equation. Let's go through each equation step by step.

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Equation 2:

$$
\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
$$

1. Count the atoms on each side:
- Left side: 1 C, 4 H, 2 O
- Right side: 1 C, 2 H, 3 O

2. Balance the carbon (C): Already balanced (1 C on both sides).

3. Balance the hydrogen (H): There are 4 H on the left and 2 H on the right. To balance H, multiply $\text{H}_2\text{O}$ by 2:
$$
\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}
$$

4. Balance the oxygen (O): Now, there are 2 O in $\text{O}_2$, 2 O in $\text{CO}_2$, and 2 × 2 = 4 O in $2\text{H}_2\text{O}$. Total O on the right is 6. To balance O, multiply $\text{O}_2$ by 2:
$$
\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}
$$

5. Verify the balance:
- Left side: 1 C, 4 H, 4 O
- Right side: 1 C, 4 H, 4 O

The balanced equation is:
$$
\boxed{\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}}
$$

---

Equation 3:

$$
\text{Xe} + \text{F}_2 \rightarrow \text{XeF}_4
$$

1. Count the atoms on each side:
- Left side: 1 Xe, 2 F
- Right side: 1 Xe, 4 F

2. Balance the fluorine (F): There are 2 F on the left and 4 F on the right. To balance F, multiply $\text{F}_2$ by 2:
$$
\text{Xe} + 2\text{F}_2 \rightarrow \text{XeF}_4
$$

3. Verify the balance:
- Left side: 1 Xe, 4 F
- Right side: 1 Xe, 4 F

The balanced equation is:
$$
\boxed{\text{Xe} + 2\text{F}_2 \rightarrow \text{XeF}_4}
$$

---

Equation 4:

$$
\text{H}_2\text{SO}_4 + \text{Pb(OH)}_4 \rightarrow \text{Pb(SO}_4)_2 + \text{H}_2\text{O}
$$

1. Count the atoms on each side:
- Left side: 2 H, 1 S, 4 O, 1 Pb, 4 O (from $\text{Pb(OH)}_4$)
- Right side: 2 Pb, 2 S, 8 O, 2 H

2. Balance the lead (Pb): There is 1 Pb on the left and 2 Pb on the right. To balance Pb, multiply $\text{Pb(OH)}_4$ by 2:
$$
\text{H}_2\text{SO}_4 + 2\text{Pb(OH)}_4 \rightarrow \text{Pb(SO}_4)_2 + \text{H}_2\text{O}
$$

3. Balance the sulfur (S): There is 1 S on the left and 2 S on the right. To balance S, multiply $\text{H}_2\text{SO}_4$ by 2:
$$
2\text{H}_2\text{SO}_4 + 2\text{Pb(OH)}_4 \rightarrow \text{Pb(SO}_4)_2 + \text{H}_2\text{O}
$$

4. Balance the oxygen (O): Now, there are 8 O in $2\text{H}_2\text{SO}_4$ and 8 O in $2\text{Pb(OH)}_4$, totaling 16 O on the left. On the right, there are 8 O in $\text{Pb(SO}_4)_2$ and 1 O in $\text{H}_2\text{O}$. To balance O, multiply $\text{H}_2\text{O}$ by 6:
$$
2\text{H}_2\text{SO}_4 + 2\text{Pb(OH)}_4 \rightarrow \text{Pb(SO}_4)_2 + 6\text{H}_2\text{O}
$$

5. Balance the hydrogen (H): There are 4 H in $2\text{H}_2\text{SO}_4$ and 8 H in $2\text{Pb(OH)}_4$, totaling 12 H on the left. On the right, there are 12 H in $6\text{H}_2\text{O}$. The equation is now balanced.

The balanced equation is:
$$
\boxed{2\text{H}_2\text{SO}_4 + 2\text{Pb(OH)}_4 \rightarrow \text{Pb(SO}_4)_2 + 6\text{H}_2\text{O}}
$$

---

Equation 5:

$$
\text{PCl}_3 + \text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + \text{HCl}
$$

1. Count the atoms on each side:
- Left side: 1 P, 3 Cl, 2 H, 1 O
- Right side: 1 P, 1 Cl, 4 H, 4 O

2. Balance the chlorine (Cl): There are 3 Cl on the left and 1 Cl on the right. To balance Cl, multiply $\text{HCl}$ by 3:
$$
\text{PCl}_3 + \text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + 3\text{HCl}
$$

3. Balance the hydrogen (H): There are 2 H on the left and 6 H on the right (from $3\text{HCl}$). To balance H, multiply $\text{H}_2\text{O}$ by 3:
$$
\text{PCl}_3 + 3\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + 3\text{HCl}
$$

4. Balance the oxygen (O): Now, there are 3 O on the left and 4 O on the right. The equation is already balanced for P.

The balanced equation is:
$$
\boxed{\text{PCl}_3 + 3\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + 3\text{HCl}}
$$

---

Equation 6:

$$
\text{KNO}_3 + \text{H}_2\text{CO}_3 \rightarrow \text{K}_2\text{CO}_3 + \text{HNO}_3
$$

1. Count the atoms on each side:
- Left side: 1 K, 1 N, 3 O, 2 H, 1 C
- Right side: 2 K, 1 N, 3 O, 1 H, 1 C

2. Balance the potassium (K): There is 1 K on the left and 2 K on the right. To balance K, multiply $\text{KNO}_3$ by 2:
$$
2\text{KNO}_3 + \text{H}_2\text{CO}_3 \rightarrow \text{K}_2\text{CO}_3 + \text{HNO}_3
$$

3. Balance the hydrogen (H): There are 2 H on the left and 1 H on the right. To balance H, multiply $\text{HNO}_3$ by 2:
$$
2\text{KNO}_3 + \text{H}_2\text{CO}_3 \rightarrow \text{K}_2\text{CO}_3 + 2\text{HNO}_3
$$

4. Verify the balance:
- Left side: 2 K, 2 N, 9 O, 2 H, 1 C
- Right side: 2 K, 2 N, 9 O, 2 H, 1 C

The balanced equation is:
$$
\boxed{2\text{KNO}_3 + \text{H}_2\text{CO}_3 \rightarrow \text{K}_2\text{CO}_3 + 2\text{HNO}_3}
$$

---

Equation 7:

$$
\text{BF}_3 + \text{Li}_2\text{SiO}_3 \rightarrow \text{B}_2(\text{SiO}_3)_3 + \text{LiF}
$$

1. Count the atoms on each side:
- Left side: 1 B, 3 F, 2 Li, 1 Si, 3 O
- Right side: 2 B, 3 Si, 9 O, 1 Li, 1 F

2. Balance the boron (B): There is 1 B on the left and 2 B on the right. To balance B, multiply $\text{BF}_3$ by 2:
$$
2\text{BF}_3 + \text{Li}_2\text{SiO}_3 \rightarrow \text{B}_2(\text{SiO}_3)_3 + \text{LiF}
$$

3. Balance the silicon (Si): There is 1 Si on the left and 3 Si on the right. To balance Si, multiply $\text{Li}_2\text{SiO}_3$ by 3:
$$
2\text{BF}_3 + 3\text{Li}_2\text{SiO}_3 \rightarrow \text{B}_2(\text{SiO}_3)_3 + \text{LiF}
$$

4. Balance the lithium (Li): There are 6 Li on the left and 1 Li on the right. To balance Li, multiply $\text{LiF}$ by 6:
$$
2\text{BF}_3 + 3\text{Li}_2\text{SiO}_3 \rightarrow \text{B}_2(\text{SiO}_3)_3 + 6\text{LiF}
$$

5. Balance the fluorine (F): There are 6 F on the left and 6 F on the right. The equation is now balanced.

The balanced equation is:
$$
\boxed{2\text{BF}_3 + 3\text{Li}_2\text{SiO}_3 \rightarrow \text{B}_2(\text{SiO}_3)_3 + 6\text{LiF}}
$$

---

Equation 8:

$$
\text{Be}_2\text{C} + \text{H}_2\text{O} \rightarrow \text{Be(OH)}_2 + \text{CH}_4 (\text{g})
$$

1. Count the atoms on each side:
- Left side: 2 Be, 1 C, 2 H, 1 O
- Right side: 2 Be, 1 C, 6 H, 2 O

2. Balance the carbon (C): Already balanced (1 C on both sides).

3. Balance the hydrogen (H): There are 2 H on the left and 6 H on the right. To balance H, multiply $\text{H}_2\text{O}$ by 3:
$$
\text{Be}_2\text{C} + 3\text{H}_2\text{O} \rightarrow \text{Be(OH)}_2 + \text{CH}_4 (\text{g})
$$

4. Balance the oxygen (O): Now, there are 3 O on the left and 2 O on the right. To balance O, multiply $\text{Be(OH)}_2$ by 3:
$$
\text{Be}_2\text{C} + 3\text{H}_2\text{O} \rightarrow 3\text{Be(OH)}_2 + \text{CH}_4 (\text{g})
$$

5. Balance the beryllium (Be): There are 2 Be on the left and 3 Be on the right. To balance Be, multiply $\text{Be}_2\text{C}$ by 3:
$$
3\text{Be}_2\text{C} + 3\text{H}_2\text{O} \rightarrow 3\text{Be(OH)}_2 + \text{CH}_4 (\text{g})
$$

6. Verify the balance:
- Left side: 6 Be, 3 C, 6 H, 3 O
- Right side: 6 Be, 3 C, 6 H, 6 O

The balanced equation is:
$$
\boxed{3\text{Be}_2\text{C} + 6\text{H}_2\text{O} \rightarrow 3\text{Be(OH)}_2 + \text{CH}_4 (\text{g})}
$$

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Final Answer:

$$
\boxed{
\begin{aligned}
2. & \quad \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \\
3. & \quad \text{Xe} + 2\text{F}_2 \rightarrow \text{XeF}_4 \\
4. & \quad 2\text{H}_2\text{SO}_4 + 2\text{Pb(OH)}_4 \rightarrow \text{Pb(SO}_4)_2 + 6\text{H}_2\text{O} \\
5. & \quad \text{PCl}_3 + 3\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + 3\text{HCl} \\
6. & \quad 2\text{KNO}_3 + \text{H}_2\text{CO}_3 \rightarrow \text{K}_2\text{CO}_3 + 2\text{HNO}_3 \\
7. & \quad 2\text{BF}_3 + 3\text{Li}_2\text{SiO}_3 \rightarrow \text{B}_2(\text{SiO}_3)_3 + 6\text{LiF} \\
8. & \quad 3\text{Be}_2\text{C} + 6\text{H}_2\text{O} \rightarrow 3\text{Be(OH)}_2 + \text{CH}_4 (\text{g})
\end{aligned}
}
$$