Writing and Balancing Chemical Equations 25-Question Worksheet - 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. Let's solve each equation step by step.

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

- Reactants: \( \text{N}_2 \) and \( \text{H}_2 \)
- Products: \( \text{NH}_3 \)

#### Step 1: Count the atoms
- Left side: 2 N atoms, 2 H atoms
- Right side: 1 N atom, 3 H atoms

#### Step 2: Balance nitrogen (N)
- To balance N, we need 2 N atoms on the right side. So, we place a coefficient of 2 in front of \( \text{NH}_3 \):
\[
\text{N}_2 + \_ \text{H}_2 \rightarrow 2 \text{NH}_3
\]

#### Step 3: Balance hydrogen (H)
- Now, the right side has \( 2 \times 3 = 6 \) H atoms.
- To balance H, we need 6 H atoms on the left side. So, we place a coefficient of 3 in front of \( \text{H}_2 \):
\[
\text{N}_2 + 3 \text{H}_2 \rightarrow 2 \text{NH}_3
\]

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

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2. \( \_ \text{H}_2\text{O} \rightarrow \_ \text{H}_2 + \_ \text{O}_2 \)

- Reactants: \( \text{H}_2\text{O} \)
- Products: \( \text{H}_2 \) and \( \text{O}_2 \)

#### Step 1: Count the atoms
- Left side: 2 H atoms, 1 O atom
- Right side: 2 H atoms, 2 O atoms

#### Step 2: Balance oxygen (O)
- To balance O, we need 2 O atoms on the left side. So, we place a coefficient of 2 in front of \( \text{H}_2\text{O} \):
\[
2 \text{H}_2\text{O} \rightarrow \_ \text{H}_2 + \text{O}_2
\]

#### Step 3: Balance hydrogen (H)
- Now, the left side has \( 2 \times 2 = 4 \) H atoms.
- To balance H, we need 4 H atoms on the right side. So, we place a coefficient of 2 in front of \( \text{H}_2 \):
\[
2 \text{H}_2\text{O} \rightarrow 2 \text{H}_2 + \text{O}_2
\]

#### Final Equation:
\[
\boxed{2 \text{H}_2\text{O} \rightarrow 2 \text{H}_2 + 1 \text{O}_2}
\]

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

- Reactants: \( \text{CH}_4 \) and \( \text{O}_2 \)
- Products: \( \text{CO}_2 \) and \( \text{H}_2\text{O} \)

#### Step 1: Count the atoms
- Left side: 1 C atom, 4 H atoms, 2 O atoms
- Right side: 1 C atom, 2 H atoms, 3 O atoms

#### Step 2: Balance carbon (C)
- C is already balanced with 1 C atom on both sides.

#### Step 3: Balance hydrogen (H)
- To balance H, we need 4 H atoms on the right side. So, we place a coefficient of 2 in front of \( \text{H}_2\text{O} \):
\[
\text{CH}_4 + \_ \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O}
\]

#### Step 4: Balance oxygen (O)
- Now, the right side has \( 2 + 2 \times 1 = 4 \) O atoms.
- To balance O, we need 4 O atoms on the left side. So, we place a coefficient of 2 in front of \( \text{O}_2 \):
\[
\text{CH}_4 + 2 \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O}
\]

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

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4. \( \_ \text{CO}_2 \rightarrow \_ \text{CO} + \_ \text{O}_2 \)

- Reactants: \( \text{CO}_2 \)
- Products: \( \text{CO} \) and \( \text{O}_2 \)

#### Step 1: Count the atoms
- Left side: 1 C atom, 2 O atoms
- Right side: 1 C atom, 3 O atoms

#### Step 2: Balance carbon (C)
- C is already balanced with 1 C atom on both sides.

#### Step 3: Balance oxygen (O)
- To balance O, we need 2 O atoms on the right side. So, we place a coefficient of 2 in front of \( \text{CO} \) and adjust \( \text{O}_2 \):
\[
\text{CO}_2 \rightarrow 2 \text{CO} + \_ \text{O}_2
\]
- Now, the right side has \( 2 \times 1 + 2 = 4 \) O atoms.
- To balance O, we need 2 O atoms from \( \text{O}_2 \). So, we place a coefficient of 1 in front of \( \text{O}_2 \):
\[
\text{CO}_2 \rightarrow 2 \text{CO} + \frac{1}{2} \text{O}_2
\]
- To avoid fractions, multiply the entire equation by 2:
\[
2 \text{CO}_2 \rightarrow 4 \text{CO} + \text{O}_2
\]

#### Final Equation:
\[
\boxed{2 \text{CO}_2 \rightarrow 4 \text{CO} + 1 \text{O}_2}
\]

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5. \( \_ \text{CH}_2\text{O} + \_ \text{H}_2 \rightarrow \_ \text{CH}_3\text{OH} \)

- Reactants: \( \text{CH}_2\text{O} \) and \( \text{H}_2 \)
- Products: \( \text{CH}_3\text{OH} \)

#### Step 1: Count the atoms
- Left side: 1 C atom, 4 H atoms, 1 O atom
- Right side: 1 C atom, 4 H atoms, 1 O atom

#### Step 2: Balance carbon (C)
- C is already balanced with 1 C atom on both sides.

#### Step 3: Balance hydrogen (H)
- H is already balanced with 4 H atoms on both sides.

#### Step 4: Balance oxygen (O)
- O is already balanced with 1 O atom on both sides.

#### Final Equation:
\[
\boxed{1 \text{CH}_2\text{O} + 1 \text{H}_2 \rightarrow 1 \text{CH}_3\text{OH}}
\]

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6. \( \_ \text{P}_4 + \_ \text{H}_2 \rightarrow \_ \text{PH}_3 \)

- Reactants: \( \text{P}_4 \) and \( \text{H}_2 \)
- Products: \( \text{PH}_3 \)

#### Step 1: Count the atoms
- Left side: 4 P atoms, 2 H atoms
- Right side: 1 P atom, 3 H atoms

#### Step 2: Balance phosphorus (P)
- To balance P, we need 4 P atoms on the right side. So, we place a coefficient of 4 in front of \( \text{PH}_3 \):
\[
\text{P}_4 + \_ \text{H}_2 \rightarrow 4 \text{PH}_3
\]

#### Step 3: Balance hydrogen (H)
- Now, the right side has \( 4 \times 3 = 12 \) H atoms.
- To balance H, we need 12 H atoms on the left side. So, we place a coefficient of 6 in front of \( \text{H}_2 \):
\[
\text{P}_4 + 6 \text{H}_2 \rightarrow 4 \text{PH}_3
\]

#### Final Equation:
\[
\boxed{1 \text{P}_4 + 6 \text{H}_2 \rightarrow 4 \text{PH}_3}
\]

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7. \( \_ \text{C} + \_ \text{O}_2 \rightarrow \_ \text{CO}_2 \)

- Reactants: \( \text{C} \) and \( \text{O}_2 \)
- Products: \( \text{CO}_2 \)

#### Step 1: Count the atoms
- Left side: 1 C atom, 2 O atoms
- Right side: 1 C atom, 2 O atoms

#### Step 2: Balance carbon (C)
- C is already balanced with 1 C atom on both sides.

#### Step 3: Balance oxygen (O)
- O is already balanced with 2 O atoms on both sides.

#### Final Equation:
\[
\boxed{1 \text{C} + 1 \text{O}_2 \rightarrow 1 \text{CO}_2}
\]

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8. \( \_ \text{HCl} \rightarrow \_ \text{H}_2 + \_ \text{Cl}_2 \)

- Reactants: \( \text{HCl} \)
- Products: \( \text{H}_2 \) and \( \text{Cl}_2 \)

#### Step 1: Count the atoms
- Left side: 1 H atom, 1 Cl atom
- Right side: 2 H atoms, 2 Cl atoms

#### Step 2: Balance hydrogen (H)
- To balance H, we need 2 H atoms on the left side. So, we place a coefficient of 2 in front of \( \text{HCl} \):
\[
2 \text{HCl} \rightarrow \_ \text{H}_2 + \_ \text{Cl}_2
\]

#### Step 3: Balance chlorine (Cl)
- Now, the left side has 2 Cl atoms.
- To balance Cl, we need 2 Cl atoms on the right side. So, we place a coefficient of 1 in front of \( \text{Cl}_2 \):
\[
2 \text{HCl} \rightarrow \text{H}_2 + \text{Cl}_2
\]

#### Final Equation:
\[
\boxed{2 \text{HCl} \rightarrow 1 \text{H}_2 + 1 \text{Cl}_2}
\]

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Final Answer:

\[
\boxed{
\begin{aligned}
1. & \quad 1 \text{N}_2 + 3 \text{H}_2 \rightarrow 2 \text{NH}_3 \\
2. & \quad 2 \text{H}_2\text{O} \rightarrow 2 \text{H}_2 + 1 \text{O}_2 \\
3. & \quad 1 \text{CH}_4 + 2 \text{O}_2 \rightarrow 1 \text{CO}_2 + 2 \text{H}_2\text{O} \\
4. & \quad 2 \text{CO}_2 \rightarrow 4 \text{CO} + 1 \text{O}_2 \\
5. & \quad 1 \text{CH}_2\text{O} + 1 \text{H}_2 \rightarrow 1 \text{CH}_3\text{OH} \\
6. & \quad 1 \text{P}_4 + 6 \text{H}_2 \rightarrow 4 \text{PH}_3 \\
7. & \quad 1 \text{C} + 1 \text{O}_2 \rightarrow 1 \text{CO}_2 \\
8. & \quad 2 \text{HCl} \rightarrow 1 \text{H}_2 + 1 \text{Cl}_2 \\
\end{aligned}
}
\]