49 Balancing Chemical Equations Worksheets [with Answers] - Free Printable

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 worksheet:

Worksheet 1: Balancing Equations

#### a. \( \text{NO} + \text{O}_2 \rightarrow \text{NO}_2 \)
- Start with the unbalanced equation:
\[
\text{NO} + \text{O}_2 \rightarrow \text{NO}_2
\]
- On the left side, there is 1 N and 3 O atoms (1 from NO and 2 from O₂).
- On the right side, there is 1 N and 2 O atoms.
- To balance oxygen, add a coefficient of 2 to NO:
\[
2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2
\]
- Now, there are 2 N and 4 O atoms on both sides.

Balanced Equation:
\[
2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2
\]

#### b. \( \text{Co} + \text{O}_2 \rightarrow \text{Co}_2\text{O}_3 \)
- Start with the unbalanced equation:
\[
\text{Co} + \text{O}_2 \rightarrow \text{Co}_2\text{O}_3
\]
- On the left side, there is 1 Co and 2 O atoms.
- On the right side, there are 2 Co and 3 O atoms.
- To balance cobalt, add a coefficient of 2 to Co:
\[
2\text{Co} + \text{O}_2 \rightarrow \text{Co}_2\text{O}_3
\]
- Now, there are 2 Co and 5 O atoms on the right side. Add a coefficient of 1.5 to O₂ (or multiply through by 2 to avoid fractions):
\[
4\text{Co} + 3\text{O}_2 \rightarrow 2\text{Co}_2\text{O}_3
\]

Balanced Equation:
\[
4\text{Co} + 3\text{O}_2 \rightarrow 2\text{Co}_2\text{O}_3
\]

#### c. \( \text{Al} + \text{Cl}_2 \rightarrow \text{Al}_2\text{Cl}_6 \)
- Start with the unbalanced equation:
\[
\text{Al} + \text{Cl}_2 \rightarrow \text{Al}_2\text{Cl}_6
\]
- On the left side, there is 1 Al and 2 Cl atoms.
- On the right side, there are 2 Al and 6 Cl atoms.
- To balance aluminum, add a coefficient of 2 to Al:
\[
2\text{Al} + \text{Cl}_2 \rightarrow \text{Al}_2\text{Cl}_6
\]
- Now, there are 2 Al and 8 Cl atoms on the right side. Add a coefficient of 3 to Cl₂:
\[
2\text{Al} + 3\text{Cl}_2 \rightarrow \text{Al}_2\text{Cl}_6
\]

Balanced Equation:
\[
2\text{Al} + 3\text{Cl}_2 \rightarrow \text{Al}_2\text{Cl}_6
\]

#### d. \( \text{C}_2\text{H}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \)
- Start with the unbalanced equation:
\[
\text{C}_2\text{H}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
\]
- On the left side, there are 2 C, 6 H, and 2 O atoms.
- On the right side, there are 1 C, 2 H, and 3 O atoms.
- Balance carbon first by adding a coefficient of 2 to CO₂:
\[
\text{C}_2\text{H}_6 + \text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O}
\]
- Now, there are 2 C, 6 H, and 5 O atoms on the right side. Balance hydrogen by adding a coefficient of 3 to H₂O:
\[
\text{C}_2\text{H}_6 + \text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}
\]
- Now, there are 2 C, 6 H, and 7 O atoms on the right side. Balance oxygen by adding a coefficient of 7/2 to O₂ (or multiply through by 2 to avoid fractions):
\[
2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O}
\]

Balanced Equation:
\[
2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O}
\]

#### e. \( \text{TiCl}_4 + \text{Na} \rightarrow \text{Ti} + \text{NaCl} \)
- Start with the unbalanced equation:
\[
\text{TiCl}_4 + \text{Na} \rightarrow \text{Ti} + \text{NaCl}
\]
- On the left side, there is 1 Ti, 4 Cl, and 1 Na.
- On the right side, there is 1 Ti, 1 Cl, and 1 Na.
- Balance chlorine by adding a coefficient of 4 to NaCl:
\[
\text{TiCl}_4 + \text{Na} \rightarrow \text{Ti} + 4\text{NaCl}
\]
- Now, there are 4 Cl atoms on the right side. Balance sodium by adding a coefficient of 4 to Na:
\[
\text{TiCl}_4 + 4\text{Na} \rightarrow \text{Ti} + 4\text{NaCl}
\]

Balanced Equation:
\[
\text{TiCl}_4 + 4\text{Na} \rightarrow \text{Ti} + 4\text{NaCl}
\]

Worksheet 2: Balancing Equations

#### a. \( \text{Sr} + \text{O}_2 \rightarrow \text{SrO} \)
- Start with the unbalanced equation:
\[
\text{Sr} + \text{O}_2 \rightarrow \text{SrO}
\]
- On the left side, there is 1 Sr and 2 O atoms.
- On the right side, there is 1 Sr and 1 O atom.
- Balance oxygen by adding a coefficient of 2 to SrO:
\[
\text{Sr} + \text{O}_2 \rightarrow 2\text{SrO}
\]

Balanced Equation:
\[
2\text{Sr} + \text{O}_2 \rightarrow 2\text{SrO}
\]

#### b. \( \text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{O} + \text{O}_2 \)
- Start with the unbalanced equation:
\[
\text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{O} + \text{O}_2
\]
- On the left side, there are 2 H and 2 O atoms.
- On the right side, there are 2 H and 3 O atoms.
- Balance oxygen by adding a coefficient of 2 to H₂O₂ and 2 to H₂O:
\[
2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2
\]

Balanced Equation:
\[
2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2
\]

#### c. \( \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{H}_2\text{O} + \text{CO}_2 \)
- Start with the unbalanced equation:
\[
\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{H}_2\text{O} + \text{CO}_2
\]
- On the left side, there are 3 C, 8 H, and 2 O atoms.
- On the right side, there are 1 C, 2 H, and 3 O atoms.
- Balance carbon by adding a coefficient of 3 to CO₂:
\[
\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{H}_2\text{O} + 3\text{CO}_2
\]
- Now, there are 3 C, 2 H, and 9 O atoms on the right side. Balance hydrogen by adding a coefficient of 4 to H₂O:
\[
\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 4\text{H}_2\text{O} + 3\text{CO}_2
\]
- Now, there are 3 C, 8 H, and 10 O atoms on the right side. Balance oxygen by adding a coefficient of 5 to O₂:
\[
\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 4\text{H}_2\text{O} + 3\text{CO}_2
\]

Balanced Equation:
\[
\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 4\text{H}_2\text{O} + 3\text{CO}_2
\]

#### d. \( \text{C}_2\text{H}_5\text{OH} + \text{O}_2 \rightarrow \text{H}_2\text{O} + \text{CO}_2 \)
- Start with the unbalanced equation:
\[
\text{C}_2\text{H}_5\text{OH} + \text{O}_2 \rightarrow \text{H}_2\text{O} + \text{CO}_2
\]
- On the left side, there are 2 C, 6 H, 1 O, and 2 O atoms.
- On the right side, there are 1 C, 2 H, 2 O, and 2 O atoms.
- Balance carbon by adding a coefficient of 2 to CO₂:
\[
\text{C}_2\text{H}_5\text{OH} + \text{O}_2 \rightarrow \text{H}_2\text{O} + 2\text{CO}_2
\]
- Now, there are 2 C, 2 H, 5 O atoms on the right side. Balance hydrogen by adding a coefficient of 3 to H₂O:
\[
\text{C}_2\text{H}_5\text{OH} + \text{O}_2 \rightarrow 3\text{H}_2\text{O} + 2\text{CO}_2
\]
- Now, there are 2 C, 6 H, 7 O atoms on the right side. Balance oxygen by adding a coefficient of 3 to O₂:
\[
\text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 3\text{H}_2\text{O} + 2\text{CO}_2
\]

Balanced Equation:
\[
\text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 3\text{H}_2\text{O} + 2\text{CO}_2
\]

#### e. \( \text{NH}_3 + \text{Cl}_2 \rightarrow \text{NH}_4\text{Cl} + \text{NCl}_3 \)
- Start with the unbalanced equation:
\[
\text{NH}_3 + \text{Cl}_2 \rightarrow \text{NH}_4\text{Cl} + \text{NCl}_3
\]
- On the left side, there are 1 N, 3 H, and 2 Cl atoms.
- On the right side, there are 2 N, 4 H, and 4 Cl atoms.
- Balance nitrogen by adding a coefficient of 2 to NH₃:
\[
2\text{NH}_3 + \text{Cl}_2 \rightarrow \text{NH}_4\text{Cl} + \text{NCl}_3
\]
- Now, there are 2 N, 6 H, and 2 Cl atoms on the left side. Balance hydrogen by adding a coefficient of 2 to NH₄Cl:
\[
2\text{NH}_3 + \text{Cl}_2 \rightarrow 2\text{NH}_4\text{Cl} + \text{NCl}_3
\]
- Now, there are 2 N, 10 H, and 5 Cl atoms on the right side. Balance chlorine by adding a coefficient of 3 to Cl₂:
\[
2\text{NH}_3 + 3\text{Cl}_2 \rightarrow 2\text{NH}_4\text{Cl} + \text{NCl}_3
\]

Balanced Equation:
\[
2\text{NH}_3 + 3\text{Cl}_2 \rightarrow 2\text{NH}_4\text{Cl} + \text{NCl}_3
\]

Worksheet 3: Balancing Equations

#### a. \( \text{Cu} + \text{FeCl}_3 \rightarrow \text{CuCl}_2 + \text{Fe} \)
- Start with the unbalanced equation:
\[
\text{Cu} + \text{FeCl}_3 \rightarrow \text{CuCl}_2 + \text{Fe}
\]
- On the left side, there is 1 Cu, 1 Fe, and 3 Cl atoms.
- On the right side, there is 1 Cu, 1 Fe, and 2 Cl atoms.
- Balance chlorine by adding a coefficient of 3 to CuCl₂:
\[
\text{Cu} + \text{FeCl}_3 \rightarrow 3\text{CuCl}_2 + \text{Fe}
\]
- Now, there are 3 Cu, 1 Fe, and 6 Cl atoms on the right side. Balance copper by adding a coefficient of 3 to Cu:
\[
3\text{Cu} + \text{FeCl}_3 \rightarrow 3\text{CuCl}_2 + \text{Fe}
\]

Balanced Equation:
\[
3\text{Cu} + 2\text{FeCl}_3 \rightarrow 3\text{CuCl}_2 + 2\text{Fe}
\]

#### b. \( \text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 \)
- Start with the unbalanced equation:
\[
\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3
\]
- On the left side, there is 1 Fe and 2 O atoms.
- On the right side, there are 2 Fe and 3 O atoms.
- Balance iron by adding a coefficient of 2 to Fe:
\[
2\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3
\]
- Now, there are 2 Fe and 5 O atoms on the right side. Balance oxygen by adding a coefficient of 1.5 to O₂ (or multiply through by 2 to avoid fractions):
\[
4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3
\]

Balanced Equation:
\[
4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3
\]

#### c. \( \text{C}_2\text{H}_2 + \text{O}_2 \rightarrow \text{H}_2\text{O} + \text{CO}_2 \)
- Start with the unbalanced equation:
\[
\text{C}_2\text{H}_2 + \text{O}_2 \rightarrow \text{H}_2\text{O} + \text{CO}_2
\]
- On the left side, there are 2 C, 2 H, and 2 O atoms.
- On the right side, there are 1 C, 2 H, and 3 O atoms.
- Balance carbon by adding a coefficient of 2 to CO₂:
\[
\text{C}_2\text{H}_2 + \text{O}_2 \rightarrow \text{H}_2\text{O} + 2\text{CO}_2
\]
- Now, there are 2 C, 2 H, and 5 O atoms on the right side. Balance oxygen by adding a coefficient of 2.5 to O₂ (or multiply through by 2 to avoid fractions):
\[
2\text{C}_2\text{H}_2 + 5\text{O}_2 \rightarrow 2\text{H}_2\text{O} + 4\text{CO}_2
\]

Balanced Equation:
\[
2\text{C}_2\text{H}_2 + 5\text{O}_2 \rightarrow 4\text{H}_2\text{O} + 4\text{CO}_2
\]

#### d. \( \text{KClO}_3 \rightarrow \text{KCl} + \text{O}_2 \)
- Start with the unbalanced equation:
\[
\text{KClO}_3 \rightarrow \text{KCl} + \text{O}_2
\]
- On the left side, there is 1 K, 1 Cl, and 3 O atoms.
- On the right side, there is 1 K, 1 Cl, and 2 O atoms.
- Balance oxygen by adding a coefficient of 3 to O₂:
\[
2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2
\]

Balanced Equation:
\[
2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2
\]

Final Answer:

\[
\boxed{
\begin{aligned}
&\text{Worksheet 1:} \\
&\text{a. } 2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2 \\
&\text{b. } 4\text{Co} + 3\text{O}_2 \rightarrow 2\text{Co}_2\text{O}_3 \\
&\text{c. } 2\text{Al} + 3\text{Cl}_2 \rightarrow \text{Al}_2\text{Cl}_6 \\
&\text{d. } 2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O} \\
&\text{e. } \text{TiCl}_4 + 4\text{Na} \rightarrow \text{Ti} + 4\text{NaCl} \\
\\
&\text{Worksheet 2:} \\
&\text{a. } 2\text{Sr} + \text{O}_2 \rightarrow 2\text{SrO} \\
&\text{b. } 2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2 \\
&\text{c. } \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 4\text{H}_2\text{O} + 3\text{CO}_2 \\
&\text{d. } \text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 3\text{H}_2\text{O} + 2\text{CO}_2 \\
&\text{e. } 2\text{NH}_3 + 3\text{Cl}_2 \rightarrow 2\text{NH}_4\text{Cl} + \text{NCl}_3 \\
\\
&\text{Worksheet 3:} \\
&\text{a. } 3\text{Cu} + 2\text{FeCl}_3 \rightarrow 3\text{CuCl}_2 + 2\text{Fe} \\
&\text{b. } 4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 \\
&\text{c. } 2\text{C}_2\text{H}_2 + 5\text{O}_2 \rightarrow 4\text{H}_2\text{O} + 4\text{CO}_2 \\
&\text{d. } 2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2 \\
\end{aligned}
}
\]