Balancing Chemical Equation activity - Free Printable
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Step-by-step solution for: Balancing Chemical Equation activity
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Show Answer Key & Explanations
Step-by-step solution for: Balancing Chemical Equation activity
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 step by step:
---
- Step 1: Start with nitrogen (N). There are 2 N atoms in \( \text{N}_2 \) and 1 N atom in \( \text{NH}_3 \). To balance N, place a coefficient of 2 in front of \( \text{NH}_3 \):
\[
\text{N}_2 + \_\_ \text{H}_2 \rightarrow 2 \text{NH}_3
\]
- Step 2: Now balance hydrogen (H). There are 6 H atoms in \( 2 \text{NH}_3 \) (since \( 2 \times 3 = 6 \)). To balance H, place a coefficient of 3 in front of \( \text{H}_2 \):
\[
\text{N}_2 + 3 \text{H}_2 \rightarrow 2 \text{NH}_3
\]
Balanced Equation:
\[
\boxed{1 \text{N}_2 + 3 \text{H}_2 \rightarrow 2 \text{NH}_3}
\]
---
- Step 1: Start with oxygen (O). There are 2 O atoms in \( \text{H}_2\text{O} \) and 2 O atoms in \( \text{O}_2 \). To balance O, place a coefficient of 2 in front of \( \text{H}_2\text{O} \):
\[
2 \text{H}_2\text{O} \rightarrow \_\_ \text{H}_2 + 1 \text{O}_2
\]
- Step 2: Now balance hydrogen (H). There are 4 H atoms in \( 2 \text{H}_2\text{O} \) (since \( 2 \times 2 = 4 \)). To balance H, place a coefficient of 2 in front of \( \text{H}_2 \):
\[
2 \text{H}_2\text{O} \rightarrow 2 \text{H}_2 + 1 \text{O}_2
\]
Balanced Equation:
\[
\boxed{2 \text{H}_2\text{O} \rightarrow 2 \text{H}_2 + 1 \text{O}_2}
\]
---
- Step 1: Start with carbon (C). There is 1 C atom in \( \text{CH}_4 \) and 1 C atom in \( \text{CO}_2 \). Place a coefficient of 1 in front of \( \text{CO}_2 \):
\[
\text{CH}_4 + \_\_ \text{O}_2 \rightarrow 1 \text{CO}_2 + \_\_ \text{H}_2\text{O}
\]
- Step 2: 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, place a coefficient of 2 in front of \( \text{H}_2\text{O} \):
\[
\text{CH}_4 + \_\_ \text{O}_2 \rightarrow 1 \text{CO}_2 + 2 \text{H}_2\text{O}
\]
- Step 3: Now balance oxygen (O). There are 2 O atoms in \( \text{CO}_2 \) and 2 O atoms in \( 2 \text{H}_2\text{O} \) (since \( 2 \times 1 = 2 \)), totaling 4 O atoms on the right. To balance O, place a coefficient of 2 in front of \( \text{O}_2 \):
\[
\text{CH}_4 + 2 \text{O}_2 \rightarrow 1 \text{CO}_2 + 2 \text{H}_2\text{O}
\]
Balanced Equation:
\[
\boxed{1 \text{CH}_4 + 2 \text{O}_2 \rightarrow 1 \text{CO}_2 + 2 \text{H}_2\text{O}}
\]
---
- Step 1: Start with carbon (C). There is 1 C atom in \( \text{CO}_2 \) and 1 C atom in \( \text{CO} \). Place a coefficient of 2 in front of \( \text{CO} \):
\[
\text{CO}_2 \rightarrow 2 \text{CO} + \_\_ \text{O}_2
\]
- Step 2: Now balance oxygen (O). There are 2 O atoms in \( \text{CO}_2 \) and 2 O atoms in \( 2 \text{CO} \) (since \( 2 \times 1 = 2 \)). To balance O, place a coefficient of 0.5 in front of \( \text{O}_2 \). However, coefficients must be whole numbers, so multiply the entire equation by 2:
\[
2 \text{CO}_2 \rightarrow 4 \text{CO} + 1 \text{O}_2
\]
Balanced Equation:
\[
\boxed{2 \text{CO}_2 \rightarrow 4 \text{CO} + 1 \text{O}_2}
\]
---
- Step 1: Start with carbon (C). There is 1 C atom in \( \text{CH}_2\text{O} \) and 1 C atom in \( \text{CH}_3\text{OH} \). Place a coefficient of 1 in front of \( \text{CH}_3\text{OH} \):
\[
\text{CH}_2\text{O} + \_\_ \text{H}_2 \rightarrow 1 \text{CH}_3\text{OH}
\]
- Step 2: Balance hydrogen (H). There are 2 H atoms in \( \text{CH}_2\text{O} \) and 4 H atoms in \( \text{CH}_3\text{OH} \). To balance H, place a coefficient of 1 in front of \( \text{H}_2 \):
\[
\text{CH}_2\text{O} + 1 \text{H}_2 \rightarrow 1 \text{CH}_3\text{OH}
\]
- Step 3: Balance oxygen (O). There is 1 O atom in \( \text{CH}_2\text{O} \) and 1 O atom in \( \text{CH}_3\text{OH} \). The equation is already balanced.
Balanced Equation:
\[
\boxed{1 \text{CH}_2\text{O} + 1 \text{H}_2 \rightarrow 1 \text{CH}_3\text{OH}}
\]
---
- Step 1: Start with phosphorus (P). There are 4 P atoms in \( \text{P}_4 \) and 1 P atom in \( \text{PH}_3 \). To balance P, place a coefficient of 4 in front of \( \text{PH}_3 \):
\[
\text{P}_4 + \_\_ \text{H}_2 \rightarrow 4 \text{PH}_3
\]
- Step 2: Now balance hydrogen (H). There are 12 H atoms in \( 4 \text{PH}_3 \) (since \( 4 \times 3 = 12 \)). To balance H, place a coefficient of 6 in front of \( \text{H}_2 \):
\[
\text{P}_4 + 6 \text{H}_2 \rightarrow 4 \text{PH}_3
\]
Balanced Equation:
\[
\boxed{1 \text{P}_4 + 6 \text{H}_2 \rightarrow 4 \text{PH}_3}
\]
---
- Step 1: Start with carbon (C). There is 1 C atom in \( \text{C} \) and 1 C atom in \( \text{CO}_2 \). Place a coefficient of 1 in front of \( \text{CO}_2 \):
\[
\text{C} + \_\_ \text{O}_2 \rightarrow 1 \text{CO}_2
\]
- Step 2: Now balance oxygen (O). There are 2 O atoms in \( \text{O}_2 \) and 2 O atoms in \( \text{CO}_2 \). The equation is already balanced.
Balanced Equation:
\[
\boxed{1 \text{C} + 1 \text{O}_2 \rightarrow 1 \text{CO}_2}
\]
---
- Step 1: Start with hydrogen (H). There is 1 H atom in \( \text{HCl} \) and 2 H atoms in \( \text{H}_2 \). To balance H, place a coefficient of 2 in front of \( \text{HCl} \):
\[
2 \text{HCl} \rightarrow 1 \text{H}_2 + \_\_ \text{Cl}_2
\]
- Step 2: Now balance chlorine (Cl). There are 2 Cl atoms in \( 2 \text{HCl} \) and 2 Cl atoms in \( \text{Cl}_2 \). The equation is already balanced.
Balanced Equation:
\[
\boxed{2 \text{HCl} \rightarrow 1 \text{H}_2 + 1 \text{Cl}_2}
\]
---
\[
\boxed{
\begin{aligned}
1. & \quad 1 \text{N}_2 + 3 \text{H}_2 \rightarrow 2 \text{NH}_3 \\
2. & \quad 2 \text{H}_2\text{O} \rightarrow 2 \text{H}_2 + 1 \text{O}_2 \\
3. & \quad 1 \text{CH}_4 + 2 \text{O}_2 \rightarrow 1 \text{CO}_2 + 2 \text{H}_2\text{O} \\
4. & \quad 2 \text{CO}_2 \rightarrow 4 \text{CO} + 1 \text{O}_2 \\
5. & \quad 1 \text{CH}_2\text{O} + 1 \text{H}_2 \rightarrow 1 \text{CH}_3\text{OH} \\
6. & \quad 1 \text{P}_4 + 6 \text{H}_2 \rightarrow 4 \text{PH}_3 \\
7. & \quad 1 \text{C} + 1 \text{O}_2 \rightarrow 1 \text{CO}_2 \\
8. & \quad 2 \text{HCl} \rightarrow 1 \text{H}_2 + 1 \text{Cl}_2 \\
\end{aligned}
}
\]
---
1. \( \_\_ \text{N}_2 + \_\_ \text{H}_2 \rightarrow \_\_ \text{NH}_3 \)
- Step 1: Start with nitrogen (N). There are 2 N atoms in \( \text{N}_2 \) and 1 N atom in \( \text{NH}_3 \). To balance N, place a coefficient of 2 in front of \( \text{NH}_3 \):
\[
\text{N}_2 + \_\_ \text{H}_2 \rightarrow 2 \text{NH}_3
\]
- Step 2: Now balance hydrogen (H). There are 6 H atoms in \( 2 \text{NH}_3 \) (since \( 2 \times 3 = 6 \)). To balance H, place a coefficient of 3 in front of \( \text{H}_2 \):
\[
\text{N}_2 + 3 \text{H}_2 \rightarrow 2 \text{NH}_3
\]
Balanced Equation:
\[
\boxed{1 \text{N}_2 + 3 \text{H}_2 \rightarrow 2 \text{NH}_3}
\]
---
2. \( \_\_ \text{H}_2\text{O} \rightarrow \_\_ \text{H}_2 + \_\_ \text{O}_2 \)
- Step 1: Start with oxygen (O). There are 2 O atoms in \( \text{H}_2\text{O} \) and 2 O atoms in \( \text{O}_2 \). To balance O, place a coefficient of 2 in front of \( \text{H}_2\text{O} \):
\[
2 \text{H}_2\text{O} \rightarrow \_\_ \text{H}_2 + 1 \text{O}_2
\]
- Step 2: Now balance hydrogen (H). There are 4 H atoms in \( 2 \text{H}_2\text{O} \) (since \( 2 \times 2 = 4 \)). To balance H, place a coefficient of 2 in front of \( \text{H}_2 \):
\[
2 \text{H}_2\text{O} \rightarrow 2 \text{H}_2 + 1 \text{O}_2
\]
Balanced Equation:
\[
\boxed{2 \text{H}_2\text{O} \rightarrow 2 \text{H}_2 + 1 \text{O}_2}
\]
---
3. \( \_\_ \text{CH}_4 + \_\_ \text{O}_2 \rightarrow \_\_ \text{CO}_2 + \_\_ \text{H}_2\text{O} \)
- Step 1: Start with carbon (C). There is 1 C atom in \( \text{CH}_4 \) and 1 C atom in \( \text{CO}_2 \). Place a coefficient of 1 in front of \( \text{CO}_2 \):
\[
\text{CH}_4 + \_\_ \text{O}_2 \rightarrow 1 \text{CO}_2 + \_\_ \text{H}_2\text{O}
\]
- Step 2: 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, place a coefficient of 2 in front of \( \text{H}_2\text{O} \):
\[
\text{CH}_4 + \_\_ \text{O}_2 \rightarrow 1 \text{CO}_2 + 2 \text{H}_2\text{O}
\]
- Step 3: Now balance oxygen (O). There are 2 O atoms in \( \text{CO}_2 \) and 2 O atoms in \( 2 \text{H}_2\text{O} \) (since \( 2 \times 1 = 2 \)), totaling 4 O atoms on the right. To balance O, place a coefficient of 2 in front of \( \text{O}_2 \):
\[
\text{CH}_4 + 2 \text{O}_2 \rightarrow 1 \text{CO}_2 + 2 \text{H}_2\text{O}
\]
Balanced Equation:
\[
\boxed{1 \text{CH}_4 + 2 \text{O}_2 \rightarrow 1 \text{CO}_2 + 2 \text{H}_2\text{O}}
\]
---
4. \( \_\_ \text{CO}_2 \rightarrow \_\_ \text{CO} + \_\_ \text{O}_2 \)
- Step 1: Start with carbon (C). There is 1 C atom in \( \text{CO}_2 \) and 1 C atom in \( \text{CO} \). Place a coefficient of 2 in front of \( \text{CO} \):
\[
\text{CO}_2 \rightarrow 2 \text{CO} + \_\_ \text{O}_2
\]
- Step 2: Now balance oxygen (O). There are 2 O atoms in \( \text{CO}_2 \) and 2 O atoms in \( 2 \text{CO} \) (since \( 2 \times 1 = 2 \)). To balance O, place a coefficient of 0.5 in front of \( \text{O}_2 \). However, coefficients must be whole numbers, so multiply the entire equation by 2:
\[
2 \text{CO}_2 \rightarrow 4 \text{CO} + 1 \text{O}_2
\]
Balanced Equation:
\[
\boxed{2 \text{CO}_2 \rightarrow 4 \text{CO} + 1 \text{O}_2}
\]
---
5. \( \_\_ \text{CH}_2\text{O} + \_\_ \text{H}_2 \rightarrow \_\_ \text{CH}_3\text{OH} \)
- Step 1: Start with carbon (C). There is 1 C atom in \( \text{CH}_2\text{O} \) and 1 C atom in \( \text{CH}_3\text{OH} \). Place a coefficient of 1 in front of \( \text{CH}_3\text{OH} \):
\[
\text{CH}_2\text{O} + \_\_ \text{H}_2 \rightarrow 1 \text{CH}_3\text{OH}
\]
- Step 2: Balance hydrogen (H). There are 2 H atoms in \( \text{CH}_2\text{O} \) and 4 H atoms in \( \text{CH}_3\text{OH} \). To balance H, place a coefficient of 1 in front of \( \text{H}_2 \):
\[
\text{CH}_2\text{O} + 1 \text{H}_2 \rightarrow 1 \text{CH}_3\text{OH}
\]
- Step 3: Balance oxygen (O). There is 1 O atom in \( \text{CH}_2\text{O} \) and 1 O atom in \( \text{CH}_3\text{OH} \). The equation is already balanced.
Balanced Equation:
\[
\boxed{1 \text{CH}_2\text{O} + 1 \text{H}_2 \rightarrow 1 \text{CH}_3\text{OH}}
\]
---
6. \( \_\_ \text{P}_4 + \_\_ \text{H}_2 \rightarrow \_\_ \text{PH}_3 \)
- Step 1: Start with phosphorus (P). There are 4 P atoms in \( \text{P}_4 \) and 1 P atom in \( \text{PH}_3 \). To balance P, place a coefficient of 4 in front of \( \text{PH}_3 \):
\[
\text{P}_4 + \_\_ \text{H}_2 \rightarrow 4 \text{PH}_3
\]
- Step 2: Now balance hydrogen (H). There are 12 H atoms in \( 4 \text{PH}_3 \) (since \( 4 \times 3 = 12 \)). To balance H, place a coefficient of 6 in front of \( \text{H}_2 \):
\[
\text{P}_4 + 6 \text{H}_2 \rightarrow 4 \text{PH}_3
\]
Balanced Equation:
\[
\boxed{1 \text{P}_4 + 6 \text{H}_2 \rightarrow 4 \text{PH}_3}
\]
---
7. \( \_\_ \text{C} + \_\_ \text{O}_2 \rightarrow \_\_ \text{CO}_2 \)
- Step 1: Start with carbon (C). There is 1 C atom in \( \text{C} \) and 1 C atom in \( \text{CO}_2 \). Place a coefficient of 1 in front of \( \text{CO}_2 \):
\[
\text{C} + \_\_ \text{O}_2 \rightarrow 1 \text{CO}_2
\]
- Step 2: Now balance oxygen (O). There are 2 O atoms in \( \text{O}_2 \) and 2 O atoms in \( \text{CO}_2 \). The equation is already balanced.
Balanced Equation:
\[
\boxed{1 \text{C} + 1 \text{O}_2 \rightarrow 1 \text{CO}_2}
\]
---
8. \( \_\_ \text{HCl} \rightarrow \_\_ \text{H}_2 + \_\_ \text{Cl}_2 \)
- Step 1: Start with hydrogen (H). There is 1 H atom in \( \text{HCl} \) and 2 H atoms in \( \text{H}_2 \). To balance H, place a coefficient of 2 in front of \( \text{HCl} \):
\[
2 \text{HCl} \rightarrow 1 \text{H}_2 + \_\_ \text{Cl}_2
\]
- Step 2: Now balance chlorine (Cl). There are 2 Cl atoms in \( 2 \text{HCl} \) and 2 Cl atoms in \( \text{Cl}_2 \). The equation is already balanced.
Balanced Equation:
\[
\boxed{2 \text{HCl} \rightarrow 1 \text{H}_2 + 1 \text{Cl}_2}
\]
---
Final Answer:
\[
\boxed{
\begin{aligned}
1. & \quad 1 \text{N}_2 + 3 \text{H}_2 \rightarrow 2 \text{NH}_3 \\
2. & \quad 2 \text{H}_2\text{O} \rightarrow 2 \text{H}_2 + 1 \text{O}_2 \\
3. & \quad 1 \text{CH}_4 + 2 \text{O}_2 \rightarrow 1 \text{CO}_2 + 2 \text{H}_2\text{O} \\
4. & \quad 2 \text{CO}_2 \rightarrow 4 \text{CO} + 1 \text{O}_2 \\
5. & \quad 1 \text{CH}_2\text{O} + 1 \text{H}_2 \rightarrow 1 \text{CH}_3\text{OH} \\
6. & \quad 1 \text{P}_4 + 6 \text{H}_2 \rightarrow 4 \text{PH}_3 \\
7. & \quad 1 \text{C} + 1 \text{O}_2 \rightarrow 1 \text{CO}_2 \\
8. & \quad 2 \text{HCl} \rightarrow 1 \text{H}_2 + 1 \text{Cl}_2 \\
\end{aligned}
}
\]
Parent Tip: Review the logic above to help your child master the concept of balancing equations worksheet grade 10.
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