Balancing Chemical Equations Worksheet featuring 15 problems to practice balancing chemical reactions.
Balancing Chemical Equations Worksheet with 15 unbalanced chemical equations to be balanced.
PNG
375×667
24 KB
Free · Personal Use
Quality Assured by Worksheets Library Team
Reviewed for educational accuracy and age-appropriateness
ID: #303807
⭐
Show Answer Key & Explanations
Step-by-step solution for: Solved) - Balancing Chemical Equations - Lab 7 Balancing Equations ...
▼
Show Answer Key & Explanations
Step-by-step solution for: Solved) - Balancing Chemical Equations - Lab 7 Balancing Equations ...
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 for each problem, along with explanations:
---
- Start by balancing bromine (\(\text{Br}\)):
- There are 2 Br atoms on the left and 1 Br atom on the right.
- Multiply \(\text{LiBr}\) by 2: \( \text{Br}_2 + \text{LiF} \rightarrow 2\text{LiBr} + \text{F}_2 \).
- Now balance lithium (\(\text{Li}\)):
- There are 2 Li atoms on the right and 0 on the left.
- Multiply \(\text{LiF}\) by 2: \( \text{Br}_2 + 2\text{LiF} \rightarrow 2\text{LiBr} + \text{F}_2 \).
- Finally, check fluorine (\(\text{F}\)):
- There are 2 F atoms on the left and 2 F atoms on the right.
Balanced Equation:
\[ \boxed{\text{Br}_2 + 2\text{LiF} \rightarrow 2\text{LiBr} + \text{F}_2} \]
---
- Start by balancing iron (\(\text{Fe}\)):
- There is 1 Fe atom on the left and 3 Fe atoms on the right.
- Multiply \(\text{Fe(OH)}_3\) by 2: \( \text{H}_3\text{PO}_4 + 2\text{Fe(OH)}_3 \rightarrow \text{H}_2\text{O} + \text{Fe}_3(\text{PO}_4)_2 \).
- Now balance phosphorus (\(\text{P}\)):
- There is 1 P atom on the left and 2 P atoms on the right.
- Multiply \(\text{H}_3\text{PO}_4\) by 2: \( 2\text{H}_3\text{PO}_4 + 2\text{Fe(OH)}_3 \rightarrow \text{H}_2\text{O} + \text{Fe}_3(\text{PO}_4)_2 \).
- Balance oxygen (\(\text{O}\)):
- There are \(8\) O atoms on the left (\(4\) from \(\text{H}_3\text{PO}_4\) and \(6\) from \(\text{Fe(OH)}_3\)) and \(14\) O atoms on the right (\(8\) from \(\text{Fe}_3(\text{PO}_4)_2\) and \(2\) from \(\text{H}_2\text{O}\)).
- Multiply \(\text{H}_2\text{O}\) by 6: \( 2\text{H}_3\text{PO}_4 + 2\text{Fe(OH)}_3 \rightarrow 6\text{H}_2\text{O} + \text{Fe}_3(\text{PO}_4)_2 \).
- Check hydrogen (\(\text{H}\)):
- There are \(6\) H atoms from \(\text{H}_3\text{PO}_4\) and \(6\) H atoms from \(\text{Fe(OH)}_3\) (total \(12\)), and \(12\) H atoms on the right from \(6\text{H}_2\text{O}\).
Balanced Equation:
\[ \boxed{2\text{H}_3\text{PO}_4 + 2\text{Fe(OH)}_3 \rightarrow 6\text{H}_2\text{O} + \text{Fe}_3(\text{PO}_4)_2} \]
---
- Start by balancing carbon (\(\text{C}\)):
- There are 2 C atoms on the left and 1 C atom on the right.
- Multiply \(\text{CO}_2\) by 2: \( \text{C}_2\text{H}_5\text{OH} + \text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O} \).
- Now balance hydrogen (\(\text{H}\)):
- There are 6 H atoms on the left and 2 H atoms on the right.
- Multiply \(\text{H}_2\text{O}\) by 3: \( \text{C}_2\text{H}_5\text{OH} + \text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \).
- Finally, balance oxygen (\(\text{O}\)):
- There are 7 O atoms on the right (\(4\) from \(\text{CO}_2\) and \(3\) from \(\text{H}_2\text{O}\)) and 2 O atoms on the left.
- Multiply \(\text{O}_2\) by 3: \( \text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \).
Balanced Equation:
\[ \boxed{\text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}} \]
---
- This equation is not balanced as written because nickel oxide (\(\text{NiO}_3\)) does not exist. The correct product should be \(\text{NiO}\).
- Corrected equation: \( \text{Ni(OH)}_2 \rightarrow \text{NiO} + \text{H}_2\text{O} \).
- Balance nickel (\(\text{Ni}\)):
- There is 1 Ni atom on both sides.
- Balance oxygen (\(\text{O}\)):
- There are 2 O atoms on the left and 1 O atom on the right.
- Multiply \(\text{NiO}\) by 2: \( \text{Ni(OH)}_2 \rightarrow 2\text{NiO} + \text{H}_2\text{O} \).
- Balance hydrogen (\(\text{H}\)):
- There are 2 H atoms on the left and 2 H atoms on the right.
Balanced Equation:
\[ \boxed{\text{Ni(OH)}_2 \rightarrow \text{NiO} + \text{H}_2\text{O}} \]
---
- Balance potassium (\(\text{K}\)):
- There are 2 K atoms on the left and 1 K atom on the right.
- Multiply \(\text{KOH}\) by 2: \( \text{K}_2\text{SO}_3 + \text{Mn(OH)}_2 \rightarrow 2\text{KOH} + \text{MnSO}_3 \).
- Balance manganese (\(\text{Mn}\)):
- There is 1 Mn atom on both sides.
- Balance sulfur (\(\text{S}\)):
- There is 1 S atom on both sides.
- Balance oxygen (\(\text{O}\)):
- There are 5 O atoms on the left (\(3\) from \(\text{K}_2\text{SO}_3\) and \(2\) from \(\text{Mn(OH)}_2\)) and 5 O atoms on the right (\(3\) from \(\text{MnSO}_3\) and \(2\) from \(2\text{KOH}\)).
- Balance hydrogen (\(\text{H}\)):
- There are 2 H atoms on the left and 2 H atoms on the right.
Balanced Equation:
\[ \boxed{\text{K}_2\text{SO}_3 + \text{Mn(OH)}_2 \rightarrow 2\text{KOH} + \text{MnSO}_3} \]
---
- Balance sodium (\(\text{Na}\)):
- There is 1 Na atom on the left and 2 Na atoms on the right.
- Multiply \(\text{NaOH}\) by 2: \( 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{O} + \text{Na}_2\text{SO}_4 \).
- Balance hydrogen (\(\text{H}\)):
- There are 4 H atoms on the left and 2 H atoms on the right.
- Multiply \(\text{H}_2\text{O}\) by 2: \( 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow 2\text{H}_2\text{O} + \text{Na}_2\text{SO}_4 \).
- Balance sulfur (\(\text{S}\)):
- There is 1 S atom on both sides.
- Balance oxygen (\(\text{O}\)):
- There are 6 O atoms on the left (\(1\) from \(\text{NaOH}\) and \(4\) from \(\text{H}_2\text{SO}_4\)) and 6 O atoms on the right (\(2\) from \(2\text{H}_2\text{O}\) and \(4\) from \(\text{Na}_2\text{SO}_4\)).
Balanced Equation:
\[ \boxed{2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow 2\text{H}_2\text{O} + \text{Na}_2\text{SO}_4} \]
---
- Balance lead (\(\text{Pb}\)):
- There is 1 Pb atom on both sides.
- Balance lithium (\(\text{Li}\)):
- There is 1 Li atom on the left and 1 Li atom on the right.
- Balance oxygen (\(\text{O}\)):
- There are 2 O atoms on the left and 1 O atom on the right.
- Multiply \(\text{LiOH}\) by 2: \( \text{Li} + \text{Pb(OH)}_2 \rightarrow \text{Pb} + 2\text{LiOH} \).
- Balance hydrogen (\(\text{H}\)):
- There are 2 H atoms on the left and 2 H atoms on the right.
Balanced Equation:
\[ \boxed{\text{Li} + \text{Pb(OH)}_2 \rightarrow \text{Pb} + 2\text{LiOH}} \]
---
- Balance carbon (\(\text{C}\)):
- There are 2 C atoms on the left and 1 C atom on the right.
- Multiply \(\text{CO}_2\) by 2: \( \text{C}_2\text{H}_6 + \text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O} \).
- Balance hydrogen (\(\text{H}\)):
- There are 6 H atoms on the left and 2 H atoms on the right.
- Multiply \(\text{H}_2\text{O}\) by 3: \( \text{C}_2\text{H}_6 + \text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \).
- Balance oxygen (\(\text{O}\)):
- There are 7 O atoms on the right (\(4\) from \(\text{CO}_2\) and \(3\) from \(\text{H}_2\text{O}\)) and 2 O atoms on the left.
- Multiply \(\text{O}_2\) by \(\frac{7}{2}\). Since coefficients must be integers, multiply the entire equation by 2:
\( 2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O} \).
Balanced Equation:
\[ \boxed{2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O}} \]
---
- Balance gallium (\(\text{Ga}\)):
- There is 1 Ga atom on both sides.
- Balance potassium (\(\text{K}\)):
- There is 1 K atom on the right and 0 on the left.
- Multiply \(\text{KF}\) by 3: \( \text{Ga(OH)}_3 + 3\text{KF} \rightarrow \text{KOH} + \text{GaF}_3 \).
- Balance fluorine (\(\text{F}\)):
- There are 3 F atoms on the left and 3 F atoms on the right.
- Balance oxygen (\(\text{O}\)):
- There are 3 O atoms on the left and 1 O atom on the right.
- Multiply \(\text{KOH}\) by 3: \( \text{Ga(OH)}_3 + 3\text{KF} \rightarrow 3\text{KOH} + \text{GaF}_3 \).
Balanced Equation:
\[ \boxed{\text{Ga(OH)}_3 + 3\text{KF} \rightarrow 3\text{KOH} + \text{GaF}_3} \]
---
- This equation is not balanced as written because \(\text{VBrs}\) is not a valid compound. The correct product should be \(\text{VBr}_2\).
- Corrected equation: \( \text{V} + \text{ZnBr}_2 \rightarrow \text{VBr}_2 + \text{Zn} \).
- Balance vanadium (\(\text{V}\)):
- There is 1 V atom on both sides.
- Balance zinc (\(\text{Zn}\)):
- There is 1 Zn atom on both sides.
- Balance bromine (\(\text{Br}\)):
- There are 2 Br atoms on the left and 2 Br atoms on the right.
Balanced Equation:
\[ \boxed{\text{V} + \text{ZnBr}_2 \rightarrow \text{VBr}_2 + \text{Zn}} \]
---
- Balance arsenic (\(\text{As}\)):
- There are 2 As atoms on the left and 1 As atom on the right.
- Multiply \(\text{H}_3\text{AsO}_3\) by 2: \( \text{As}_2\text{O}_3 + \text{H}_2\text{O} \rightarrow 2\text{H}_3\text{AsO}_3 \).
- Balance oxygen (\(\text{O}\)):
- There are 3 O atoms on the left and 6 O atoms on the right.
- Multiply \(\text{H}_2\text{O}\) by 3: \( \text{As}_2\text{O}_3 + 3\text{H}_2\text{O} \rightarrow 2\text{H}_3\text{AsO}_3 \).
- Balance hydrogen (\(\text{H}\)):
- There are 6 H atoms on the left and 6 H atoms on the right.
Balanced Equation:
\[ \boxed{\text{As}_2\text{O}_3 + 3\text{H}_2\text{O} \rightarrow 2\text{H}_3\text{AsO}_3} \]
---
- Balance nitrogen (\(\text{N}\)):
- There is 1 N atom on both sides.
- Balance hydrogen (\(\text{H}\)):
- There are 3 H atoms on the left and 2 H atoms on the right.
- Multiply \(\text{H}_2\text{O}\) by \(\frac{3}{2}\). Since coefficients must be integers, multiply the entire equation by 2:
\( 2\text{NH}_3 + \text{O}_2 \rightarrow 2\text{NO} + 3\text{H}_2\text{O} \).
- Balance oxygen (\(\text{O}\)):
- There are 5 O atoms on the right (\(2\) from \(\text{NO}\) and \(3\) from \(\text{H}_2\text{O}\)) and 2 O atoms on the left.
- Multiply \(\text{O}_2\) by \(\frac{5}{2}\). Since coefficients must be integers, multiply the entire equation by 2:
\( 4\text{NH}_3 + 5\text{O}_2 \rightarrow 4\text{NO} + 6\text{H}_2\text{O} \).
Balanced Equation:
\[ \boxed{4\text{NH}_3 + 5\text{O}_2 \rightarrow 4\text{NO} + 6\text{H}_2\text{O}} \]
---
- Balance carbon (\(\text{C}\)):
- There are 2 C atoms on the left and 1 C atom on the right.
- Multiply \(\text{CO}_2\) by 2: \( \text{C}_2\text{H}_4 + \text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O} \).
- Balance hydrogen (\(\text{H}\)):
- There are 4 H atoms on the left and 2 H atoms on the right.
- Multiply \(\text{H}_2\text{O}\) by 2: \( \text{C}_2\text{H}_4 + \text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} \).
- Balance oxygen (\(\text{O}\)):
- There are 6 O atoms on the right (\(4\) from \(\text{CO}_2\) and \(2\) from \(\text{H}_2\text{O}\)) and 2 O atoms on the left.
- Multiply \(\text{O}_2\) by 3: \( \text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} \).
Balanced Equation:
\[ \boxed{\text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O}} \]
---
- Balance sodium (\(\text{Na}\)):
- There is 1 Na atom on both sides.
- Balance chlorine (\(\text{Cl}\)):
- There is 1 Cl atom on both sides.
- Balance oxygen (\(\text{O}\)):
- There are 3 O atoms on the left and 2 O atoms on the right.
- Multiply \(\text{O}_2\) by \(\frac{3}{2}\). Since coefficients must be integers, multiply the entire equation by 2:
\( 2\text{NaClO}_3 \rightarrow 2\text{NaCl} + 3\text{O}_2 \).
Balanced Equation:
\[ \boxed{2\text{NaClO}_3 \rightarrow 2\text{NaCl} + 3\text{O}_2} \]
---
- Balance calcium (\(\text{Ca}\)):
- There is 1 Ca atom on both sides.
- Balance oxygen (\(\text{O}\)):
- There are 2 O atoms on the left and 1 O atom on the right.
- Multiply \(\text{CaO}\) by 2: \( \text{Ca} + \text{O}_2 \rightarrow 2\text{CaO} \).
Balanced Equation:
\[ \boxed{2\text{Ca} + \text{O}_2 \rightarrow 2\text{CaO}} \]
---
\[ \boxed{
\begin{aligned}
1. & \quad \text{Br}_2 + 2\text{LiF} \rightarrow 2\text{LiBr} + \text{F}_2 \\
2. & \quad 2\text{H}_3\text{PO}_4 + 2\text{Fe(OH)}_3 \rightarrow 6\text{H}_2\text{O} + \text{Fe}_3(\text{PO}_4)_2 \\
3. & \quad \text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \\
4. & \quad \text{Ni(OH)}_2 \rightarrow \text{NiO} + \text{H}_2\text{O} \\
5. & \quad \text{K}_2\text{SO}_3 + \text{Mn(OH)}_2 \rightarrow 2\text{KOH} + \text{MnSO}_3 \\
6. & \quad 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow 2\text{H}_2\text{O} + \text{Na}_2\text{SO}_4 \\
7. & \quad \text{Li} + \text{Pb(OH)}_2 \rightarrow \text{Pb} + 2\text{LiOH} \\
8. & \quad 2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O} \\
9. & \quad \text{Ga(OH)}_3 + 3\text{KF} \rightarrow 3\text{KOH} + \text{GaF}_3 \\
10. & \quad \text{V} + \text{ZnBr}_2 \rightarrow \text{VBr}_2 + \text{Zn} \\
11. & \quad \text{As}_2\text{O}_3 + 3\text{H}_2\text{O} \rightarrow 2\text{H}_3\text{AsO}_3 \\
12. & \quad 4\text{NH}_3 + 5\text{O}_2 \rightarrow 4\text{NO} + 6\text{H}_2\text{O} \\
13. & \quad \text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} \\
14. & \quad 2\text{NaClO}_3 \rightarrow 2\text{NaCl} + 3\text{O}_2 \\
15. & \quad 2\text{Ca} + \text{O}_2 \rightarrow 2\text{CaO} \\
\end{aligned}
} \]
---
1. \( \text{Br}_2 + \text{LiF} \rightarrow \text{LiBr} + \text{F}_2 \)
- Start by balancing bromine (\(\text{Br}\)):
- There are 2 Br atoms on the left and 1 Br atom on the right.
- Multiply \(\text{LiBr}\) by 2: \( \text{Br}_2 + \text{LiF} \rightarrow 2\text{LiBr} + \text{F}_2 \).
- Now balance lithium (\(\text{Li}\)):
- There are 2 Li atoms on the right and 0 on the left.
- Multiply \(\text{LiF}\) by 2: \( \text{Br}_2 + 2\text{LiF} \rightarrow 2\text{LiBr} + \text{F}_2 \).
- Finally, check fluorine (\(\text{F}\)):
- There are 2 F atoms on the left and 2 F atoms on the right.
Balanced Equation:
\[ \boxed{\text{Br}_2 + 2\text{LiF} \rightarrow 2\text{LiBr} + \text{F}_2} \]
---
2. \( \text{H}_3\text{PO}_4 + \text{Fe(OH)}_3 \rightarrow \text{H}_2\text{O} + \text{Fe}_3(\text{PO}_4)_2 \)
- Start by balancing iron (\(\text{Fe}\)):
- There is 1 Fe atom on the left and 3 Fe atoms on the right.
- Multiply \(\text{Fe(OH)}_3\) by 2: \( \text{H}_3\text{PO}_4 + 2\text{Fe(OH)}_3 \rightarrow \text{H}_2\text{O} + \text{Fe}_3(\text{PO}_4)_2 \).
- Now balance phosphorus (\(\text{P}\)):
- There is 1 P atom on the left and 2 P atoms on the right.
- Multiply \(\text{H}_3\text{PO}_4\) by 2: \( 2\text{H}_3\text{PO}_4 + 2\text{Fe(OH)}_3 \rightarrow \text{H}_2\text{O} + \text{Fe}_3(\text{PO}_4)_2 \).
- Balance oxygen (\(\text{O}\)):
- There are \(8\) O atoms on the left (\(4\) from \(\text{H}_3\text{PO}_4\) and \(6\) from \(\text{Fe(OH)}_3\)) and \(14\) O atoms on the right (\(8\) from \(\text{Fe}_3(\text{PO}_4)_2\) and \(2\) from \(\text{H}_2\text{O}\)).
- Multiply \(\text{H}_2\text{O}\) by 6: \( 2\text{H}_3\text{PO}_4 + 2\text{Fe(OH)}_3 \rightarrow 6\text{H}_2\text{O} + \text{Fe}_3(\text{PO}_4)_2 \).
- Check hydrogen (\(\text{H}\)):
- There are \(6\) H atoms from \(\text{H}_3\text{PO}_4\) and \(6\) H atoms from \(\text{Fe(OH)}_3\) (total \(12\)), and \(12\) H atoms on the right from \(6\text{H}_2\text{O}\).
Balanced Equation:
\[ \boxed{2\text{H}_3\text{PO}_4 + 2\text{Fe(OH)}_3 \rightarrow 6\text{H}_2\text{O} + \text{Fe}_3(\text{PO}_4)_2} \]
---
3. \( \text{C}_2\text{H}_5\text{OH} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \)
- Start by balancing carbon (\(\text{C}\)):
- There are 2 C atoms on the left and 1 C atom on the right.
- Multiply \(\text{CO}_2\) by 2: \( \text{C}_2\text{H}_5\text{OH} + \text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O} \).
- Now balance hydrogen (\(\text{H}\)):
- There are 6 H atoms on the left and 2 H atoms on the right.
- Multiply \(\text{H}_2\text{O}\) by 3: \( \text{C}_2\text{H}_5\text{OH} + \text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \).
- Finally, balance oxygen (\(\text{O}\)):
- There are 7 O atoms on the right (\(4\) from \(\text{CO}_2\) and \(3\) from \(\text{H}_2\text{O}\)) and 2 O atoms on the left.
- Multiply \(\text{O}_2\) by 3: \( \text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \).
Balanced Equation:
\[ \boxed{\text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}} \]
---
4. \( \text{Ni(OH)}_2 \rightarrow \text{NiO}_3 + \text{H}_2\text{O} \)
- This equation is not balanced as written because nickel oxide (\(\text{NiO}_3\)) does not exist. The correct product should be \(\text{NiO}\).
- Corrected equation: \( \text{Ni(OH)}_2 \rightarrow \text{NiO} + \text{H}_2\text{O} \).
- Balance nickel (\(\text{Ni}\)):
- There is 1 Ni atom on both sides.
- Balance oxygen (\(\text{O}\)):
- There are 2 O atoms on the left and 1 O atom on the right.
- Multiply \(\text{NiO}\) by 2: \( \text{Ni(OH)}_2 \rightarrow 2\text{NiO} + \text{H}_2\text{O} \).
- Balance hydrogen (\(\text{H}\)):
- There are 2 H atoms on the left and 2 H atoms on the right.
Balanced Equation:
\[ \boxed{\text{Ni(OH)}_2 \rightarrow \text{NiO} + \text{H}_2\text{O}} \]
---
5. \( \text{K}_2\text{SO}_3 + \text{Mn(OH)}_2 \rightarrow \text{KOH} + \text{MnSO}_3 \)
- Balance potassium (\(\text{K}\)):
- There are 2 K atoms on the left and 1 K atom on the right.
- Multiply \(\text{KOH}\) by 2: \( \text{K}_2\text{SO}_3 + \text{Mn(OH)}_2 \rightarrow 2\text{KOH} + \text{MnSO}_3 \).
- Balance manganese (\(\text{Mn}\)):
- There is 1 Mn atom on both sides.
- Balance sulfur (\(\text{S}\)):
- There is 1 S atom on both sides.
- Balance oxygen (\(\text{O}\)):
- There are 5 O atoms on the left (\(3\) from \(\text{K}_2\text{SO}_3\) and \(2\) from \(\text{Mn(OH)}_2\)) and 5 O atoms on the right (\(3\) from \(\text{MnSO}_3\) and \(2\) from \(2\text{KOH}\)).
- Balance hydrogen (\(\text{H}\)):
- There are 2 H atoms on the left and 2 H atoms on the right.
Balanced Equation:
\[ \boxed{\text{K}_2\text{SO}_3 + \text{Mn(OH)}_2 \rightarrow 2\text{KOH} + \text{MnSO}_3} \]
---
6. \( \text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{O} + \text{Na}_2\text{SO}_4 \)
- Balance sodium (\(\text{Na}\)):
- There is 1 Na atom on the left and 2 Na atoms on the right.
- Multiply \(\text{NaOH}\) by 2: \( 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{O} + \text{Na}_2\text{SO}_4 \).
- Balance hydrogen (\(\text{H}\)):
- There are 4 H atoms on the left and 2 H atoms on the right.
- Multiply \(\text{H}_2\text{O}\) by 2: \( 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow 2\text{H}_2\text{O} + \text{Na}_2\text{SO}_4 \).
- Balance sulfur (\(\text{S}\)):
- There is 1 S atom on both sides.
- Balance oxygen (\(\text{O}\)):
- There are 6 O atoms on the left (\(1\) from \(\text{NaOH}\) and \(4\) from \(\text{H}_2\text{SO}_4\)) and 6 O atoms on the right (\(2\) from \(2\text{H}_2\text{O}\) and \(4\) from \(\text{Na}_2\text{SO}_4\)).
Balanced Equation:
\[ \boxed{2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow 2\text{H}_2\text{O} + \text{Na}_2\text{SO}_4} \]
---
7. \( \text{Li} + \text{Pb(OH)}_2 \rightarrow \text{Pb} + \text{LiOH} \)
- Balance lead (\(\text{Pb}\)):
- There is 1 Pb atom on both sides.
- Balance lithium (\(\text{Li}\)):
- There is 1 Li atom on the left and 1 Li atom on the right.
- Balance oxygen (\(\text{O}\)):
- There are 2 O atoms on the left and 1 O atom on the right.
- Multiply \(\text{LiOH}\) by 2: \( \text{Li} + \text{Pb(OH)}_2 \rightarrow \text{Pb} + 2\text{LiOH} \).
- Balance hydrogen (\(\text{H}\)):
- There are 2 H atoms on the left and 2 H atoms on the right.
Balanced Equation:
\[ \boxed{\text{Li} + \text{Pb(OH)}_2 \rightarrow \text{Pb} + 2\text{LiOH}} \]
---
8. \( \text{C}_2\text{H}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \)
- Balance carbon (\(\text{C}\)):
- There are 2 C atoms on the left and 1 C atom on the right.
- Multiply \(\text{CO}_2\) by 2: \( \text{C}_2\text{H}_6 + \text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O} \).
- Balance hydrogen (\(\text{H}\)):
- There are 6 H atoms on the left and 2 H atoms on the right.
- Multiply \(\text{H}_2\text{O}\) by 3: \( \text{C}_2\text{H}_6 + \text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \).
- Balance oxygen (\(\text{O}\)):
- There are 7 O atoms on the right (\(4\) from \(\text{CO}_2\) and \(3\) from \(\text{H}_2\text{O}\)) and 2 O atoms on the left.
- Multiply \(\text{O}_2\) by \(\frac{7}{2}\). Since coefficients must be integers, multiply the entire equation by 2:
\( 2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O} \).
Balanced Equation:
\[ \boxed{2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O}} \]
---
9. \( \text{Ga(OH)}_3 + \text{KF} \rightarrow \text{KOH} + \text{GaF}_3 \)
- Balance gallium (\(\text{Ga}\)):
- There is 1 Ga atom on both sides.
- Balance potassium (\(\text{K}\)):
- There is 1 K atom on the right and 0 on the left.
- Multiply \(\text{KF}\) by 3: \( \text{Ga(OH)}_3 + 3\text{KF} \rightarrow \text{KOH} + \text{GaF}_3 \).
- Balance fluorine (\(\text{F}\)):
- There are 3 F atoms on the left and 3 F atoms on the right.
- Balance oxygen (\(\text{O}\)):
- There are 3 O atoms on the left and 1 O atom on the right.
- Multiply \(\text{KOH}\) by 3: \( \text{Ga(OH)}_3 + 3\text{KF} \rightarrow 3\text{KOH} + \text{GaF}_3 \).
Balanced Equation:
\[ \boxed{\text{Ga(OH)}_3 + 3\text{KF} \rightarrow 3\text{KOH} + \text{GaF}_3} \]
---
10. \( \text{V} + \text{ZnBr}_2 \rightarrow \text{VBrs} + \text{Zn} \)
- This equation is not balanced as written because \(\text{VBrs}\) is not a valid compound. The correct product should be \(\text{VBr}_2\).
- Corrected equation: \( \text{V} + \text{ZnBr}_2 \rightarrow \text{VBr}_2 + \text{Zn} \).
- Balance vanadium (\(\text{V}\)):
- There is 1 V atom on both sides.
- Balance zinc (\(\text{Zn}\)):
- There is 1 Zn atom on both sides.
- Balance bromine (\(\text{Br}\)):
- There are 2 Br atoms on the left and 2 Br atoms on the right.
Balanced Equation:
\[ \boxed{\text{V} + \text{ZnBr}_2 \rightarrow \text{VBr}_2 + \text{Zn}} \]
---
11. \( \text{As}_2\text{O}_3 + \text{H}_2\text{O} \rightarrow \text{H}_3\text{AsO}_3 \)
- Balance arsenic (\(\text{As}\)):
- There are 2 As atoms on the left and 1 As atom on the right.
- Multiply \(\text{H}_3\text{AsO}_3\) by 2: \( \text{As}_2\text{O}_3 + \text{H}_2\text{O} \rightarrow 2\text{H}_3\text{AsO}_3 \).
- Balance oxygen (\(\text{O}\)):
- There are 3 O atoms on the left and 6 O atoms on the right.
- Multiply \(\text{H}_2\text{O}\) by 3: \( \text{As}_2\text{O}_3 + 3\text{H}_2\text{O} \rightarrow 2\text{H}_3\text{AsO}_3 \).
- Balance hydrogen (\(\text{H}\)):
- There are 6 H atoms on the left and 6 H atoms on the right.
Balanced Equation:
\[ \boxed{\text{As}_2\text{O}_3 + 3\text{H}_2\text{O} \rightarrow 2\text{H}_3\text{AsO}_3} \]
---
12. \( \text{NH}_3 + \text{O}_2 \rightarrow \text{NO} + \text{H}_2\text{O} \)
- Balance nitrogen (\(\text{N}\)):
- There is 1 N atom on both sides.
- Balance hydrogen (\(\text{H}\)):
- There are 3 H atoms on the left and 2 H atoms on the right.
- Multiply \(\text{H}_2\text{O}\) by \(\frac{3}{2}\). Since coefficients must be integers, multiply the entire equation by 2:
\( 2\text{NH}_3 + \text{O}_2 \rightarrow 2\text{NO} + 3\text{H}_2\text{O} \).
- Balance oxygen (\(\text{O}\)):
- There are 5 O atoms on the right (\(2\) from \(\text{NO}\) and \(3\) from \(\text{H}_2\text{O}\)) and 2 O atoms on the left.
- Multiply \(\text{O}_2\) by \(\frac{5}{2}\). Since coefficients must be integers, multiply the entire equation by 2:
\( 4\text{NH}_3 + 5\text{O}_2 \rightarrow 4\text{NO} + 6\text{H}_2\text{O} \).
Balanced Equation:
\[ \boxed{4\text{NH}_3 + 5\text{O}_2 \rightarrow 4\text{NO} + 6\text{H}_2\text{O}} \]
---
13. \( \text{C}_2\text{H}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \)
- Balance carbon (\(\text{C}\)):
- There are 2 C atoms on the left and 1 C atom on the right.
- Multiply \(\text{CO}_2\) by 2: \( \text{C}_2\text{H}_4 + \text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O} \).
- Balance hydrogen (\(\text{H}\)):
- There are 4 H atoms on the left and 2 H atoms on the right.
- Multiply \(\text{H}_2\text{O}\) by 2: \( \text{C}_2\text{H}_4 + \text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} \).
- Balance oxygen (\(\text{O}\)):
- There are 6 O atoms on the right (\(4\) from \(\text{CO}_2\) and \(2\) from \(\text{H}_2\text{O}\)) and 2 O atoms on the left.
- Multiply \(\text{O}_2\) by 3: \( \text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} \).
Balanced Equation:
\[ \boxed{\text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O}} \]
---
14. \( \text{NaClO}_3 \rightarrow \text{NaCl} + \text{O}_2 \)
- Balance sodium (\(\text{Na}\)):
- There is 1 Na atom on both sides.
- Balance chlorine (\(\text{Cl}\)):
- There is 1 Cl atom on both sides.
- Balance oxygen (\(\text{O}\)):
- There are 3 O atoms on the left and 2 O atoms on the right.
- Multiply \(\text{O}_2\) by \(\frac{3}{2}\). Since coefficients must be integers, multiply the entire equation by 2:
\( 2\text{NaClO}_3 \rightarrow 2\text{NaCl} + 3\text{O}_2 \).
Balanced Equation:
\[ \boxed{2\text{NaClO}_3 \rightarrow 2\text{NaCl} + 3\text{O}_2} \]
---
15. \( \text{Ca} + \text{O}_2 \rightarrow \text{CaO} \)
- Balance calcium (\(\text{Ca}\)):
- There is 1 Ca atom on both sides.
- Balance oxygen (\(\text{O}\)):
- There are 2 O atoms on the left and 1 O atom on the right.
- Multiply \(\text{CaO}\) by 2: \( \text{Ca} + \text{O}_2 \rightarrow 2\text{CaO} \).
Balanced Equation:
\[ \boxed{2\text{Ca} + \text{O}_2 \rightarrow 2\text{CaO}} \]
---
Final Answer:
\[ \boxed{
\begin{aligned}
1. & \quad \text{Br}_2 + 2\text{LiF} \rightarrow 2\text{LiBr} + \text{F}_2 \\
2. & \quad 2\text{H}_3\text{PO}_4 + 2\text{Fe(OH)}_3 \rightarrow 6\text{H}_2\text{O} + \text{Fe}_3(\text{PO}_4)_2 \\
3. & \quad \text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \\
4. & \quad \text{Ni(OH)}_2 \rightarrow \text{NiO} + \text{H}_2\text{O} \\
5. & \quad \text{K}_2\text{SO}_3 + \text{Mn(OH)}_2 \rightarrow 2\text{KOH} + \text{MnSO}_3 \\
6. & \quad 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow 2\text{H}_2\text{O} + \text{Na}_2\text{SO}_4 \\
7. & \quad \text{Li} + \text{Pb(OH)}_2 \rightarrow \text{Pb} + 2\text{LiOH} \\
8. & \quad 2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O} \\
9. & \quad \text{Ga(OH)}_3 + 3\text{KF} \rightarrow 3\text{KOH} + \text{GaF}_3 \\
10. & \quad \text{V} + \text{ZnBr}_2 \rightarrow \text{VBr}_2 + \text{Zn} \\
11. & \quad \text{As}_2\text{O}_3 + 3\text{H}_2\text{O} \rightarrow 2\text{H}_3\text{AsO}_3 \\
12. & \quad 4\text{NH}_3 + 5\text{O}_2 \rightarrow 4\text{NO} + 6\text{H}_2\text{O} \\
13. & \quad \text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} \\
14. & \quad 2\text{NaClO}_3 \rightarrow 2\text{NaCl} + 3\text{O}_2 \\
15. & \quad 2\text{Ca} + \text{O}_2 \rightarrow 2\text{CaO} \\
\end{aligned}
} \]
Parent Tip: Review the logic above to help your child master the concept of balancing chemical equations chapter 7 worksheet 1 answers.
![49 Balancing Chemical Equations Worksheets [with Answers]](/images2/balancing-chemical-equations-chapter-7-worksheet-1-answers/balancing-chemical-equations-chapter-7-worksheet-1-answers-32.jpg)
