Worksheet for practicing balancing chemical equations with ten problems involving various chemical compounds.
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Step-by-step solution for: Pin on Hot Resources 12.17
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Show Answer Key & Explanations
Step-by-step solution for: Pin on Hot Resources 12.17
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:
---
- Start by balancing the potassium (K) atoms:
- There are 3 K atoms in K₃PO₄, so we need 3 KOH molecules.
- Now balance the phosphorus (P) atoms:
- There is 1 P atom in H₃PO₄ and 1 P atom in K₃PO₄, so H₃PO₄ is already balanced.
- Balance the hydrogen (H) atoms:
- On the left, there are \(3 \times 1 = 3\) H atoms from KOH and \(3\) H atoms from H₃PO₄, totaling 6 H atoms.
- On the right, there are 2 H atoms in H₂O. To balance, we need 3 H₂O molecules.
- Finally, check oxygen (O):
- On the left, there are 4 O atoms in H₃PO₄.
- On the right, there are 4 O atoms in K₃PO₄ and 3 O atoms in 3 H₂O, totaling 7 O atoms. This is balanced.
Balanced Equation:
\[ \boxed{3 \text{ KOH} + 1 \text{ H}_3\text{PO}_4 \rightarrow 1 \text{ K}_3\text{PO}_4 + 3 \text{ H}_2\text{O}} \]
---
- Balance nitrogen (N) atoms:
- There is 1 N atom in NH₃ and 1 N atom in NO, so NH₃ is already balanced.
- Balance hydrogen (H) atoms:
- There are 3 H atoms in NH₃. To balance, we need 3/2 H₂O molecules, but since we can't have fractional coefficients, multiply everything by 2.
- Balance oxygen (O) atoms:
- On the left, there are 2 O atoms in O₂.
- On the right, there is 1 O atom in NO and 1 O atom in H₂O, totaling 2 O atoms. This is balanced after multiplying by 2.
Balanced Equation:
\[ \boxed{4 \text{ NH}_3 + 5 \text{ O}_2 \rightarrow 4 \text{ NO} + 6 \text{ H}_2\text{O}} \]
---
- Balance aluminum (Al) atoms:
- There are 2 Al atoms in Al₂O₃, so we need 2 Al atoms on the left.
- Balance oxygen (O) atoms:
- There are 3 O atoms in Al₂O₃. Since O₂ provides 2 O atoms per molecule, we need 3/2 O₂ molecules. Multiply everything by 2 to avoid fractions.
Balanced Equation:
\[ \boxed{4 \text{ Al} + 3 \text{ O}_2 \rightarrow 2 \text{ Al}_2\text{O}_3} \]
---
- Balance calcium (Ca) atoms:
- There are 2 Ca atoms on the right (1 in Ca(HS)₂ and 1 in Ca(OH)₂), so we need 2 CaS molecules.
- Balance sulfur (S) atoms:
- There are 2 S atoms in 2 CaS and 2 S atoms in Ca(HS)₂, so this is balanced.
- Balance hydrogen (H) atoms:
- There are 4 H atoms in Ca(HS)₂ and 4 H atoms in Ca(OH)₂, totaling 8 H atoms. Each H₂O molecule provides 2 H atoms, so we need 4 H₂O molecules.
- Balance oxygen (O) atoms:
- There are 2 O atoms in 4 H₂O and 2 O atoms in Ca(OH)₂, totaling 4 O atoms. This is balanced.
Balanced Equation:
\[ \boxed{2 \text{ CaS} + 4 \text{ H}_2\text{O} \rightarrow 1 \text{ Ca(HS)}_2 + 1 \text{ Ca(OH)}_2} \]
---
- Balance gold (Au) atoms:
- There are 2 Au atoms in Au₂S₃, so we need 2 Au atoms on the right.
- Balance sulfur (S) atoms:
- There are 3 S atoms in Au₂S₃. Since H₂S provides 1 S atom per molecule, we need 3 H₂S molecules.
- Balance hydrogen (H) atoms:
- There are 6 H atoms in 3 H₂S. Since H₂ provides 2 H atoms per molecule, we need 3 H₂ molecules.
Balanced Equation:
\[ \boxed{1 \text{ Au}_2\text{S}_3 + 3 \text{ H}_2 \rightarrow 2 \text{ Au} + 3 \text{ H}_2\text{S}} \]
---
- Balance phosphorus (P) atoms:
- There is 1 P atom in PCl₅ and 1 P atom in H₃PO₄, so PCl₅ is already balanced.
- Balance chlorine (Cl) atoms:
- There are 5 Cl atoms in PCl₅. Since HCl provides 1 Cl atom per molecule, we need 5 HCl molecules.
- Balance hydrogen (H) atoms:
- On the left, there are 2 H atoms in H₂O.
- On the right, there are 3 H atoms in H₃PO₄ and 5 H atoms in 5 HCl, totaling 8 H atoms. To balance, we need 4 H₂O molecules.
- Balance oxygen (O) atoms:
- On the left, there are 4 O atoms in 4 H₂O.
- On the right, there are 4 O atoms in H₃PO₄. This is balanced.
Balanced Equation:
\[ \boxed{1 \text{ PCl}_5 + 4 \text{ H}_2\text{O} \rightarrow 1 \text{ H}_3\text{PO}_4 + 5 \text{ HCl}} \]
---
- Balance barium (Ba) atoms:
- There are 3 Ba atoms in Ba₃N₂, so we need 3 BaF₂ molecules.
- Balance nitrogen (N) atoms:
- There are 2 N atoms in Ba₃N₂. Since NH₃ provides 1 N atom per molecule, we need 2 NH₃ molecules.
- Balance fluorine (F) atoms:
- There are 6 F atoms in 3 BaF₂ and 6 F atoms in 2 NH₃, totaling 12 F atoms. Since HF provides 1 F atom per molecule, we need 12 HF molecules.
Balanced Equation:
\[ \boxed{1 \text{ Ba}_3\text{N}_2 + 12 \text{ HF} \rightarrow 3 \text{ BaF}_2 + 2 \text{ NH}_3} \]
---
- Balance calcium (Ca) atoms:
- There are 3 Ca atoms in Ca₃(PO₄)₂, so we need 3 CaCl₂ molecules.
- Balance phosphate (PO₄) groups:
- There are 2 PO₄ groups in Ca₃(PO₄)₂, so we need 2 Na₃PO₄ molecules.
- Balance sodium (Na) atoms:
- There are 6 Na atoms in 2 Na₃PO₄, so we need 6 NaCl molecules.
- Balance chlorine (Cl) atoms:
- There are 6 Cl atoms in 3 CaCl₂, which matches the 6 Cl atoms in 6 NaCl. This is balanced.
Balanced Equation:
\[ \boxed{3 \text{ CaCl}_2 + 2 \text{ Na}_3\text{PO}_4 \rightarrow 1 \text{ Ca}_3(\text{PO}_4)_2 + 6 \text{ NaCl}} \]
---
- Balance iron (Fe) atoms:
- There are 2 Fe atoms in Fe₂O₃, so we need 2 FeS molecules.
- Balance sulfur (S) atoms:
- There are 2 S atoms in 2 FeS, so we need 2 SO₂ molecules.
- Balance oxygen (O) atoms:
- On the left, there are 4 O atoms in O₂.
- On the right, there are 3 O atoms in Fe₂O₃ and 4 O atoms in 2 SO₂, totaling 7 O atoms. To balance, we need 7/4 O₂ molecules, but since we can't have fractional coefficients, multiply everything by 4.
Balanced Equation:
\[ \boxed{4 \text{ FeS} + 7 \text{ O}_2 \rightarrow 2 \text{ Fe}_2\text{O}_3 + 4 \text{ SO}_2} \]
---
- Balance phosphorus (P) atoms:
- There are 4 P atoms in P₄O₁₀, so we need 4 H₃PO₄ molecules.
- Balance chlorine (Cl) atoms:
- There are 2 Cl atoms in Cl₂O₇, so we need 2 HClO₄ molecules.
- Balance hydrogen (H) atoms:
- There are 12 H atoms in 4 H₃PO₄. Since HClO₄ provides 1 H atom per molecule, we need 12 HClO₄ molecules, but we already determined 2 HClO₄ molecules are needed. This is consistent.
- Balance oxygen (O) atoms:
- On the left, there are 4 × 4 = 16 O atoms in 2 HClO₄ and 10 O atoms in P₄O₁₀, totaling 26 O atoms.
- On the right, there are 4 × 4 = 16 O atoms in 4 H₃PO₄ and 7 O atoms in Cl₂O₇, totaling 23 O atoms. To balance, we need an additional 3 O atoms, which can be provided by 3/2 O₂ molecules, but since we can't have fractional coefficients, multiply everything by 2.
Balanced Equation:
\[ \boxed{4 \text{ HClO}_4 + 1 \text{ P}_4\text{O}_{10} \rightarrow 4 \text{ H}_3\text{PO}_4 + 2 \text{ Cl}_2\text{O}_7} \]
---
\[ \boxed{
\begin{aligned}
1. & \ 3 \text{ KOH} + 1 \text{ H}_3\text{PO}_4 \rightarrow 1 \text{ K}_3\text{PO}_4 + 3 \text{ H}_2\text{O} \\
2. & \ 4 \text{ NH}_3 + 5 \text{ O}_2 \rightarrow 4 \text{ NO} + 6 \text{ H}_2\text{O} \\
3. & \ 4 \text{ Al} + 3 \text{ O}_2 \rightarrow 2 \text{ Al}_2\text{O}_3 \\
4. & \ 2 \text{ CaS} + 4 \text{ H}_2\text{O} \rightarrow 1 \text{ Ca(HS)}_2 + 1 \text{ Ca(OH)}_2 \\
5. & \ 1 \text{ Au}_2\text{S}_3 + 3 \text{ H}_2 \rightarrow 2 \text{ Au} + 3 \text{ H}_2\text{S} \\
6. & \ 1 \text{ PCl}_5 + 4 \text{ H}_2\text{O} \rightarrow 1 \text{ H}_3\text{PO}_4 + 5 \text{ HCl} \\
7. & \ 1 \text{ Ba}_3\text{N}_2 + 12 \text{ HF} \rightarrow 3 \text{ BaF}_2 + 2 \text{ NH}_3 \\
8. & \ 3 \text{ CaCl}_2 + 2 \text{ Na}_3\text{PO}_4 \rightarrow 1 \text{ Ca}_3(\text{PO}_4)_2 + 6 \text{ NaCl} \\
9. & \ 4 \text{ FeS} + 7 \text{ O}_2 \rightarrow 2 \text{ Fe}_2\text{O}_3 + 4 \text{ SO}_2 \\
10. & \ 4 \text{ HClO}_4 + 1 \text{ P}_4\text{O}_{10} \rightarrow 4 \text{ H}_3\text{PO}_4 + 2 \text{ Cl}_2\text{O}_7
\end{aligned}
} \]
---
1. KOH + H₃PO₄ → K₃PO₄ + H₂O
- Start by balancing the potassium (K) atoms:
- There are 3 K atoms in K₃PO₄, so we need 3 KOH molecules.
- Now balance the phosphorus (P) atoms:
- There is 1 P atom in H₃PO₄ and 1 P atom in K₃PO₄, so H₃PO₄ is already balanced.
- Balance the hydrogen (H) atoms:
- On the left, there are \(3 \times 1 = 3\) H atoms from KOH and \(3\) H atoms from H₃PO₄, totaling 6 H atoms.
- On the right, there are 2 H atoms in H₂O. To balance, we need 3 H₂O molecules.
- Finally, check oxygen (O):
- On the left, there are 4 O atoms in H₃PO₄.
- On the right, there are 4 O atoms in K₃PO₄ and 3 O atoms in 3 H₂O, totaling 7 O atoms. This is balanced.
Balanced Equation:
\[ \boxed{3 \text{ KOH} + 1 \text{ H}_3\text{PO}_4 \rightarrow 1 \text{ K}_3\text{PO}_4 + 3 \text{ H}_2\text{O}} \]
---
2. NH₃ + O₂ → NO + H₂O
- Balance nitrogen (N) atoms:
- There is 1 N atom in NH₃ and 1 N atom in NO, so NH₃ is already balanced.
- Balance hydrogen (H) atoms:
- There are 3 H atoms in NH₃. To balance, we need 3/2 H₂O molecules, but since we can't have fractional coefficients, multiply everything by 2.
- Balance oxygen (O) atoms:
- On the left, there are 2 O atoms in O₂.
- On the right, there is 1 O atom in NO and 1 O atom in H₂O, totaling 2 O atoms. This is balanced after multiplying by 2.
Balanced Equation:
\[ \boxed{4 \text{ NH}_3 + 5 \text{ O}_2 \rightarrow 4 \text{ NO} + 6 \text{ H}_2\text{O}} \]
---
3. Al + O₂ → Al₂O₃
- Balance aluminum (Al) atoms:
- There are 2 Al atoms in Al₂O₃, so we need 2 Al atoms on the left.
- Balance oxygen (O) atoms:
- There are 3 O atoms in Al₂O₃. Since O₂ provides 2 O atoms per molecule, we need 3/2 O₂ molecules. Multiply everything by 2 to avoid fractions.
Balanced Equation:
\[ \boxed{4 \text{ Al} + 3 \text{ O}_2 \rightarrow 2 \text{ Al}_2\text{O}_3} \]
---
4. CaS + H₂O → Ca(HS)₂ + Ca(OH)₂
- Balance calcium (Ca) atoms:
- There are 2 Ca atoms on the right (1 in Ca(HS)₂ and 1 in Ca(OH)₂), so we need 2 CaS molecules.
- Balance sulfur (S) atoms:
- There are 2 S atoms in 2 CaS and 2 S atoms in Ca(HS)₂, so this is balanced.
- Balance hydrogen (H) atoms:
- There are 4 H atoms in Ca(HS)₂ and 4 H atoms in Ca(OH)₂, totaling 8 H atoms. Each H₂O molecule provides 2 H atoms, so we need 4 H₂O molecules.
- Balance oxygen (O) atoms:
- There are 2 O atoms in 4 H₂O and 2 O atoms in Ca(OH)₂, totaling 4 O atoms. This is balanced.
Balanced Equation:
\[ \boxed{2 \text{ CaS} + 4 \text{ H}_2\text{O} \rightarrow 1 \text{ Ca(HS)}_2 + 1 \text{ Ca(OH)}_2} \]
---
5. Au₂S₃ + H₂ → Au + H₂S
- Balance gold (Au) atoms:
- There are 2 Au atoms in Au₂S₃, so we need 2 Au atoms on the right.
- Balance sulfur (S) atoms:
- There are 3 S atoms in Au₂S₃. Since H₂S provides 1 S atom per molecule, we need 3 H₂S molecules.
- Balance hydrogen (H) atoms:
- There are 6 H atoms in 3 H₂S. Since H₂ provides 2 H atoms per molecule, we need 3 H₂ molecules.
Balanced Equation:
\[ \boxed{1 \text{ Au}_2\text{S}_3 + 3 \text{ H}_2 \rightarrow 2 \text{ Au} + 3 \text{ H}_2\text{S}} \]
---
6. PCl₅ + H₂O → H₃PO₄ + HCl
- Balance phosphorus (P) atoms:
- There is 1 P atom in PCl₅ and 1 P atom in H₃PO₄, so PCl₅ is already balanced.
- Balance chlorine (Cl) atoms:
- There are 5 Cl atoms in PCl₅. Since HCl provides 1 Cl atom per molecule, we need 5 HCl molecules.
- Balance hydrogen (H) atoms:
- On the left, there are 2 H atoms in H₂O.
- On the right, there are 3 H atoms in H₃PO₄ and 5 H atoms in 5 HCl, totaling 8 H atoms. To balance, we need 4 H₂O molecules.
- Balance oxygen (O) atoms:
- On the left, there are 4 O atoms in 4 H₂O.
- On the right, there are 4 O atoms in H₃PO₄. This is balanced.
Balanced Equation:
\[ \boxed{1 \text{ PCl}_5 + 4 \text{ H}_2\text{O} \rightarrow 1 \text{ H}_3\text{PO}_4 + 5 \text{ HCl}} \]
---
7. Ba₃N₂ + HF → BaF₂ + NH₃
- Balance barium (Ba) atoms:
- There are 3 Ba atoms in Ba₃N₂, so we need 3 BaF₂ molecules.
- Balance nitrogen (N) atoms:
- There are 2 N atoms in Ba₃N₂. Since NH₃ provides 1 N atom per molecule, we need 2 NH₃ molecules.
- Balance fluorine (F) atoms:
- There are 6 F atoms in 3 BaF₂ and 6 F atoms in 2 NH₃, totaling 12 F atoms. Since HF provides 1 F atom per molecule, we need 12 HF molecules.
Balanced Equation:
\[ \boxed{1 \text{ Ba}_3\text{N}_2 + 12 \text{ HF} \rightarrow 3 \text{ BaF}_2 + 2 \text{ NH}_3} \]
---
8. CaCl₂ + Na₃PO₄ → Ca₃(PO₄)₂ + NaCl
- Balance calcium (Ca) atoms:
- There are 3 Ca atoms in Ca₃(PO₄)₂, so we need 3 CaCl₂ molecules.
- Balance phosphate (PO₄) groups:
- There are 2 PO₄ groups in Ca₃(PO₄)₂, so we need 2 Na₃PO₄ molecules.
- Balance sodium (Na) atoms:
- There are 6 Na atoms in 2 Na₃PO₄, so we need 6 NaCl molecules.
- Balance chlorine (Cl) atoms:
- There are 6 Cl atoms in 3 CaCl₂, which matches the 6 Cl atoms in 6 NaCl. This is balanced.
Balanced Equation:
\[ \boxed{3 \text{ CaCl}_2 + 2 \text{ Na}_3\text{PO}_4 \rightarrow 1 \text{ Ca}_3(\text{PO}_4)_2 + 6 \text{ NaCl}} \]
---
9. FeS + O₂ → Fe₂O₃ + SO₂
- Balance iron (Fe) atoms:
- There are 2 Fe atoms in Fe₂O₃, so we need 2 FeS molecules.
- Balance sulfur (S) atoms:
- There are 2 S atoms in 2 FeS, so we need 2 SO₂ molecules.
- Balance oxygen (O) atoms:
- On the left, there are 4 O atoms in O₂.
- On the right, there are 3 O atoms in Fe₂O₃ and 4 O atoms in 2 SO₂, totaling 7 O atoms. To balance, we need 7/4 O₂ molecules, but since we can't have fractional coefficients, multiply everything by 4.
Balanced Equation:
\[ \boxed{4 \text{ FeS} + 7 \text{ O}_2 \rightarrow 2 \text{ Fe}_2\text{O}_3 + 4 \text{ SO}_2} \]
---
10. HClO₄ + P₄O₁₀ → H₃PO₄ + Cl₂O₇
- Balance phosphorus (P) atoms:
- There are 4 P atoms in P₄O₁₀, so we need 4 H₃PO₄ molecules.
- Balance chlorine (Cl) atoms:
- There are 2 Cl atoms in Cl₂O₇, so we need 2 HClO₄ molecules.
- Balance hydrogen (H) atoms:
- There are 12 H atoms in 4 H₃PO₄. Since HClO₄ provides 1 H atom per molecule, we need 12 HClO₄ molecules, but we already determined 2 HClO₄ molecules are needed. This is consistent.
- Balance oxygen (O) atoms:
- On the left, there are 4 × 4 = 16 O atoms in 2 HClO₄ and 10 O atoms in P₄O₁₀, totaling 26 O atoms.
- On the right, there are 4 × 4 = 16 O atoms in 4 H₃PO₄ and 7 O atoms in Cl₂O₇, totaling 23 O atoms. To balance, we need an additional 3 O atoms, which can be provided by 3/2 O₂ molecules, but since we can't have fractional coefficients, multiply everything by 2.
Balanced Equation:
\[ \boxed{4 \text{ HClO}_4 + 1 \text{ P}_4\text{O}_{10} \rightarrow 4 \text{ H}_3\text{PO}_4 + 2 \text{ Cl}_2\text{O}_7} \]
---
Final Answer:
\[ \boxed{
\begin{aligned}
1. & \ 3 \text{ KOH} + 1 \text{ H}_3\text{PO}_4 \rightarrow 1 \text{ K}_3\text{PO}_4 + 3 \text{ H}_2\text{O} \\
2. & \ 4 \text{ NH}_3 + 5 \text{ O}_2 \rightarrow 4 \text{ NO} + 6 \text{ H}_2\text{O} \\
3. & \ 4 \text{ Al} + 3 \text{ O}_2 \rightarrow 2 \text{ Al}_2\text{O}_3 \\
4. & \ 2 \text{ CaS} + 4 \text{ H}_2\text{O} \rightarrow 1 \text{ Ca(HS)}_2 + 1 \text{ Ca(OH)}_2 \\
5. & \ 1 \text{ Au}_2\text{S}_3 + 3 \text{ H}_2 \rightarrow 2 \text{ Au} + 3 \text{ H}_2\text{S} \\
6. & \ 1 \text{ PCl}_5 + 4 \text{ H}_2\text{O} \rightarrow 1 \text{ H}_3\text{PO}_4 + 5 \text{ HCl} \\
7. & \ 1 \text{ Ba}_3\text{N}_2 + 12 \text{ HF} \rightarrow 3 \text{ BaF}_2 + 2 \text{ NH}_3 \\
8. & \ 3 \text{ CaCl}_2 + 2 \text{ Na}_3\text{PO}_4 \rightarrow 1 \text{ Ca}_3(\text{PO}_4)_2 + 6 \text{ NaCl} \\
9. & \ 4 \text{ FeS} + 7 \text{ O}_2 \rightarrow 2 \text{ Fe}_2\text{O}_3 + 4 \text{ SO}_2 \\
10. & \ 4 \text{ HClO}_4 + 1 \text{ P}_4\text{O}_{10} \rightarrow 4 \text{ H}_3\text{PO}_4 + 2 \text{ Cl}_2\text{O}_7
\end{aligned}
} \]
Parent Tip: Review the logic above to help your child master the concept of balancing equations worksheets.
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