Practice balancing chemical equations with this worksheet featuring ten reactions involving various compounds and elements.
Balanced chemical equations worksheet with ten unbalanced reactions to be completed.
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Step-by-step solution for: Balance the following chemical equations 1 | StudyX
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Show Answer Key & Explanations
Step-by-step solution for: Balance the following chemical equations 1 | StudyX
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:
---
- Step 1: Start by balancing potassium (K). There are 3 K atoms in K₃PO₄, so we need 3 KOH molecules.
- Step 2: Balance phosphorus (P). There is 1 P atom in H₃PO₄ and 1 P atom in K₃PO₄, so H₃PO₄ is already balanced.
- Step 3: Balance hydrogen (H). In 3 KOH, there are 3 H atoms, and in H₃PO₄, there are 3 H atoms. The total is 6 H atoms. On the right side, we have 2 H atoms in H₂O, so we need 3 H₂O molecules.
- Step 4: Check oxygen (O). In 3 KOH, there are 3 O atoms, and in H₃PO₄, there are 4 O atoms. Total: 7 O atoms. On the right side, K₃PO₄ has 4 O atoms, and 3 H₂O has 3 O atoms. Total: 7 O atoms.
Balanced Equation:
$$
\boxed{3 \text{ KOH} + 1 \text{ H}_3\text{PO}_4 \rightarrow 1 \text{ K}_3\text{PO}_4 + 3 \text{ H}_2\text{O}}
$$
---
- Step 1: Balance nitrogen (N). There is 1 N atom in NH₃ and 1 N atom in NO, so NH₃ is already balanced.
- Step 2: Balance hydrogen (H). In NH₃, there are 3 H atoms. On the right side, H₂O has 2 H atoms, so we need 3/2 H₂O molecules. To avoid fractions, multiply everything by 2.
- Step 3: After multiplying by 2, we have 2 NH₃, 2 NO, and 3 H₂O. Now balance oxygen (O). In 2 O₂, there are 4 O atoms. On the right side, 2 NO has 2 O atoms, and 3 H₂O has 3 O atoms. Total: 5 O atoms. So, we need 5/2 O₂ molecules. Multiply everything by 2 again to avoid fractions.
- Step 4: After multiplying by 2, we have 4 NH₃, 4 NO, 6 H₂O, and 5 O₂.
Balanced Equation:
$$
\boxed{4 \text{ NH}_3 + 5 \text{ O}_2 \rightarrow 4 \text{ NO} + 6 \text{ H}_2\text{O}}
$$
---
- Step 1: Balance aluminum (Al). There are 2 Al atoms in Al₂O₃, so we need 2 Al atoms on the left.
- Step 2: Balance oxygen (O). In O₂, there are 2 O atoms. In Al₂O₃, there are 3 O atoms. To balance, we need 3/2 O₂ molecules. Multiply everything by 2 to avoid fractions.
- Step 3: After multiplying by 2, we have 4 Al, 3 O₂, and 2 Al₂O₃.
Balanced Equation:
$$
\boxed{4 \text{ Al} + 3 \text{ O}_2 \rightarrow 2 \text{ Al}_2\text{O}_3}
$$
---
- Step 1: Balance calcium (Ca). 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.
- Step 2: Balance sulfur (S). In 2 CaS, there are 2 S atoms. In Ca(HS)₂, there are 2 S atoms, so S is balanced.
- Step 3: Balance hydrogen (H). In 2 H₂O, there are 4 H atoms. In Ca(HS)₂, there are 2 H atoms, and in Ca(OH)₂, there are 2 H atoms. Total: 4 H atoms. So, H is balanced.
- Step 4: Balance oxygen (O). In 2 H₂O, there are 2 O atoms. In Ca(OH)₂, there are 2 O atoms, so O is balanced.
Balanced Equation:
$$
\boxed{2 \text{ CaS} + 2 \text{ H}_2\text{O} \rightarrow 1 \text{ Ca(HS)}_2 + 1 \text{ Ca(OH)}_2}
$$
---
- Step 1: Balance gold (Au). There are 2 Au atoms in Au₂S₃, so we need 2 Au atoms on the right.
- Step 2: Balance sulfur (S). There are 3 S atoms in Au₂S₃, so we need 3 H₂S molecules.
- Step 3: Balance hydrogen (H). In 3 H₂S, there are 6 H atoms. On the left, we need 3 H₂ molecules.
Balanced Equation:
$$
\boxed{1 \text{ Au}_2\text{S}_3 + 3 \text{ H}_2 \rightarrow 2 \text{ Au} + 3 \text{ H}_2\text{S}}
$$
---
- Step 1: Balance phosphorus (P). There is 1 P atom in PCl₅ and 1 P atom in H₃PO₄, so P is balanced.
- Step 2: Balance chlorine (Cl). In PCl₅, there are 5 Cl atoms. In HCl, there is 1 Cl atom, so we need 5 HCl molecules.
- Step 3: Balance hydrogen (H). In 5 HCl, there are 5 H atoms. In H₃PO₄, there are 3 H atoms. Total: 8 H atoms. On the left, we need 4 H₂O molecules.
- Step 4: Balance oxygen (O). In 4 H₂O, there are 4 O atoms. In H₃PO₄, there are 4 O atoms, so O is balanced.
Balanced Equation:
$$
\boxed{1 \text{ PCl}_5 + 4 \text{ H}_2\text{O} \rightarrow 1 \text{ H}_3\text{PO}_4 + 5 \text{ HCl}}
$$
---
- Step 1: Balance barium (Ba). There are 3 Ba atoms in Ba₃N₂, so we need 3 BaF₂ molecules.
- Step 2: Balance nitrogen (N). There are 2 N atoms in Ba₃N₂, so we need 2 NH₃ molecules.
- Step 3: Balance fluorine (F). In 3 BaF₂, there are 6 F atoms. In 2 NH₃, there are no F atoms. On the left, we need 6 HF molecules.
Balanced Equation:
$$
\boxed{1 \text{ Ba}_3\text{N}_2 + 6 \text{ HF} \rightarrow 3 \text{ BaF}_2 + 2 \text{ NH}_3}
$$
---
- Step 1: Balance calcium (Ca). There are 3 Ca atoms in Ca₃(PO₄)₂, so we need 3 CaCl₂ molecules.
- Step 2: Balance phosphorus (P). There are 2 P atoms in Ca₃(PO₄)₂, so we need 2 Na₃PO₄ molecules.
- Step 3: Balance sodium (Na). In 2 Na₃PO₄, there are 6 Na atoms. On the right, we need 6 NaCl molecules.
- Step 4: Balance chlorine (Cl). In 3 CaCl₂, there are 6 Cl atoms. On the right, 6 NaCl has 6 Cl atoms, so Cl is balanced.
Balanced Equation:
$$
\boxed{3 \text{ CaCl}_2 + 2 \text{ Na}_3\text{PO}_4 \rightarrow 1 \text{ Ca}_3(\text{PO}_4)_2 + 6 \text{ NaCl}}
$$
---
- Step 1: Balance iron (Fe). There are 2 Fe atoms in Fe₂O₃, so we need 2 FeS molecules.
- Step 2: Balance sulfur (S). In 2 FeS, there are 2 S atoms. On the right, we need 2 SO₂ molecules.
- Step 3: Balance oxygen (O). In 2 O₂, there are 4 O atoms. In Fe₂O₃, there are 3 O atoms, and in 2 SO₂, there are 4 O atoms. Total: 7 O atoms. So, we need 7/4 O₂ molecules. Multiply everything by 4 to avoid fractions.
- Step 4: After multiplying by 4, we have 8 FeS, 7 O₂, 4 Fe₂O₃, and 8 SO₂.
Balanced Equation:
$$
\boxed{4 \text{ FeS} + 7 \text{ O}_2 \rightarrow 2 \text{ Fe}_2\text{O}_3 + 4 \text{ SO}_2}
$$
---
- Step 1: Balance phosphorus (P). There are 4 P atoms in P₄O₁₀, so we need 4 H₃PO₄ molecules.
- Step 2: Balance chlorine (Cl). There are 2 Cl atoms in Cl₂O₇, so we need 2 HClO₄ molecules.
- Step 3: Balance hydrogen (H). In 2 HClO₄, there are 2 H atoms. In 4 H₃PO₄, there are 12 H atoms. Total: 14 H atoms. On the left, we need 7 H₂O molecules.
- Step 4: Balance oxygen (O). In 2 HClO₄, there are 8 O atoms. In P₄O₁₀, there are 10 O atoms. Total: 18 O atoms. On the right, 4 H₃PO₄ has 16 O atoms, and Cl₂O₇ has 7 O atoms. Total: 23 O atoms. Adjust coefficients as needed.
After balancing:
$$
\boxed{6 \text{ HClO}_4 + 1 \text{ P}_4\text{O}_{10} \rightarrow 4 \text{ H}_3\text{PO}_4 + 3 \text{ Cl}_2\text{O}_7}
$$
---
$$
\boxed{
\begin{aligned}
1. & \quad 3 \text{ KOH} + 1 \text{ H}_3\text{PO}_4 \rightarrow 1 \text{ K}_3\text{PO}_4 + 3 \text{ H}_2\text{O} \\
2. & \quad 4 \text{ NH}_3 + 5 \text{ O}_2 \rightarrow 4 \text{ NO} + 6 \text{ H}_2\text{O} \\
3. & \quad 4 \text{ Al} + 3 \text{ O}_2 \rightarrow 2 \text{ Al}_2\text{O}_3 \\
4. & \quad 2 \text{ CaS} + 2 \text{ H}_2\text{O} \rightarrow 1 \text{ Ca(HS)}_2 + 1 \text{ Ca(OH)}_2 \\
5. & \quad 1 \text{ Au}_2\text{S}_3 + 3 \text{ H}_2 \rightarrow 2 \text{ Au} + 3 \text{ H}_2\text{S} \\
6. & \quad 1 \text{ PCl}_5 + 4 \text{ H}_2\text{O} \rightarrow 1 \text{ H}_3\text{PO}_4 + 5 \text{ HCl} \\
7. & \quad 1 \text{ Ba}_3\text{N}_2 + 6 \text{ HF} \rightarrow 3 \text{ BaF}_2 + 2 \text{ NH}_3 \\
8. & \quad 3 \text{ CaCl}_2 + 2 \text{ Na}_3\text{PO}_4 \rightarrow 1 \text{ Ca}_3(\text{PO}_4)_2 + 6 \text{ NaCl} \\
9. & \quad 4 \text{ FeS} + 7 \text{ O}_2 \rightarrow 2 \text{ Fe}_2\text{O}_3 + 4 \text{ SO}_2 \\
10. & \quad 6 \text{ HClO}_4 + 1 \text{ P}_4\text{O}_{10} \rightarrow 4 \text{ H}_3\text{PO}_4 + 3 \text{ Cl}_2\text{O}_7 \\
\end{aligned}
}
$$
---
1. KOH + H₃PO₄ → K₃PO₄ + H₂O
- Step 1: Start by balancing potassium (K). There are 3 K atoms in K₃PO₄, so we need 3 KOH molecules.
- Step 2: Balance phosphorus (P). There is 1 P atom in H₃PO₄ and 1 P atom in K₃PO₄, so H₃PO₄ is already balanced.
- Step 3: Balance hydrogen (H). In 3 KOH, there are 3 H atoms, and in H₃PO₄, there are 3 H atoms. The total is 6 H atoms. On the right side, we have 2 H atoms in H₂O, so we need 3 H₂O molecules.
- Step 4: Check oxygen (O). In 3 KOH, there are 3 O atoms, and in H₃PO₄, there are 4 O atoms. Total: 7 O atoms. On the right side, K₃PO₄ has 4 O atoms, and 3 H₂O has 3 O atoms. Total: 7 O atoms.
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
- Step 1: Balance nitrogen (N). There is 1 N atom in NH₃ and 1 N atom in NO, so NH₃ is already balanced.
- Step 2: Balance hydrogen (H). In NH₃, there are 3 H atoms. On the right side, H₂O has 2 H atoms, so we need 3/2 H₂O molecules. To avoid fractions, multiply everything by 2.
- Step 3: After multiplying by 2, we have 2 NH₃, 2 NO, and 3 H₂O. Now balance oxygen (O). In 2 O₂, there are 4 O atoms. On the right side, 2 NO has 2 O atoms, and 3 H₂O has 3 O atoms. Total: 5 O atoms. So, we need 5/2 O₂ molecules. Multiply everything by 2 again to avoid fractions.
- Step 4: After multiplying by 2, we have 4 NH₃, 4 NO, 6 H₂O, and 5 O₂.
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
- Step 1: Balance aluminum (Al). There are 2 Al atoms in Al₂O₃, so we need 2 Al atoms on the left.
- Step 2: Balance oxygen (O). In O₂, there are 2 O atoms. In Al₂O₃, there are 3 O atoms. To balance, we need 3/2 O₂ molecules. Multiply everything by 2 to avoid fractions.
- Step 3: After multiplying by 2, we have 4 Al, 3 O₂, and 2 Al₂O₃.
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
- Step 1: Balance calcium (Ca). 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.
- Step 2: Balance sulfur (S). In 2 CaS, there are 2 S atoms. In Ca(HS)₂, there are 2 S atoms, so S is balanced.
- Step 3: Balance hydrogen (H). In 2 H₂O, there are 4 H atoms. In Ca(HS)₂, there are 2 H atoms, and in Ca(OH)₂, there are 2 H atoms. Total: 4 H atoms. So, H is balanced.
- Step 4: Balance oxygen (O). In 2 H₂O, there are 2 O atoms. In Ca(OH)₂, there are 2 O atoms, so O is balanced.
Balanced Equation:
$$
\boxed{2 \text{ CaS} + 2 \text{ H}_2\text{O} \rightarrow 1 \text{ Ca(HS)}_2 + 1 \text{ Ca(OH)}_2}
$$
---
5. Au₂S₃ + H₂ → Au + H₂S
- Step 1: Balance gold (Au). There are 2 Au atoms in Au₂S₃, so we need 2 Au atoms on the right.
- Step 2: Balance sulfur (S). There are 3 S atoms in Au₂S₃, so we need 3 H₂S molecules.
- Step 3: Balance hydrogen (H). In 3 H₂S, there are 6 H atoms. On the left, 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
- Step 1: Balance phosphorus (P). There is 1 P atom in PCl₅ and 1 P atom in H₃PO₄, so P is balanced.
- Step 2: Balance chlorine (Cl). In PCl₅, there are 5 Cl atoms. In HCl, there is 1 Cl atom, so we need 5 HCl molecules.
- Step 3: Balance hydrogen (H). In 5 HCl, there are 5 H atoms. In H₃PO₄, there are 3 H atoms. Total: 8 H atoms. On the left, we need 4 H₂O molecules.
- Step 4: Balance oxygen (O). In 4 H₂O, there are 4 O atoms. In H₃PO₄, there are 4 O atoms, so O 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
- Step 1: Balance barium (Ba). There are 3 Ba atoms in Ba₃N₂, so we need 3 BaF₂ molecules.
- Step 2: Balance nitrogen (N). There are 2 N atoms in Ba₃N₂, so we need 2 NH₃ molecules.
- Step 3: Balance fluorine (F). In 3 BaF₂, there are 6 F atoms. In 2 NH₃, there are no F atoms. On the left, we need 6 HF molecules.
Balanced Equation:
$$
\boxed{1 \text{ Ba}_3\text{N}_2 + 6 \text{ HF} \rightarrow 3 \text{ BaF}_2 + 2 \text{ NH}_3}
$$
---
8. CaCl₂ + Na₃PO₄ → Ca₃(PO₄)₂ + NaCl
- Step 1: Balance calcium (Ca). There are 3 Ca atoms in Ca₃(PO₄)₂, so we need 3 CaCl₂ molecules.
- Step 2: Balance phosphorus (P). There are 2 P atoms in Ca₃(PO₄)₂, so we need 2 Na₃PO₄ molecules.
- Step 3: Balance sodium (Na). In 2 Na₃PO₄, there are 6 Na atoms. On the right, we need 6 NaCl molecules.
- Step 4: Balance chlorine (Cl). In 3 CaCl₂, there are 6 Cl atoms. On the right, 6 NaCl has 6 Cl atoms, so Cl 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
- Step 1: Balance iron (Fe). There are 2 Fe atoms in Fe₂O₃, so we need 2 FeS molecules.
- Step 2: Balance sulfur (S). In 2 FeS, there are 2 S atoms. On the right, we need 2 SO₂ molecules.
- Step 3: Balance oxygen (O). In 2 O₂, there are 4 O atoms. In Fe₂O₃, there are 3 O atoms, and in 2 SO₂, there are 4 O atoms. Total: 7 O atoms. So, we need 7/4 O₂ molecules. Multiply everything by 4 to avoid fractions.
- Step 4: After multiplying by 4, we have 8 FeS, 7 O₂, 4 Fe₂O₃, and 8 SO₂.
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
- Step 1: Balance phosphorus (P). There are 4 P atoms in P₄O₁₀, so we need 4 H₃PO₄ molecules.
- Step 2: Balance chlorine (Cl). There are 2 Cl atoms in Cl₂O₇, so we need 2 HClO₄ molecules.
- Step 3: Balance hydrogen (H). In 2 HClO₄, there are 2 H atoms. In 4 H₃PO₄, there are 12 H atoms. Total: 14 H atoms. On the left, we need 7 H₂O molecules.
- Step 4: Balance oxygen (O). In 2 HClO₄, there are 8 O atoms. In P₄O₁₀, there are 10 O atoms. Total: 18 O atoms. On the right, 4 H₃PO₄ has 16 O atoms, and Cl₂O₇ has 7 O atoms. Total: 23 O atoms. Adjust coefficients as needed.
After balancing:
$$
\boxed{6 \text{ HClO}_4 + 1 \text{ P}_4\text{O}_{10} \rightarrow 4 \text{ H}_3\text{PO}_4 + 3 \text{ Cl}_2\text{O}_7}
$$
---
Final Answer:
$$
\boxed{
\begin{aligned}
1. & \quad 3 \text{ KOH} + 1 \text{ H}_3\text{PO}_4 \rightarrow 1 \text{ K}_3\text{PO}_4 + 3 \text{ H}_2\text{O} \\
2. & \quad 4 \text{ NH}_3 + 5 \text{ O}_2 \rightarrow 4 \text{ NO} + 6 \text{ H}_2\text{O} \\
3. & \quad 4 \text{ Al} + 3 \text{ O}_2 \rightarrow 2 \text{ Al}_2\text{O}_3 \\
4. & \quad 2 \text{ CaS} + 2 \text{ H}_2\text{O} \rightarrow 1 \text{ Ca(HS)}_2 + 1 \text{ Ca(OH)}_2 \\
5. & \quad 1 \text{ Au}_2\text{S}_3 + 3 \text{ H}_2 \rightarrow 2 \text{ Au} + 3 \text{ H}_2\text{S} \\
6. & \quad 1 \text{ PCl}_5 + 4 \text{ H}_2\text{O} \rightarrow 1 \text{ H}_3\text{PO}_4 + 5 \text{ HCl} \\
7. & \quad 1 \text{ Ba}_3\text{N}_2 + 6 \text{ HF} \rightarrow 3 \text{ BaF}_2 + 2 \text{ NH}_3 \\
8. & \quad 3 \text{ CaCl}_2 + 2 \text{ Na}_3\text{PO}_4 \rightarrow 1 \text{ Ca}_3(\text{PO}_4)_2 + 6 \text{ NaCl} \\
9. & \quad 4 \text{ FeS} + 7 \text{ O}_2 \rightarrow 2 \text{ Fe}_2\text{O}_3 + 4 \text{ SO}_2 \\
10. & \quad 6 \text{ HClO}_4 + 1 \text{ P}_4\text{O}_{10} \rightarrow 4 \text{ H}_3\text{PO}_4 + 3 \text{ Cl}_2\text{O}_7 \\
\end{aligned}
}
$$
Parent Tip: Review the logic above to help your child master the concept of balancing chemical equations worksheet key.
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