SOLUTION: Worksheet balancing equations - Studypool - Free Printable

Problem Overview:

The task involves writing and balancing chemical equations for various chemical reactions. Below, I will solve each part step by step.

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Part 1: Writing Chemical Equations

#### Problem 1:
Description: When lithium hydroxide pellets are added to a solution of sulfuric acid (H₂SO₄), lithium sulfate and water are formed.

- Reactants: Lithium hydroxide (LiOH) and sulfuric acid (H₂SO₄)
- Products: Lithium sulfate (Li₂SO₄) and water (H₂O)

Unbalanced Equation:
\[ \text{LiOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Li}_2\text{SO}_4 + \text{H}_2\text{O} \]

Balancing:
- Start with Li: There are 2 Li atoms in Li₂SO₄, so we need 2 LiOH.
- Balance H: With 2 LiOH, there are 2 H atoms on the left. H₂SO₄ provides 2 more H atoms, totaling 4 H atoms. The product side has 2 H atoms in H₂O, so we need 2 H₂O.
- Check O: On the left, there are 4 O atoms (1 from each LiOH and 4 from H₂SO₄). On the right, there are 4 O atoms (4 from Li₂SO₄ and 1 from each H₂O).

Balanced Equation:
\[ 2\text{LiOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Li}_2\text{SO}_4 + 2\text{H}_2\text{O} \]

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#### Problem 2:
Description: Magnesium reacts with sodium fluoride to produce magnesium fluoride and elemental sodium.

- Reactants: Magnesium (Mg) and sodium fluoride (NaF)
- Products: Magnesium fluoride (MgF₂) and sodium (Na)

Unbalanced Equation:
\[ \text{Mg} + \text{NaF} \rightarrow \text{MgF}_2 + \text{Na} \]

Balancing:
- Mg is already balanced with 1 atom on both sides.
- F: There are 2 F atoms in MgF₂, so we need 2 NaF.
- Na: With 2 NaF, there are 2 Na atoms, so we need 2 Na on the product side.

Balanced Equation:
\[ \text{Mg} + 2\text{NaF} \rightarrow \text{MgF}_2 + 2\text{Na} \]

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#### Problem 3:
Description: If a copper coil is placed into a solution of silver nitrate, silver crystals form and copper (I) nitrate is generated.

- Reactants: Copper (Cu) and silver nitrate (AgNO₃)
- Products: Silver (Ag) and copper (I) nitrate [Cu(NO₃)₂]

Unbalanced Equation:
\[ \text{Cu} + \text{AgNO}_3 \rightarrow \text{Ag} + \text{Cu(NO}_3)_2 \]

Balancing:
- Cu: There is 1 Cu atom on the left and 1 Cu atom in Cu(NO₃)₂ on the right.
- Ag: There is 1 Ag atom in AgNO₃ on the left and 1 Ag atom on the right.
- NO₃: There are 2 NO₃ groups in Cu(NO₃)₂ on the right, so we need 2 AgNO₃ on the left.

Balanced Equation:
\[ \text{Cu} + 2\text{AgNO}_3 \rightarrow 2\text{Ag} + \text{Cu(NO}_3)_2 \]

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#### Problem 4:
Description: When crystalline C₆H₁₂O₆ is burned in oxygen, carbon dioxide and water vapor are formed.

- Reactants: Glucose (C₆H₁₂O₆) and oxygen (O₂)
- Products: Carbon dioxide (CO₂) and water (H₂O)

Unbalanced Equation:
\[ \text{C}_6\text{H}_{12}\text{O}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]

Balancing:
- C: There are 6 C atoms in C₆H₁₂O₆, so we need 6 CO₂.
- H: There are 12 H atoms in C₆H₁₂O₆, so we need 6 H₂O.
- O: On the left, there are 6 O atoms in C₆H₁₂O₆. On the right, there are 12 O atoms in 6 CO₂ and 6 O atoms in 6 H₂O, totaling 18 O atoms. To balance, we need 6 O₂ molecules (12 O atoms).

Balanced Equation:
\[ \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O} \]

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#### Problem 5:
Description: Calcium carbonate combines with hydrochloric acid (HCl) to produce calcium chloride, water, and carbon dioxide gas.

- Reactants: Calcium carbonate (CaCO₃) and hydrochloric acid (HCl)
- Products: Calcium chloride (CaCl₂), water (H₂O), and carbon dioxide (CO₂)

Unbalanced Equation:
\[ \text{CaCO}_3 + \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \]

Balancing:
- Ca: There is 1 Ca atom in CaCO₃ and 1 Ca atom in CaCl₂.
- Cl: There are 2 Cl atoms in CaCl₂, so we need 2 HCl.
- H: With 2 HCl, there are 2 H atoms, so we need 1 H₂O.
- O: There are 3 O atoms in CaCO₃, 1 O atom in H₂O, and 2 O atoms in CO₂, which balances.

Balanced Equation:
\[ \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \]

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Part 2: Balancing Equations

#### Problem 1:
\[ \_ \text{N}_2 + \_ \text{H}_2 \rightarrow \_ \text{NH}_3 \]

- N: There are 2 N atoms in N₂, so we need 2 NH₃.
- H: With 2 NH₃, there are 6 H atoms, so we need 3 H₂.

Balanced Equation:
\[ \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \]

#### Problem 2:
\[ \_ \text{KClO}_3 \rightarrow \_ \text{KCl} + \_ \text{O}_2 \]

- K: There is 1 K atom in KClO₃, so we need 1 KCl.
- Cl: There is 1 Cl atom in KClO₃, so we need 1 KCl.
- O: There are 3 O atoms in KClO₃. In KCl, there are no O atoms, so all 3 O atoms go to O₂. We need 1.5 O₂, but since coefficients must be whole numbers, multiply everything by 2.

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

#### Problem 3:
\[ \_ \text{NaCl} + \_ \text{F}_2 \rightarrow \_ \text{NaF} + \_ \text{Cl}_2 \]

- Na: There is 1 Na atom in NaCl, so we need 1 NaF.
- Cl: There is 1 Cl atom in NaCl, so we need 1 Cl₂.
- F: There are 2 F atoms in F₂, so we need 2 NaF.

Balanced Equation:
\[ 2\text{NaCl} + \text{F}_2 \rightarrow 2\text{NaF} + \text{Cl}_2 \]

#### Problem 4:
\[ \_ \text{H}_2 + \_ \text{O}_2 \rightarrow \_ \text{H}_2\text{O} \]

- H: There are 2 H atoms in H₂, so we need 2 H₂O.
- O: With 2 H₂O, there are 2 O atoms, so we need 1 O₂.

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

#### Problem 5:
\[ \_ \text{Pb(OH)}_2 + \_ \text{HCl} \rightarrow \_ \text{H}_2\text{O} + \_ \text{PbCl}_2 \]

- Pb: There is 1 Pb atom in Pb(OH)₂, so we need 1 PbCl₂.
- OH: There are 2 OH groups in Pb(OH)₂, so we need 2 H₂O.
- H: With 2 H₂O, there are 4 H atoms, so we need 4 HCl.
- Cl: With 4 HCl, there are 4 Cl atoms, so we need 2 Cl atoms in PbCl₂.

Balanced Equation:
\[ \text{Pb(OH)}_2 + 4\text{HCl} \rightarrow 2\text{H}_2\text{O} + \text{PbCl}_2 \]

#### Problem 6:
\[ \_ \text{AlBr}_3 + \_ \text{K}_2\text{SO}_4 \rightarrow \_ \text{KBr} + \_ \text{Al}_2(\text{SO}_4)_3 \]

- Al: There is 1 Al atom in AlBr₃, so we need 2 AlBr₃ to get 2 Al atoms in Al₂(SO₄)₃.
- Br: With 2 AlBr₃, there are 6 Br atoms, so we need 6 KBr.
- K: There are 2 K atoms in K₂SO₄, so we need 6 K₂SO₄ to get 12 K atoms in 6 KBr.
- SO₄: With 6 K₂SO₄, there are 6 SO₄ groups, so we need 3 SO₄ groups in Al₂(SO₄)₃.

Balanced Equation:
\[ 2\text{AlBr}_3 + 3\text{K}_2\text{SO}_4 \rightarrow 6\text{KBr} + \text{Al}_2(\text{SO}_4)_3 \]

#### Problem 7:
\[ \_ \text{CH}_4 + \_ \text{O}_2 \rightarrow \_ \text{CO}_2 + \_ \text{H}_2\text{O} \]

- C: There is 1 C atom in CH₄, so we need 1 CO₂.
- H: There are 4 H atoms in CH₄, so we need 2 H₂O.
- O: With 1 CO₂, there are 2 O atoms, and with 2 H₂O, there are 2 O atoms, totaling 4 O atoms. We need 2 O₂.

Balanced Equation:
\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \]

#### Problem 8:
\[ \_ \text{C}_3\text{H}_8 + \_ \text{O}_2 \rightarrow \_ \text{CO}_2 + \_ \text{H}_2\text{O} \]

- C: There are 3 C atoms in C₃H₈, so we need 3 CO₂.
- H: There are 8 H atoms in C₃H₈, so we need 4 H₂O.
- O: With 3 CO₂, there are 6 O atoms, and with 4 H₂O, there are 4 O atoms, totaling 10 O atoms. We need 5 O₂.

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

#### Problem 9:
\[ \_ \text{C}_8\text{H}_{18} + \_ \text{O}_2 \rightarrow \_ \text{CO}_2 + \_ \text{H}_2\text{O} \]

- C: There are 8 C atoms in C₈H₁₈, so we need 8 CO₂.
- H: There are 18 H atoms in C₈H₁₈, so we need 9 H₂O.
- O: With 8 CO₂, there are 16 O atoms, and with 9 H₂O, there are 9 O atoms, totaling 25 O atoms. We need 12.5 O₂, but since coefficients must be whole numbers, multiply everything by 2.

Balanced Equation:
\[ 2\text{C}_8\text{H}_{18} + 25\text{O}_2 \rightarrow 16\text{CO}_2 + 18\text{H}_2\text{O} \]

#### Problem 10:
\[ \_ \text{FeCl}_3 + \_ \text{NaOH} \rightarrow \_ \text{Fe(OH)}_3 + \_ \text{NaCl} \]

- Fe: There is 1 Fe atom in FeCl₃, so we need 1 Fe(OH)₃.
- Cl: There are 3 Cl atoms in FeCl₃, so we need 3 NaCl.
- Na: With 3 NaCl, there are 3 Na atoms, so we need 3 NaOH.
- OH: With 3 NaOH, there are 3 OH groups, so we need 3 OH groups in Fe(OH)₃.

Balanced Equation:
\[ \text{FeCl}_3 + 3\text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3\text{NaCl} \]

#### Problem 11:
\[ \_ \text{P} + \_ \text{O}_2 \rightarrow \_ \text{P}_2\text{O}_5 \]

- P: There are 2 P atoms in P₂O₅, so we need 2 P.
- O: With 2 P₂O₅, there are 10 O atoms, so we need 5 O₂.

Balanced Equation:
\[ 4\text{P} + 5\text{O}_2 \rightarrow 2\text{P}_2\text{O}_5 \]

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Final Answer:

\[
\boxed{
\begin{aligned}
1. & \quad 2\text{LiOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Li}_2\text{SO}_4 + 2\text{H}_2\text{O} \\
2. & \quad \text{Mg} + 2\text{NaF} \rightarrow \text{MgF}_2 + 2\text{Na} \\
3. & \quad \text{Cu} + 2\text{AgNO}_3 \rightarrow 2\text{Ag} + \text{Cu(NO}_3)_2 \\
4. & \quad \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O} \\
5. & \quad \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \\
6. & \quad \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \\
7. & \quad 2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2 \\
8. & \quad 2\text{NaCl} + \text{F}_2 \rightarrow 2\text{NaF} + \text{Cl}_2 \\
9. & \quad 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \\
10. & \quad \text{Pb(OH)}_2 + 4\text{HCl} \rightarrow 2\text{H}_2\text{O} + \text{PbCl}_2 \\
11. & \quad 2\text{AlBr}_3 + 3\text{K}_2\text{SO}_4 \rightarrow 6\text{KBr} + \text{Al}_2(\text{SO}_4)_3 \\
12. & \quad \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \\
13. & \quad \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \\
14. & \quad 2\text{C}_8\text{H}_{18} + 25\text{O}_2 \rightarrow 16\text{CO}_2 + 18\text{H}_2\text{O} \\
15. & \quad \text{FeCl}_3 + 3\text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3\text{NaCl} \\
16. & \quad 4\text{P} + 5\text{O}_2 \rightarrow 2\text{P}_2\text{O}_5
\end{aligned}
}
\]