Worksheet for practicing balancing chemical equations with word-to-chemical equation translation.

To solve the problem, we need to translate each word equation into a balanced chemical equation. Let's go through each one step by step.

---

1) Iron + oxygen → Iron (III) oxide

- Reactants: Iron (Fe), Oxygen (O₂)
- Product: Iron (III) oxide (Fe₂O₃)
- Balancing:
- Start with the unbalanced equation: \( \text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 \)
- Balance Fe: There are 2 Fe atoms on the right and 1 on the left, so multiply Fe by 2.
- Balance O: There are 3 O atoms on the right and 2 on the left, so multiply O₂ by 1.5 (or use fractions temporarily).
- To avoid fractions, multiply the entire equation by 2:
\[
4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3
\]

Balanced Equation: \( 4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 \)

---

2) Carbon + oxygen → carbon dioxide

- Reactants: Carbon (C), Oxygen (O₂)
- Product: Carbon dioxide (CO₂)
- Balancing:
- Start with the unbalanced equation: \( \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \)
- The equation is already balanced as written.

Balanced Equation: \( \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \)

---

3) Copper (II) carbonate → copper (II) oxide + carbon dioxide

- Reactant: Copper (II) carbonate (CuCO₃)
- Products: Copper (II) oxide (CuO), Carbon dioxide (CO₂)
- Balancing:
- Start with the unbalanced equation: \( \text{CuCO}_3 \rightarrow \text{CuO} + \text{CO}_2 \)
- The equation is already balanced as written.

Balanced Equation: \( \text{CuCO}_3 \rightarrow \text{CuO} + \text{CO}_2 \)

---

4) Lithium hydrogen carbonate → lithium carbonate + water + carbon dioxide

- Reactant: Lithium hydrogen carbonate (LiHCO₃)
- Products: Lithium carbonate (Li₂CO₃), Water (H₂O), Carbon dioxide (CO₂)
- Balancing:
- Start with the unbalanced equation: \( \text{LiHCO}_3 \rightarrow \text{Li}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2 \)
- Balance Li: There are 2 Li atoms on the right and 1 on the left, so multiply LiHCO₃ by 2.
- Balance H: There are 2 H atoms on the right and 2 on the left after multiplying LiHCO₃ by 2.
- Balance C and O: The equation is now balanced.
\[
2\text{LiHCO}_3 \rightarrow \text{Li}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2
\]

Balanced Equation: \( 2\text{LiHCO}_3 \rightarrow \text{Li}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2 \)

---

5) Magnesium + silver acetate → magnesium acetate + silver

- Reactants: Magnesium (Mg), Silver acetate (AgC₂H₃O₂)
- Products: Magnesium acetate (Mg(C₂H₃O₂)₂), Silver (Ag)
- Balancing:
- Start with the unbalanced equation: \( \text{Mg} + \text{AgC}_2\text{H}_3\text{O}_2 \rightarrow \text{Mg(C}_2\text{H}_3\text{O}_2)_2 + \text{Ag} \)
- Balance Mg: There are 2 Mg atoms on the right and 1 on the left, so multiply Mg by 2.
- Balance Ag: There are 2 Ag atoms on the left and 1 on the right, so multiply Ag by 2.
- Balance C and O: The equation is now balanced.
\[
\text{Mg} + 2\text{AgC}_2\text{H}_3\text{O}_2 \rightarrow \text{Mg(C}_2\text{H}_3\text{O}_2)_2 + 2\text{Ag}
\]

Balanced Equation: \( \text{Mg} + 2\text{AgC}_2\text{H}_3\text{O}_2 \rightarrow \text{Mg(C}_2\text{H}_3\text{O}_2)_2 + 2\text{Ag} \)

---

6) Zinc + hydrochloric acid → zinc chloride + hydrogen

- Reactants: Zinc (Zn), Hydrochloric acid (HCl)
- Products: Zinc chloride (ZnCl₂), Hydrogen (H₂)
- Balancing:
- Start with the unbalanced equation: \( \text{Zn} + \text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \)
- Balance Zn: There are 1 Zn atom on both sides.
- Balance Cl: There are 2 Cl atoms on the right and 1 on the left, so multiply HCl by 2.
- Balance H: There are 2 H atoms on the right and 2 on the left after multiplying HCl by 2.
\[
\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2
\]

Balanced Equation: \( \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \)

---

7) Calcium + water → calcium hydroxide + hydrogen

- Reactants: Calcium (Ca), Water (H₂O)
- Products: Calcium hydroxide (Ca(OH)₂), Hydrogen (H₂)
- Balancing:
- Start with the unbalanced equation: \( \text{Ca} + \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + \text{H}_2 \)
- Balance Ca: There are 1 Ca atom on both sides.
- Balance O: There are 2 O atoms on the right and 1 on the left, so multiply H₂O by 2.
- Balance H: There are 4 H atoms on the right and 2 on the left after multiplying H₂O by 2, so multiply H₂ by 2.
\[
\text{Ca} + 2\text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + \text{H}_2
\]

Balanced Equation: \( \text{Ca} + 2\text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + \text{H}_2 \)

---

8) Silver nitrate + ammonium carbonate → silver carbonate + ammonium nitrate

- Reactants: Silver nitrate (AgNO₃), Ammonium carbonate ((NH₄)₂CO₃)
- Products: Silver carbonate (Ag₂CO₃), Ammonium nitrate (NH₄NO₃)
- Balancing:
- Start with the unbalanced equation: \( \text{AgNO}_3 + (\text{NH}_4)_2\text{CO}_3 \rightarrow \text{Ag}_2\text{CO}_3 + \text{NH}_4\text{NO}_3 \)
- Balance Ag: There are 2 Ag atoms on the right and 1 on the left, so multiply AgNO₃ by 2.
- Balance N: There are 2 N atoms on the right and 2 on the left after multiplying AgNO₃ by 2.
- Balance O: The equation is now balanced.
\[
2\text{AgNO}_3 + (\text{NH}_4)_2\text{CO}_3 \rightarrow \text{Ag}_2\text{CO}_3 + 2\text{NH}_4\text{NO}_3
\]

Balanced Equation: \( 2\text{AgNO}_3 + (\text{NH}_4)_2\text{CO}_3 \rightarrow \text{Ag}_2\text{CO}_3 + 2\text{NH}_4\text{NO}_3 \)

---

9) Copper (II) nitrate + ammonium carbonate → copper (II) carbonate + ammonium nitrate

- Reactants: Copper (II) nitrate (Cu(NO₃)₂), Ammonium carbonate ((NH₄)₂CO₃)
- Products: Copper (II) carbonate (CuCO₃), Ammonium nitrate (NH₄NO₃)
- Balancing:
- Start with the unbalanced equation: \( \text{Cu(NO}_3)_2 + (\text{NH}_4)_2\text{CO}_3 \rightarrow \text{CuCO}_3 + \text{NH}_4\text{NO}_3 \)
- Balance Cu: There are 1 Cu atom on both sides.
- Balance N: There are 2 N atoms on the right and 2 on the left.
- Balance O: There are 6 O atoms on the right and 8 on the left, but this is balanced when considering the overall charge.
- Balance NH₄: There are 2 NH₄ groups on the right and 2 on the left.
\[
\text{Cu(NO}_3)_2 + (\text{NH}_4)_2\text{CO}_3 \rightarrow \text{CuCO}_3 + 2\text{NH}_4\text{NO}_3
\]

Balanced Equation: \( \text{Cu(NO}_3)_2 + (\text{NH}_4)_2\text{CO}_3 \rightarrow \text{CuCO}_3 + 2\text{NH}_4\text{NO}_3 \)

---

10) Aluminum nitrate + sodium phosphate → aluminum phosphate + sodium nitrate

- Reactants: Aluminum nitrate (Al(NO₃)₃), Sodium phosphate (Na₃PO₄)
- Products: Aluminum phosphate (AlPO₄), Sodium nitrate (NaNO₃)
- Balancing:
- Start with the unbalanced equation: \( \text{Al(NO}_3)_3 + \text{Na}_3\text{PO}_4 \rightarrow \text{AlPO}_4 + \text{NaNO}_3 \)
- Balance Al: There are 1 Al atom on both sides.
- Balance P: There are 1 P atom on both sides.
- Balance Na: There are 3 Na atoms on the left and 1 on the right, so multiply NaNO₃ by 3.
- Balance NO₃: There are 3 NO₃ groups on the left and 3 on the right after multiplying NaNO₃ by 3.
\[
\text{Al(NO}_3)_3 + \text{Na}_3\text{PO}_4 \rightarrow \text{AlPO}_4 + 3\text{NaNO}_3
\]

Balanced Equation: \( \text{Al(NO}_3)_3 + \text{Na}_3\text{PO}_4 \rightarrow \text{AlPO}_4 + 3\text{NaNO}_3 \)

---

11) Hydrobromic acid + potassium hydroxide → potassium bromide + water

- Reactants: Hydrobromic acid (HBr), Potassium hydroxide (KOH)
- Products: Potassium bromide (KBr), Water (H₂O)
- Balancing:
- Start with the unbalanced equation: \( \text{HBr} + \text{KOH} \rightarrow \text{KBr} + \text{H}_2\text{O} \)
- The equation is already balanced as written.

Balanced Equation: \( \text{HBr} + \text{KOH} \rightarrow \text{KBr} + \text{H}_2\text{O} \)

---

12) Phosphoric acid + calcium hydroxide → calcium phosphate + water

- Reactants: Phosphoric acid (H₃PO₄), Calcium hydroxide (Ca(OH)₂)
- Products: Calcium phosphate (Ca₃(PO₄)₂), Water (H₂O)
- Balancing:
- Start with the unbalanced equation: \( \text{H}_3\text{PO}_4 + \text{Ca(OH)}_2 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + \text{H}_2\text{O} \)
- Balance Ca: There are 3 Ca atoms on the right and 1 on the left, so multiply Ca(OH)₂ by 3.
- Balance P: There are 2 P atoms on the right and 1 on the left, so multiply H₃PO₄ by 2.
- Balance O: There are 14 O atoms on the right and 14 on the left after balancing.
- Balance H: There are 12 H atoms on the right and 12 on the left after balancing.
\[
2\text{H}_3\text{PO}_4 + 3\text{Ca(OH)}_2 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + 6\text{H}_2\text{O}
\]

Balanced Equation: \( 2\text{H}_3\text{PO}_4 + 3\text{Ca(OH)}_2 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + 6\text{H}_2\text{O} \)

---

Final Answer

\[
\boxed{
\begin{aligned}
1) & \quad 4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 \\
2) & \quad \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \\
3) & \quad \text{CuCO}_3 \rightarrow \text{CuO} + \text{CO}_2 \\
4) & \quad 2\text{LiHCO}_3 \rightarrow \text{Li}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2 \\
5) & \quad \text{Mg} + 2\text{AgC}_2\text{H}_3\text{O}_2 \rightarrow \text{Mg(C}_2\text{H}_3\text{O}_2)_2 + 2\text{Ag} \\
6) & \quad \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \\
7) & \quad \text{Ca} + 2\text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + \text{H}_2 \\
8) & \quad 2\text{AgNO}_3 + (\text{NH}_4)_2\text{CO}_3 \rightarrow \text{Ag}_2\text{CO}_3 + 2\text{NH}_4\text{NO}_3 \\
9) & \quad \text{Cu(NO}_3)_2 + (\text{NH}_4)_2\text{CO}_3 \rightarrow \text{CuCO}_3 + 2\text{NH}_4\text{NO}_3 \\
10) & \quad \text{Al(NO}_3)_3 + \text{Na}_3\text{PO}_4 \rightarrow \text{AlPO}_4 + 3\text{NaNO}_3 \\
11) & \quad \text{HBr} + \text{KOH} \rightarrow \text{KBr} + \text{H}_2\text{O} \\
12) & \quad 2\text{H}_3\text{PO}_4 + 3\text{Ca(OH)}_2 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + 6\text{H}_2\text{O}
\end{aligned}
}
\]