Free Printable Writing and Balancing Chemical Equations Worksheets - Free Printable

Let's go through each of the chemical reactions described in your worksheet. For each, I will:

1. Write the correct chemical formulas for the reactants and products.
2. Balance the chemical equation using appropriate coefficients.

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1. Nitric oxide reacts with ozone to produce nitrogen dioxide and oxygen gas.

- Reactants: Nitric oxide (NO), Ozone (O₃)
- Products: Nitrogen dioxide (NO₂), Oxygen gas (O₂)

Unbalanced Equation:
\[
\text{NO} + \text{O}_3 \rightarrow \text{NO}_2 + \text{O}_2
\]

Now balance:
- N: 1 on both sides ✔
- O: Left = 1 (from NO) + 3 (from O₃) = 4; Right = 2 (NO₂) + 2 (O₂) = 4 → already balanced

But let’s check atom counts carefully:
- Left: N=1, O=1+3=4
- Right: N=1, O=2+2=4

✔ Already balanced!

Balanced Equation:
\[
\boxed{\text{NO} + \text{O}_3 \rightarrow \text{NO}_2 + \text{O}_2}
\]

---

2. Iron burns in air to form a black solid Fe₃O₄.

- Reactant: Iron (Fe), Oxygen gas (O₂)
- Product: Iron(II,III) oxide (Fe₃O₄)

Unbalanced Equation:
\[
\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_3\text{O}_4
\]

Balance:
- Fe: 3 on right → need 3 Fe on left
- O: 4 on right → but O₂ has 2 atoms → need 2 O₂ molecules (4 O atoms)

So:
\[
3\text{Fe} + 2\text{O}_2 \rightarrow \text{Fe}_3\text{O}_4
\]

Check:
- Fe: 3 = 3 ✔
- O: 2×2 = 4 = 4 ✔

Balanced Equation:
\[
\boxed{3\text{Fe} + 2\text{O}_2 \rightarrow \text{Fe}_3\text{O}_4}
\]

---

3. Sodium metal reacts with chlorine gas to form sodium chloride.

- Reactants: Sodium (Na), Chlorine gas (Cl₂)
- Product: Sodium chloride (NaCl)

Unbalanced Equation:
\[
\text{Na} + \text{Cl}_2 \rightarrow \text{NaCl}
\]

Balance:
- Cl: 2 on left → need 2 NaCl on right
- So need 2 Na on left

\[
2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl}
\]

Check:
- Na: 2 = 2 ✔
- Cl: 2 = 2 ✔

Balanced Equation:
\[
\boxed{2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl}}
\]

---

4. Acetylene burns in air to form carbon dioxide and water.

- Acetylene: C₂H₂
- Air provides O₂
- Products: CO₂ and H₂O

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

Balance:
- C: 2 on left → need 2 CO₂ on right
- H: 2 on left → need 1 H₂O? No: H₂O has 2 H → so 1 H₂O is fine, but we have 2 H → one H₂O gives 2 H → OK
But wait: C₂H₂ has 2 H → one H₂O has 2 H → so one H₂O is enough?

Wait — actually: C₂H₂ has 2 hydrogen atoms, so one H₂O molecule has 2 H → that’s good.

But now:
- C: 2 → 2 CO₂ ✔
- H: 2 → 1 H₂O ✔
- O: Right side: 2×2 (from CO₂) + 1 (from H₂O) = 5 O atoms
- Left: O₂ → so need 5/2 O₂ → multiply entire equation by 2 to eliminate fraction

Start over:

\[
\text{C}_2\text{H}_2 + \text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O}
\]

Now:
- C: 2 = 2 ✔
- H: 2 = 2 ✔
- O: Right: 2×2 + 1 = 5; Left: 2 from O₂ → so need 5/2 O₂

So:
\[
\text{C}_2\text{H}_2 + \frac{5}{2}\text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O}
\]

Multiply all by 2:
\[
2\text{C}_2\text{H}_2 + 5\text{O}_2 \rightarrow 4\text{CO}_2 + 2\text{H}_2\text{O}
\]

Check:
- C: 4 = 4 ✔
- H: 4 = 4 ✔
- O: 5×2 = 10; Right: 4×2 + 2×1 = 8 + 2 = 10 ✔

Balanced Equation:
\[
\boxed{2\text{C}_2\text{H}_2 + 5\text{O}_2 \rightarrow 4\text{CO}_2 + 2\text{H}_2\text{O}}
\]

---

5. Hydrogen peroxide easily decomposes into water and oxygen gas.

- Reactant: H₂O₂
- Products: H₂O and O₂

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

Balance:
- H: 2 on both sides → okay
- O: Left: 2 → Right: 1 (in H₂O) + 2 (in O₂) = 3 → not balanced

Try:
- 2 H₂O₂ → 2 H₂O + O₂

Check:
- H: 4 = 4 ✔
- O: 4 = 2 (in 2 H₂O) + 2 (in O₂) = 4 ✔

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

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6. Hydrazine and hydrogen peroxide are used together as rocket fuel. The products are nitrogen gas and water.

- Hydrazine: N₂H₄
- Hydrogen peroxide: H₂O₂
- Products: N₂ (gas), H₂O (liquid)

Unbalanced Equation:
\[
\text{N}_2\text{H}_4 + \text{H}_2\text{O}_2 \rightarrow \text{N}_2 + \text{H}_2\text{O}
\]

Balance:
- N: 2 on both sides ✔
- H: Left: 4 (N₂H₄) + 2 (H₂O₂) = 6 H → Right: 2 H in H₂O → need 3 H₂O
- O: Left: 2 (from H₂O₂) → Right: 3 H₂O → 3 O → too many

So:
Try:
\[
\text{N}_2\text{H}_4 + 2\text{H}_2\text{O}_2 \rightarrow \text{N}_2 + 4\text{H}_2\text{O}
\]

Check:
- N: 2 = 2 ✔
- H: 4 (N₂H₄) + 4 (2×H₂O₂) = 8 H → Right: 4×2 = 8 H ✔
- O: 2×2 = 4 → Right: 4×1 = 4 ✔

Perfect.

Balanced Equation:
\[
\boxed{\text{N}_2\text{H}_4 + 2\text{H}_2\text{O}_2 \rightarrow \text{N}_2 + 4\text{H}_2\text{O}}
\]

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7. If strongly heated, potassium chlorate decomposes to yield oxygen gas and potassium chloride.

- Potassium chlorate: KClO₃
- Products: KCl and O₂

Unbalanced Equation:
\[
\text{KClO}_3 \rightarrow \text{KCl} + \text{O}_2
\]

Balance:
- K: 1 = 1 ✔
- Cl: 1 = 1 ✔
- O: 3 on left → O₂ has 2 → need multiple

Try 2 KClO₃ → 2 KCl + 3 O₂? Check O: 2×3 = 6 → 3×2 = 6 ✔

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

Check:
- K: 2 = 2 ✔
- Cl: 2 = 2 ✔
- O: 6 = 6 ✔

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

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8. Aqueous sodium hydroxide and gaseous carbon dioxide yield a solution of sodium carbonate and liquid water.

- Reactants: NaOH (aq), CO₂ (g)
- Products: Na₂CO₃ (aq), H₂O (l)

Unbalanced Equation:
\[
\text{NaOH} + \text{CO}_2 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O}
\]

Balance:
- Na: 1 on left, 2 on right → need 2 NaOH
- O and H will follow

Try:
\[
2\text{NaOH} + \text{CO}_2 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O}
\]

Check:
- Na: 2 = 2 ✔
- O: Left: 2 (from NaOH) + 2 (from CO₂) = 4 → Right: 3 (Na₂CO₃) + 1 (H₂O) = 4 ✔
- H: 2 = 2 ✔
- C: 1 = 1 ✔

✔ Balanced.

Balanced Equation:
\[
\boxed{2\text{NaOH} + \text{CO}_2 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O}}
\]

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✔ Final Answers Summary:

1. \( \text{NO} + \text{O}_3 \rightarrow \text{NO}_2 + \text{O}_2 \)
2. \( 3\text{Fe} + 2\text{O}_2 \rightarrow \text{Fe}_3\text{O}_4 \)
3. \( 2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl} \)
4. \( 2\text{C}_2\text{H}_2 + 5\text{O}_2 \rightarrow 4\text{CO}_2 + 2\text{H}_2\text{O} \)
5. \( 2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2 \)
6. \( \text{N}_2\text{H}_4 + 2\text{H}_2\text{O}_2 \rightarrow \text{N}_2 + 4\text{H}_2\text{O} \)
7. \( 2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2 \)
8. \( 2\text{NaOH} + \text{CO}_2 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} \)

Let me know if you'd like these equations written with states (s, l, g, aq)!