Worksheet for practicing balancing chemical equations with various reactants and products.

Let's solve each of these chemical equations by balancing them. Balancing means adjusting the coefficients (the numbers in front of the formulas) so that the number of atoms of each element is the same on both sides of the equation.

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1. \( \_\_ \text{H}_2 + \_\_ \text{O}_2 \rightarrow \_\_ \text{H}_2\text{O} \)

- Left: H = 2, O = 2
- Right: H = 2, O = 1 → Not balanced

To balance oxygen: Put 2 in front of H₂O → 2 H₂O
Now right has: H = 4, O = 2
So left needs 2 H₂ and 1 O₂

✔ Balanced:
2 H₂ + 1 O₂ → 2 H₂O

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2. \( \_\_ \text{Mg} + \_\_ \text{O}_2 \rightarrow \_\_ \text{MgO} \)

- Left: Mg = 1, O = 2
- Right: Mg = 1, O = 1 → Not balanced

Need 2 MgO to balance oxygen → 2 MgO
Then need 2 Mg on left

✔ Balanced:
2 Mg + 1 O₂ → 2 MgO

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3. \( \_\_ \text{U} + \_\_ \text{F}_2 \rightarrow \_\_ \text{UF}_4 \)

- Uranium fluoride is UF₄ (fluorine forms F⁻, U⁴⁺)
- Left: U = 1, F = 2
- Right: U = 1, F = 4 → Need 2 F₂ to get 4 F atoms

So:
1 U + 2 F₂ → 1 UF₄

✔ Balanced:
1 U + 2 F₂ → 1 UF₄

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4. \( \_\_ \text{K} + \_\_ \text{O}_2 \rightarrow \_\_ \text{K}_2\text{O} \)

- Potassium oxide is K₂O
- Left: K = 1, O = 2
- Right: K = 2, O = 1

Need 2 K on left → 2 K
But then O₂ gives 2 O atoms → need 2 K₂O to use up 2 O atoms → 2 K₂O requires 4 K

So:
4 K + 1 O₂ → 2 K₂O

✔ Balanced:
4 K + 1 O₂ → 2 K₂O

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5. \( \_\_ \text{H}_2\text{O}_2 \rightarrow \_\_ \text{H}_2\text{O} + \_\_ \text{O}_2 \)

Decomposition of hydrogen peroxide:

- Left: H = 2, O = 2
- Right: H₂O has H=2, O=1; O₂ has O=2 → total O = 3? No — unbalanced

Try:
2 H₂O₂ → 2 H₂O + 1 O₂

Left: H = 4, O = 4
Right: 2 H₂O → H=4, O=2; O₂ → O=2 → Total O = 4 ✔

✔ Balanced:
2 H₂O₂ → 2 H₂O + 1 O₂

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6. \( \_\_ \text{Al} + \_\_ \text{Cl}_2 \rightarrow \_\_ \text{AlCl}_3 \)

- Aluminum chloride is AlCl₃
- Left: Al = 1, Cl = 2
- Right: Al = 1, Cl = 3

LCM of 2 and 3 is 6 → need 3 Cl₂ and 2 AlCl₃

So:
2 Al + 3 Cl₂ → 2 AlCl₃

✔ Balanced:
2 Al + 3 Cl₂ → 2 AlCl₃

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7. \( \_\_ \text{Ag}_2\text{O} \rightarrow \_\_ \text{Ag} + \_\_ \text{O}_2 \)

Decomposition of silver oxide:

- Left: Ag = 2, O = 1
- Right: Ag = 1, O = 2

Need 2 Ag₂O → 4 Ag and 1 O₂

So:
2 Ag₂O → 4 Ag + 1 O₂

✔ Balanced:
2 Ag₂O → 4 Ag + 1 O₂

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8. \( \_\_ \text{H}_2 + \_\_ \text{N}_2 \rightarrow \_\_ \text{NH}_3 \)

- Left: H = 2, N = 2
- Right: NH₃ → N=1, H=3

LCM of H and N: Use 3 H₂ and 1 N₂ → 6 H, 2 N → make 2 NH₃

So:
3 H₂ + 1 N₂ → 2 NH₃

✔ Balanced:
3 H₂ + 1 N₂ → 2 NH₃

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9. \( \_\_ \text{Ca} + \_\_ \text{H}_2\text{O} \rightarrow \_\_ \text{Ca(OH)}_2 + \_\_ \text{H}_2 \)

Calcium reacts with water:

- Left: Ca = 1, H = 2, O = 1
- Right: Ca(OH)₂ → Ca=1, O=2, H=2; H₂ → H=2 → total H=4

So we need more water.

Try:
1 Ca + 2 H₂O → 1 Ca(OH)₂ + 1 H₂

Left: Ca=1, H=4, O=2
Right: Ca=1, O=2, H=2 (in OH) + 2 (in H₂) = 4 ✔

✔ Balanced:
1 Ca + 2 H₂O → 1 Ca(OH)₂ + 1 H₂

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10. \( \_\_ \text{SeCl}_4 + \_\_ \text{O}_2 \rightarrow \_\_ \text{SeO}_2 + \_\_ \text{Cl}_2 \)

Selenium tetrachloride reacts with oxygen:

- Left: Se = 1, Cl = 4, O = 2
- Right: Se = 1, O = 2, Cl = 2 → not balanced

We need 2 Cl₂ to account for 4 Cl atoms → so 2 Cl₂

So:
1 SeCl₄ + 1 O₂ → 1 SeO₂ + 2 Cl₂

Check atoms:
- Left: Se=1, Cl=4, O=2
- Right: Se=1, O=2, Cl=4 ✔

✔ Balanced:
1 SeCl₄ + 1 O₂ → 1 SeO₂ + 2 Cl₂

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✔ Final Balanced Equations:

1. \( 2 \text{H}_2 + 1 \text{O}_2 \rightarrow 2 \text{H}_2\text{O} \)
2. \( 2 \text{Mg} + 1 \text{O}_2 \rightarrow 2 \text{MgO} \)
3. \( 1 \text{U} + 2 \text{F}_2 \rightarrow 1 \text{UF}_4 \)
4. \( 4 \text{K} + 1 \text{O}_2 \rightarrow 2 \text{K}_2\text{O} \)
5. \( 2 \text{H}_2\text{O}_2 \rightarrow 2 \text{H}_2\text{O} + 1 \text{O}_2 \)
6. \( 2 \text{Al} + 3 \text{Cl}_2 \rightarrow 2 \text{AlCl}_3 \)
7. \( 2 \text{Ag}_2\text{O} \rightarrow 4 \text{Ag} + 1 \text{O}_2 \)
8. \( 3 \text{H}_2 + 1 \text{N}_2 \rightarrow 2 \text{NH}_3 \)
9. \( 1 \text{Ca} + 2 \text{H}_2\text{O} \rightarrow 1 \text{Ca(OH)}_2 + 1 \text{H}_2 \)
10. \( 1 \text{SeCl}_4 + 1 \text{O}_2 \rightarrow 1 \text{SeO}_2 + 2 \text{Cl}_2 \)

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