Worksheet for practicing balancing chemical equations, featuring 10 problems involving various elements and compounds.

Let’s go through each equation one by one and balance them. Balancing means making sure the number of atoms of each element is the same on both sides of the arrow.

We’ll use coefficients (numbers in front of molecules) to balance — never change the subscripts!

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1. H₂ + O₂ → H₂O

Left: 2 H, 2 O
Right: 2 H, 1 O → Need more O on right

Try putting 2 in front of H₂O:
H₂ + O₂ → 2H₂O → Now right has 4 H, 2 O
Left has 2 H → Put 2 in front of H₂:
2H₂ + O₂ → 2H₂O ✔

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2. Mg + O₂ → MgO

Left: 1 Mg, 2 O
Right: 1 Mg, 1 O → Need 2 O on right → put 2 in front of MgO:
Mg + O₂ → 2MgO → Now right has 2 Mg, 2 O
Left has 1 Mg → put 2 in front of Mg:
2Mg + O₂ → 2MgO ✔

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3. Li + F₂ → LiF

Left: 1 Li, 2 F
Right: 1 Li, 1 F → Need 2 F on right → put 2 in front of LiF:
Li + F₂ → 2LiF → Now right has 2 Li, 2 F
Left has 1 Li → put 2 in front of Li:
2Li + F₂ → 2LiF ✔

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4. K + O₂ → K₂O

Left: 1 K, 2 O
Right: 2 K, 1 O → Need 2 O on right → put 2 in front of K₂O:
K + O₂ → 2K₂O → Now right has 4 K, 2 O
Left has 1 K → put 4 in front of K:
4K + O₂ → 2K₂O ✔

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5. H₂O₂ → H₂O + O₂

Left: 2 H, 2 O
Right: 2 H, 1 O (from H₂O) + 2 O (from O₂) = 3 O total → Not balanced

Try 2H₂O₂ → ?
Left: 4 H, 4 O
Now try 2H₂O + O₂ → Right: 4 H, 2 O (from water) + 2 O (from O₂) = 4 O → Perfect!
2H₂O₂ → 2H₂O + O₂ ✔

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6. Al + Cl₂ → AlCl₃

Left: 1 Al, 2 Cl
Right: 1 Al, 3 Cl → Need common multiple for Cl → LCM of 2 and 3 is 6

So make 6 Cl on both sides:
Put 3 in front of Cl₂ → 3Cl₂ = 6 Cl
Put 2 in front of AlCl₃ → 2AlCl₃ = 2 Al, 6 Cl
Now left needs 2 Al → put 2 in front of Al:
2Al + 3Cl₂ → 2AlCl₃ ✔

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7. Ag₂O → Ag + O₂

Left: 2 Ag, 1 O
Right: 1 Ag, 2 O → Need even oxygen on left → put 2 in front of Ag₂O:
2Ag₂O → ? → Left: 4 Ag, 2 O
Right: need 4 Ag → put 4 in front of Ag
Need 2 O → already have O₂ → so:
2Ag₂O → 4Ag + O₂ ✔

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8. H₂ + N₂ → NH₃

Left: 2 H, 2 N
Right: 1 N, 3 H → Need common multiples

Make 6 H on both sides:
Put 3 in front of H₂ → 3H₂ = 6 H
Put 2 in front of NH₃ → 2NH₃ = 2 N, 6 H
Now left has 2 N → already have N₂ → perfect:
3H₂ + N₂ → 2NH₃ ✔

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9. Ca + H₂O → Ca(OH)₂ + H₂

Left: 1 Ca, 2 H, 1 O
Right: 1 Ca, 2 O, 2 H (from OH) + 2 H (from H₂) = 4 H, 2 O → Too many H and O on right

Wait — Ca(OH)₂ has 2 O and 2 H from OH, plus H₂ has 2 H → total 4 H and 2 O

Left only has 2 H and 1 O → so we need 2 H₂O on left:

Ca + 2H₂O → Ca(OH)₂ + H₂
Left: 1 Ca, 4 H, 2 O
Right: 1 Ca, 2 O, 2 H (from OH) + 2 H (from H₂) = 4 H → Balanced!
Ca + 2H₂O → Ca(OH)₂ + H₂ ✔

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10. SeCl₆ + O₂ → SeO₂ + Cl₂

Left: 1 Se, 6 Cl, 2 O
Right: 1 Se, 2 O, 2 Cl → Not balanced

Se is good. Cl: left 6, right 2 → put 3 in front of Cl₂ → 3Cl₂ = 6 Cl
O: left 2, right 2 → good? Wait — SeO₂ has 2 O, O₂ has 2 O → but we’re using O₂ as reactant.

Actually, let’s count again:

Left: SeCl₆ + O₂ → 1 Se, 6 Cl, 2 O
Right: SeO₂ + Cl₂ → 1 Se, 2 O, 2 Cl → not enough Cl or O?

Wait — if we put 3Cl₂ on right → 6 Cl
But O: right has 2 O (from SeO₂), left has 2 O (from O₂) → seems okay?

But wait — is that all? Let’s write:

SeCl₆ + O₂ → SeO₂ + 3Cl₂

Check atoms:

Left: Se=1, Cl=6, O=2
Right: Se=1, O=2, Cl=6 → YES! It’s balanced!

SeCl₆ + O₂ → SeO₂ + 3Cl₂ ✔

Wait — actually, let me double-check because sometimes oxygen might need adjustment.

In this case, it works as written. No coefficient needed for O₂ or SeO₂.

Yes — SeCl₆ + O₂ → SeO₂ + 3Cl₂ is balanced.

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Final Answer:

1. 2H₂ + O₂ → 2H₂O
2. 2Mg + O₂ → 2MgO
3. 2Li + F₂ → 2LiF
4. 4K + O₂ → 2K₂O
5. 2H₂O₂ → 2H₂O + O₂
6. 2Al + 3Cl₂ → 2AlCl₃
7. 2Ag₂O → 4Ag + O₂
8. 3H₂ + N₂ → 2NH₃
9. Ca + 2H₂O → Ca(OH)₂ + H₂
10. SeCl₆ + O₂ → SeO₂ + 3Cl₂