Solved Types of Reactions Worksheet Balance the following | Chegg.com - Free Printable

Let’s go through each reaction one by one. We’ll balance the equation first (make sure atoms on left = atoms on right), then figure out what type of reaction it is.

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1) NaBr + H₃PO₄ → Na₃PO₄ + HBr

Left: Na=1, Br=1, H=3, P=1, O=4
Right: Na=3, Br=1, H=1, P=1, O=4

We need 3 Na on left → put 3 in front of NaBr
Now Br = 3 on left → need 3 HBr on right
H on right now = 3 → matches left!

Balanced: 3NaBr + H₃PO₄ → Na₃PO₄ + 3HBr

Type: Two compounds swap partners → Double Replacement

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2) Ca(OH)₂ + Al₂(SO₄)₃ → CaSO₄ + Al(OH)₃

Left: Ca=1, O=2+12=14? Wait — let’s count carefully:

Ca(OH)₂: Ca=1, O=2, H=2
Al₂(SO₄)₃: Al=2, S=3, O=12
Total left: Ca=1, Al=2, S=3, O=14, H=2

Right: CaSO₄: Ca=1, S=1, O=4
Al(OH)₃: Al=1, O=3, H=3

To get 3 S on right → need 3 CaSO₄ → so Ca=3 on right → need 3 Ca(OH)₂ on left
Now left: Ca=3, O from hydroxide = 6, H=6
Al still 2 → need 2 Al(OH)₃ on right → that gives Al=2, O=6, H=6
Check oxygen: Left: from 3 Ca(OH)₂ → 6 O; from Al₂(SO₄)₃ → 12 O → total 18 O
Right: 3 CaSO₄ → 12 O; 2 Al(OH)₃ → 6 O → total 18 O ✔

Balanced: 3Ca(OH)₂ + Al₂(SO₄)₃ → 3CaSO₄ + 2Al(OH)₃

Type: Two compounds swap ions → Double Replacement

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3) Mg + Fe₂O₃ → Fe + MgO

Left: Mg=1, Fe=2, O=3
Right: Fe=1, Mg=1, O=1

Need 2 Fe on right → put 2 in front of Fe
Need 3 O on right → put 3 in front of MgO → now Mg=3 on right → need 3 Mg on left

Balanced: 3Mg + Fe₂O₃ → 2Fe + 3MgO

Type: One element replaces another in a compound → Single Replacement

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4) C₂H₄ + O₂ → CO₂ + H₂O

Left: C=2, H=4, O=2
Right: C=1, H=2, O=3

Put 2 in front of CO₂ → C=2
Put 2 in front of H₂O → H=4
Now O on right: 2×2 + 2×1 = 6 → need 3 O₂ on left (since O₂ has 2 O per molecule → 3×2=6)

Balanced: C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

Type: Hydrocarbon + oxygen → carbon dioxide + water → Combustion

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5) PbSO₄ → PbSO₃ + O₂

Left: Pb=1, S=1, O=4
Right: Pb=1, S=1, O=3+2=5 → too many O on right?

Wait — actually, this looks like decomposition. Let’s check if it balances as written.

PbSO₄ → PbSO₃ + ½O₂? But we want whole numbers.

Multiply everything by 2:

2PbSO₄ → 2PbSO₃ + O₂

Check: Left: Pb=2, S=2, O=8
Right: 2PbSO₃ → Pb=2, S=2, O=6; plus O₂ → O=2 → total O=8 ✔

Balanced: 2PbSO₄ → 2PbSO₃ + O₂

Type: One compound breaks into two or more → Decomposition

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6) NH₃ + I₂ → N₂I₆ + H₂

Left: N=1, H=3, I=2
Right: N=2, I=6, H=2

Need 2 N on left → 2NH₃ → now H=6 → need 3H₂ on right (since H₂ has 2 H → 3×2=6)
I on right = 6 → need 3I₂ on left (3×2=6)

Balanced: 2NH₃ + 3I₂ → N₂I₆ + 3H₂

Type: Two elements form a compound? Wait — actually, ammonia and iodine react to make nitrogen triiodide and hydrogen gas. This is a single replacement? Or redox? Actually, it’s often classified as Single Replacement because H is being replaced by I? Hmm… but better to think: NH₃ loses H, gains I? Actually, standard classification for this is Single Replacement — iodine replaces hydrogen.

But wait — some sources call this a “displacement” or even “redox”. For high school level, since one element (I₂) is replacing part of a compound, it’s usually called Single Replacement.

Actually, looking again: NH₃ + I₂ → NI₃ + H₂ — yes, iodine displaces hydrogen. So Single Replacement

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7) H₂O + SO₃ → H₂SO₄

Left: H=2, O=1+3=4, S=1
Right: H=2, S=1, O=4 → already balanced!

Balanced: H₂O + SO₃ → H₂SO₄

Type: Two substances combine to make one → Synthesis (or Combination)

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8) H₂SO₄ + NH₄OH → H₂O + (NH₄)₂SO₄

Left: H₂SO₄: H=2, S=1, O=4
NH₄OH: N=1, H=5, O=1 → total left: H=7, S=1, O=5, N=1

Right: H₂O: H=2, O=1
(NH₄)₂SO₄: N=2, H=8, S=1, O=4 → total right: H=10, O=5, S=1, N=2

Need 2 NH₄OH on left → now N=2, H from ammonium hydroxide = 10, plus H₂SO₄ has 2H → total H=12? Wait no:

Wait — NH₄OH is NH₄⁺ and OH⁻ → formula is correct.

If we put 2 NH₄OH:
Left: H₂SO₄ + 2NH₄OH → H=2 + 2*(4+1)=2+10=12? No:

NH₄OH: N=1, H=4+1=5, O=1 → so 2NH₄OH: N=2, H=10, O=2
Plus H₂SO₄: H=2, S=1, O=4 → total left: H=12, S=1, O=6, N=2

Right: (NH₄)₂SO₄: N=2, H=8, S=1, O=4
Plus H₂O: H=2, O=1 → total right: H=10, O=5 → not matching.

Ah — we need 2 H₂O on right!

Try: H₂SO₄ + 2NH₄OH → 2H₂O + (NH₄)₂SO₄

Left: H₂SO₄: H=2, S=1, O=4
2NH₄OH: N=2, H=10, O=2 → total H=12, O=6, S=1, N=2

Right: 2H₂O: H=4, O=2
(NH₄)₂SO₄: N=2, H=8, S=1, O=4 → total H=12, O=6, S=1, N=2 ✔

Balanced: H₂SO₄ + 2NH₄OH → 2H₂O + (NH₄)₂SO₄

Type: Acid + base → salt + water → Neutralization (which is a type of Double Replacement)

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Final Answers:

1) 3NaBr + H₃PO₄ → Na₃PO₄ + 3HBr — Double Replacement
2) 3Ca(OH)₂ + Al₂(SO₄)₃ → 3CaSO₄ + 2Al(OH)₃ — Double Replacement
3) 3Mg + Fe₂O₃ → 2Fe + 3MgO — Single Replacement
4) C₂H₄ + 3O₂ → 2CO₂ + 2H₂O — Combustion
5) 2PbSO₄ → 2PbSO₃ + O₂ — Decomposition
6) 2NH₃ + 3I₂ → N₂I₆ + 3H₂ — Single Replacement
7) H₂O + SO₃ → H₂SO₄ — Synthesis
8) H₂SO₄ + 2NH₄OH → 2H₂O + (NH₄)₂SO₄ — Neutralization (Double Replacement)

Note: For #6, some might argue it's not typical single replacement, but at high school level, it’s often grouped there. If your teacher says otherwise, adjust accordingly.

Final Answer:
1) 3, 1, 1, 3 — Double Replacement
2) 3, 1, 3, 2 — Double Replacement
3) 3, 1, 2, 3 — Single Replacement
4) 1, 3, 2, 2 — Combustion
5) 2, 2, 1 — Decomposition
6) 2, 3, 1, 3 — Single Replacement
7) 1, 1, 1 — Synthesis
8) 1, 2, 2, 1 — Neutralization (Double Replacement)