Chemistry worksheet focusing on balancing chemical reactions and identifying reaction types such as synthesis, decomposition, single-replacement, double-replacement, and combustion.

Let's solve this step by step.

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A. Balance the following chemical reactions:

We will balance each equation by ensuring the number of atoms of each element is equal on both sides.

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1) NaBr + Ca(OH)₂ → CaBr₂ + NaOH

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

Balance Br: 2 NaBr
Then Na becomes 2 → need 2 NaOH
Now H and O: 2 NaOH has 2 H and 2 O → matches left side (Ca(OH)₂ has 2 O and 2 H)

✔ Balanced:
2 NaBr + Ca(OH)₂ → CaBr₂ + 2 NaOH

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2) N₂ + H₂ → NH₃

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

Balance N: 2 NH₃ → now N=2, H=6
So H₂ must be 3 → 3 H₂

✔ Balanced:
N₂ + 3 H₂ → 2 NH₃

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3) NaCl + F₂ → NaF + Cl₂

- Left: Na=1, Cl=1, F=2
- Right: Na=1, F=1, Cl=2 → not balanced

Need 2 NaF → F=2, Na=2
So 2 NaCl → Na=2, Cl=2 → then Cl₂ = 1 molecule

✔ Balanced:
2 NaCl + F₂ → 2 NaF + Cl₂

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4) Pb(OH)₂ + HCl → PbCl₂ + H₂O

- Left: Pb=1, O=2, H=2+1=3, Cl=1
- Right: Pb=1, Cl=2, H=2, O=1 → not balanced

Need 2 HCl to get 2 Cl → 2 HCl
Now H: 2 from OH and 2 from HCl = 4 H; O: 2 from OH
Right: PbCl₂ has no H or O, H₂O has 2 H and 1 O → need 2 H₂O

So: 2 H₂O → 4 H and 2 O → matches left

✔ Balanced:
Pb(OH)₂ + 2 HCl → PbCl₂ + 2 H₂O

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5) CH₄ + O₂ → CO₂ + H₂O

- Left: C=1, H=4, O=2
- Right: C=1, O=2+1=3, H=2 → not balanced

H: need 2 H₂O → H=4 → good
O: 2 from CO₂ + 1 from H₂O ×2 = 2+2=4 O → so need 2 O₂

✔ Balanced:
CH₄ + 2 O₂ → CO₂ + 2 H₂O

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6) H₂SO₄ + B(OH)₃ → B₂(SO₄)₃ + H₂O

- Left: H=2, S=1, O=4, B=1, O=3, H=3 → total H=5, O=7, etc.
- Right: B=2, S=3, O=12 from SO₄, plus H₂O

Need 2 B → 2 B(OH)₃
Need 3 SO₄ → 3 H₂SO₄
Now check:

Left: 3 H₂SO₄ → H=6, S=3, O=12
2 B(OH)₃ → B=2, O=6, H=6 → total H=12, O=18

Right: B₂(SO₄)₃ → B=2, S=3, O=12
H₂O: need H=12 → 6 H₂O → O=6 → total O=12+6=18 → matches

✔ Balanced:
3 H₂SO₄ + 2 B(OH)₃ → B₂(SO₄)₃ + 6 H₂O

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7) C₅H₉O + O₂ → CO₂ + H₂O

This is a combustion reaction.

C₅H₉O → 5 C, 9 H, 1 O

Combustion: C → CO₂, H → H₂O

So: 5 CO₂ and 9/2 H₂O → but we want whole numbers.

Set:
C₅H₉O + O₂ → 5 CO₂ + (9/2) H₂O

Multiply entire equation by 2:

2 C₅H₉O + ? O₂ → 10 CO₂ + 9 H₂O

Now count O:

Left: 2×1 = 2 O from fuel, plus 2×O₂ = 2x O
Right: 10×2 = 20 O in CO₂, 9×1 = 9 O in H₂O → total 29 O

So: 2 + 2x = 29 → 2x = 27 → x = 13.5 → multiply again by 2

→ 4 C₅H₉O + 27 O₂ → 20 CO₂ + 18 H₂O

Check:
Left: C=20, H=36, O=4×1 + 27×2 = 4+54=58
Right: C=20, H=36, O=20×2 + 18×1 = 40+18=58 → ✔

✔ Balanced:
4 C₅H₉O + 27 O₂ → 20 CO₂ + 18 H₂O

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8) Li₃N + NH₄NO₃ → LiNO₃ + (NH₄)₃N

Look at products: LiNO₃ and (NH₄)₃N

Li₃N has 3 Li, 1 N
NH₄NO₃ has NH₄⁺ and NO₃⁻ → so it can provide NO₃⁻ and NH₄⁺

(NH₄)₃N is ammonium nitride → 3 NH₄⁺ and N³⁻

But Li₃N provides N³⁻, so likely:

Li₃N + NH₄NO₃ → LiNO₃ + (NH₄)₃N

Try balancing:

Suppose we want 3 LiNO₃ → need 3 Li and 3 NO₃

Li comes from Li₃N → 1 Li₃N gives 3 Li → good

So 1 Li₃N → 3 LiNO₃ → need 3 NO₃⁻ → 3 NH₄NO₃

Now: 3 NH₄NO₃ → 3 NH₄⁺ and 3 NO₃⁻

We already used 3 NO₃⁻ for LiNO₃

But we have 3 NH₄⁺ left → they can form (NH₄)₃N with the N³⁻ from Li₃N?

Wait: Li₃N has one N³⁻, but (NH₄)₃N needs one N³⁻ → so yes!

So: 1 Li₃N + 3 NH₄NO₃ → 3 LiNO₃ + (NH₄)₃N

Check atoms:

Left:
- Li: 3
- N: 1 (from Li₃N) + 3 (from NH₄NO₃) = 4 N
- H: 3×4 = 12 H
- O: 3×3 = 9 O

Right:
- 3 LiNO₃ → Li=3, N=3, O=9
- (NH₄)₃N → N=1 (central) + 3 N from NH₄? No: (NH₄)₃N means 3 NH₄⁺ and 1 N³⁻ → total N=4, H=12

Yes! So:
- N: 3 (in LiNO₃) + 4 (in (NH₄)₃N) = 7 → wait, conflict

Wait — mistake: (NH₄)₃N has one nitrogen as anion, and three ammonium ions each with one N → total 4 nitrogen atoms.

But left: Li₃N → 1 N, 3 NH₄NO₃ → 3 N (from NO₃) + 3 N (from NH₄) → total 1+3+3=7 N

Right: 3 LiNO₃ → 3 N, (NH₄)₃N → 4 N → total 7 N → OK

H: left: 3×4 = 12 H → right: (NH₄)₃N → 3×4 = 12 H → OK

O: left: 3×3 = 9 → right: 3×3 = 9 → OK

Li: 3 → 3 → OK

So ✔ Balanced:
Li₃N + 3 NH₄NO₃ → 3 LiNO₃ + (NH₄)₃N

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9) HBr + Al(OH)₃ → AlBr₃ + H₂O

Acid-base reaction.

Al(OH)₃ has 3 OH⁻ → needs 3 H⁺ → 3 HBr

So:
3 HBr + Al(OH)₃ → AlBr₃ + 3 H₂O

Check:
Left: H=3+3=6, Br=3, Al=1, O=3, H=3 → total H=6, O=3
Right: AlBr₃, 3 H₂O → H=6, O=3 → ✔

✔ Balanced:
3 HBr + Al(OH)₃ → AlBr₃ + 3 H₂O

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10) Pb + H₃PO₄ → Pb₃(PO₄)₂ + H₂

This is a single replacement.

Pb replaces H in acid.

Pb₃(PO₄)₂ requires 3 Pb and 2 PO₄

So need 2 H₃PO₄ → gives 2 PO₄ and 6 H

So H₂ produced: 3 H₂ (since 6 H)

And 3 Pb atoms

So:
3 Pb + 2 H₃PO₄ → Pb₃(PO₄)₂ + 3 H₂

Check:
Left: Pb=3, H=6, P=2, O=8
Right: Pb=3, P=2, O=8, H=6 → ✔

✔ Balanced:
3 Pb + 2 H₃PO₄ → Pb₃(PO₄)₂ + 3 H₂

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B. Identify the type of reaction

1) Na₃PO₄ + 3 KOH → 3 NaOH + K₃PO₄

- Ions swap: Na⁺ with K⁺ → double-replacement

✔ Double-replacement

2) Pb + FeSO₄ → PbSO₄ + Fe

- Pb replaces Fe → single replacement

✔ Single-replacement

3) 2 BF₃ + 3 H₂O → B₂O₃ + 6 HF

- A compound breaks down into simpler substances? Or hydration?

BF₃ reacts with water → hydrolysis → decomposition-like

But two reactants → not decomposition.

Actually: BF₃ + 3 H₂O → B(OH)₃ + 3 HF → but here it’s B₂O₃

So: 2 BF₃ + 3 H₂O → B₂O₃ + 6 HF

This is a double-replacement? But no ions clearly.

Better: It’s a decomposition if we think of BF₃ hydrolyzing, but it's not decomposition.

Alternatively, it's a combination? No.

Actually, it's a double-replacement with water acting as reactant.

But more accurately, it's a hydrolysis reaction, which is a type of double-replacement.

But let's see: BF₃ acts as Lewis acid, accepts electron pair.

But for classification: since two compounds react and exchange parts → double-replacement

✔ Double-replacement

Alternatively, some classify this as decomposition of BF₃ in water, but it's better as double-replacement.

Wait: actually, it's not double-replacement because no ions are exchanging in a typical sense.

It's a synthesis? No, multiple reactants and products.

Best answer: double-replacement or metathesis?

No, not really.

Alternative: This is a redox? Let's check oxidation states.

B in BF₃: +3, in B₂O₃: +3 → no change
H: +1 → +1, O: -2 → -2, F: -1 → -1 → no redox.

So not redox.

This is a double-replacement only if we consider it as:

BF₃ + H₂O → B(OH)₃ + HF → but here it's B₂O₃.

Actually, B₂O₃ is formed from dehydration.

So perhaps it's a combination of BF₃ and H₂O to form HF and B₂O₃.

But still, it's a double-replacement or displacement?

Better: double-replacement is acceptable.

But many sources call such reactions acid-base or hydrolysis.

But among given types: closest is double-replacement

✔ Double-replacement

4) 2 Al + 6 HCl → 2 AlCl₃ + 3 H₂

- Aluminum replaces hydrogen → single-replacement

✔ Single-replacement

5) 2 Fe + O₂ + 2 H₂O → 2 Fe(OH)₃

- Iron combines with oxygen and water to form rust → synthesis

Also, it's a combustion? Not really — no fuel burning.

But it’s oxidation of iron.

But among choices: synthesis — forming a compound from elements and water.

But synthesis typically: A + B → AB

Here: Fe + O₂ + H₂O → Fe(OH)₃ → multiple reactants.

Still, it's a synthesis reaction.

Alternatively, could be combination.

But "synthesis" includes combination.

✔ Synthesis

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✔ Final Answers:

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A. Balanced Equations:

1) 2 NaBr + Ca(OH)₂ → CaBr₂ + 2 NaOH
2) N₂ + 3 H₂ → 2 NH₃
3) 2 NaCl + F₂ → 2 NaF + Cl₂
4) Pb(OH)₂ + 2 HCl → PbCl₂ + 2 H₂O
5) CH₄ + 2 O₂ → CO₂ + 2 H₂O
6) 3 H₂SO₄ + 2 B(OH)₃ → B₂(SO₄)₃ + 6 H₂O
7) 4 C₅H₉O + 27 O₂ → 20 CO₂ + 18 H₂O
8) Li₃N + 3 NH₄NO₃ → 3 LiNO₃ + (NH₄)₃N
9) 3 HBr + Al(OH)₃ → AlBr₃ + 3 H₂O
10) 3 Pb + 2 H₃PO₄ → Pb₃(PO₄)₂ + 3 H₂

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B. Reaction Types:

1) Double-replacement
2) Single-replacement
3) Double-replacement
4) Single-replacement
5) Synthesis

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Let me know if you'd like explanations for any specific one!