Chemistry worksheet featuring practice problems for balancing equations and classifying reaction types.

Let’s go step by step to balance each reaction and identify the type. We’ll do this carefully, one at a time.

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1. NaBr + Ca(OH)₂ → CaBr₂ + NaOH

Left side:
Na: 1, Br: 1, Ca: 1, O: 2, H: 2
Right side:
Ca: 1, Br: 2, Na: 1, O: 1, H: 1 → Not balanced!

We need 2 NaOH on right to match O and H from left (since Ca(OH)₂ has 2 OH groups).
So try:
NaBr + Ca(OH)₂ → CaBr₂ + 2 NaOH

Now check atoms:
Left: Na=1, Br=1, Ca=1, O=2, H=2
Right: Ca=1, Br=2, Na=2, O=2, H=2 → Br and Na are off.

So we need 2 NaBr on left:
2 NaBr + Ca(OH)₂ → CaBr₂ + 2 NaOH

Check again:
Left: Na=2, Br=2, Ca=1, O=2, H=2
Right: Ca=1, Br=2, Na=2, O=2, H=2 ✔ Balanced!

Type: Two compounds swap partners → Double Replacement

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2. NH₃ + H₂SO₄ → (NH₄)₂SO₄

Left: N=1, H=3+2=5, S=1, O=4
Right: N=2, H=8, S=1, O=4 → Need more NH₃

Try 2 NH₃:
2 NH₃ + H₂SO₄ → (NH₄)₂SO₄

Left: N=2, H=6+2=8, S=1, O=4
Right: N=2, H=8, S=1, O=4 ✔ Balanced!

Type: Two reactants form one product → Synthesis

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3. C₅H₈O + O₂ → CO₂ + H₂O

This is combustion (hydrocarbon + oxygen → CO₂ + H₂O)

Balance C first: 5 carbons → 5 CO₂
Balance H: 8 hydrogens → 4 H₂O (since each has 2 H)
Now count O on right:
5 CO₂ = 10 O, 4 H₂O = 4 O → total 14 O atoms

Left: C₅H₈O has 1 O, so O₂ must supply 13 O → but O₂ comes in pairs → not possible? Wait — let’s write:

C₅H₈O + ? O₂ → 5 CO₂ + 4 H₂O

O on right: 5×2 + 4×1 = 10 + 4 = 14
O on left: 1 (from C₅H₈O) + 2x (from O₂) → 1 + 2x = 14 → 2x=13 → x=6.5 → not whole number!

Multiply entire equation by 2 to eliminate fraction:

2 C₅H₈O + 13 O₂ → 10 CO₂ + 8 H₂O

Check:
C: 10=10
H: 16=16
O: left: 2×1 + 13×2 = 2+26=28; right: 10×2 + 8×1 = 20+8=28 ✔

Balanced: 2 C₅H₈O + 13 O₂ → 10 CO₂ + 8 H₂O

Type: Combustion (fuel + O₂ → CO₂ + H₂O) → Combustion

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4. Pb + H₃PO₄ → H₂ + Pb₃(PO)₂

Right has Pb₃ → need 3 Pb on left
Right has PO₄ twice → need 2 H₃PO₄ on left? Let’s see:

Try: 3 Pb + 2 H₃PO₄ → ? H₂ + Pb₃(PO₄)₂

Left: Pb=3, H=6, P=2, O=8
Right: Pb=3, P=2, O=8, H=? → H₂ must be 3 molecules to get 6 H

So: 3 Pb + 2 H₃PO₄ → 3 H₂ + Pb₃(PO₄)₂

Check:
Pb: 3=3
H: 6=6
P: 2=2
O: 8=8 ✔

Type: Element replaces another in compound → Single Replacement

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5. Li₃N + NH₄NO₃ → LiNO₃ + (NH₄)₃N

Left: Li=3, N=1+1=2, H=4, O=3
Right: Li=1, N=1+3=4, H=12, O=3 → way off

Notice: (NH₄)₃N means 3 NH₄ and 1 N → total N=4, H=12
LiNO₃ has Li=1, N=1, O=3

To get 3 Li on right, need 3 LiNO₃
Then right: Li=3, N=3 (from LiNO₃) + 4 (from (NH₄)₃N) = 7? Wait no — (NH₄)₃N has 3 N from NH₄ and 1 N from N → total 4 N? Actually, (NH₄)₃N is triammonium nitride — it has 3 nitrogen atoms from ammonium and 1 from nitride? No — actually, formula is correct as written: (NH₄)₃N contains 3 N from NH₄ and 1 N from N? That would be 4 N total? But that doesn’t make sense chemically — wait, perhaps it's a typo? Or maybe it's correct.

Actually, let’s treat it as given.

Assume: Li₃N + NH₄NO₃ → LiNO₃ + (NH₄)₃N

We need 3 Li on right → 3 LiNO₃
Then right: Li=3, N=3 (from LiNO₃) + 4 (from (NH₄)₃N) = 7 N? But (NH₄)₃N has 3 N from NH₄ and 1 N from N → yes, 4 N total? Actually, no — in (NH₄)₃N, the “N” at end is separate, so total N atoms: 3 (in NH₄) + 1 (the N ion) = 4 N atoms.

But left: Li₃N has 1 N, NH₄NO₃ has 2 N (one in NH₄, one in NO₃) → total 3 N on left.

Not matching. Maybe coefficient for NH₄NO₃ is 3?

Try: Li₃N + 3 NH₄NO₃ → 3 LiNO₃ + (NH₄)₃N

Left: Li=3, N=1 + 3×2 = 7, H=3×4=12, O=3×3=9
Right: Li=3, N=3 (from LiNO₃) + 4 (from (NH₄)₃N) = 7, H=12, O=9 ✔

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

Type: Two compounds exchange ions → Double Replacement

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6. HBr + Al(OH)₃ → H₂O + AlBr₃

AlBr₃ needs 3 Br → so 3 HBr
Al(OH)₃ has 3 OH → makes 3 H₂O

So: 3 HBr + Al(OH)₃ → 3 H₂O + AlBr₃

Check:
H: 3+3=6 left, 6 right
Br: 3=3
Al: 1=1
O: 3=3 ✔

Type: Acid + base → water + salt → Double Replacement (also called neutralization)

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Now questions 7 and 8 ask to identify type only (already balanced):

7. Na₃PO₄ + 3 KOH → 3 NaOH + K₃PO₄

Two compounds swap cations → Double Replacement

8. MgCl₂ + Li₂CO₃ → MgCO₃ + 2 LiCl

Again, swapping partners → Double Replacement

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Final Answers:

1. 2 NaBr + Ca(OH)₂ → CaBr₂ + 2 NaOH — Double Replacement
2. 2 NH₃ + H₂SO₄ → (NH₄)₂SO₄ — Synthesis
3. 2 C₅H₈O + 13 O₂ → 10 CO₂ + 8 H₂O — Combustion
4. 3 Pb + 2 H₃PO₄ → 3 H₂ + Pb₃(PO₄)₂ — Single Replacement
5. Li₃N + 3 NH₄NO₃ → 3 LiNO₃ + (NH₄)₃N — Double Replacement
6. 3 HBr + Al(OH)₃ → 3 H₂O + AlBr₃ — Double Replacement
7. Double Replacement
8. Double Replacement

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Final Answer:
1. 2 NaBr + Ca(OH)₂ → CaBr₂ + 2 NaOH — Double Replacement
2. 2 NH₃ + H₂SO₄ → (NH₄)₂SO₄ — Synthesis
3. 2 C₅H₈O + 13 O₂ → 10 CO₂ + 8 H₂O — Combustion
4. 3 Pb + 2 H₃PO₄ → 3 H₂ + Pb₃(PO₄)₂ — Single Replacement
5. Li₃N + 3 NH₄NO₃ → 3 LiNO₃ + (NH₄)₃N — Double Replacement
6. 3 HBr + Al(OH)₃ → 3 H₂O + AlBr₃ — Double Replacement
7. Double Replacement
8. Double Replacement