Chemistry worksheet focusing on balancing and identifying single and double displacement reactions.

Let’s go through each problem one by one. We’ll balance the chemical equations first, then decide if it’s a single or double displacement reaction.

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What’s the difference?

- Single Displacement: One element replaces another in a compound.
Example: A + BC → AC + B

- Double Displacement: Two compounds swap partners.
Example: AB + CD → AD + CB

We’ll use that to classify each.

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## Problem 1:
__ Cl₂ + __ KI → __ KCl + __ I₂

Balance atoms:

Left: Cl=2, K=1, I=1
Right: K=1, Cl=1, I=2 → need to fix I and Cl

Try:
Cl₂ + 2KI → 2KCl + I₂
Check:
Left: Cl=2, K=2, I=2
Right: K=2, Cl=2, I=2 ✔

Type: Chlorine (element) kicks out iodine from KI → SINGLE

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## Problem 2:
__ K₂CO₃ + __ BaCl₂ → __ KCl + __ BaCO₃

Left: K=2, C=1, O=3, Ba=1, Cl=2
Right: K=1, Cl=1, Ba=1, C=1, O=3 → need to fix K and Cl

Try:
K₂CO₃ + BaCl₂ → 2KCl + BaCO₃
Check:
Left: K=2, C=1, O=3, Ba=1, Cl=2
Right: K=2, Cl=2, Ba=1, C=1, O=3 ✔

Type: Two compounds swapping ions → DOUBLE

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## Problem 3:
__ Na + __ MgCl₂ → __ NaCl + __ Mg

Left: Na=1, Mg=1, Cl=2
Right: Na=1, Cl=1, Mg=1 → need more Cl on right

Try:
2Na + MgCl₂ → 2NaCl + Mg
Check:
Left: Na=2, Mg=1, Cl=2
Right: Na=2, Cl=2, Mg=1 ✔

Type: Sodium (element) replaces magnesium → SINGLE

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## Problem 4:
__ Al + __ CuCl₂ → __ AlCl₃ + __ Cu

Left: Al=1, Cu=1, Cl=2
Right: Al=1, Cl=3, Cu=1 → Cl doesn’t match

Find LCM of 2 and 3 → 6

So:
2Al + 3CuCl₂ → 2AlCl₃ + 3Cu
Check:
Left: Al=2, Cu=3, Cl=6
Right: Al=2, Cl=6, Cu=3 ✔

Type: Aluminum replaces copper → SINGLE

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## Problem 5:
__ Al + __ Pb(NO₃)₂ → __ Al(NO₃)₃ + __ Pb

Left: Al=1, Pb=1, N=2, O=6
Right: Al=1, N=3, O=9, Pb=1 → not balanced

LCM for NO₃ groups: 2 and 3 → 6

So:
2Al + 3Pb(NO₃)₂ → 2Al(NO₃)₃ + 3Pb
Check:
Left: Al=2, Pb=3, N=6, O=18
Right: Al=2, N=6, O=18, Pb=3 ✔

Type: Aluminum replaces lead → SINGLE

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## Problem 6:
__ AgNO₃ + __ CaBr₂ → __ AgBr + __ Ca(NO₃)₂

Left: Ag=1, N=1, O=3, Ca=1, Br=2
Right: Ag=1, Br=1, Ca=1, N=2, O=6 → not balanced

Need 2 AgNO₃ and 2 AgBr:

2AgNO₃ + CaBr₂ → 2AgBr + Ca(NO₃)₂
Check:
Left: Ag=2, N=2, O=6, Ca=1, Br=2
Right: Ag=2, Br=2, Ca=1, N=2, O=6 ✔

Type: Swap between two compounds → DOUBLE

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## Problem 7:
__ Al₂(SO₄)₃ + __ Ca₃(PO)₂ → __ AlPO + __ CaSO₄

This looks tricky. Let’s count atoms.

Left: Al=2, S=3, O=12 from sulfate + O=8 from phosphate? Wait — better to treat polyatomic ions as units.

Actually, let’s write with ion counts:

Al₂(SO₄)₃ has 2 Al³⁺ and 3 SO₄²⁻
Ca₃(PO₄)₂ has 3 Ca²⁺ and 2 PO₄³⁻

Products: AlPO₄ and CaSO₄

To balance, we need same number of each ion.

Try:
Al₂(SO₄)₃ + Ca₃(PO₄)₂ → 2AlPO₄ + 3CaSO₄
Check atoms:

Left: Al=2, S=3, O=12+8=20? Wait — better to count per atom.

Actually:

Left:
Al=2, S=3, O=12 (from SO₄) + 8 (from PO₄) = 20? No — wait, Ca₃(PO₄)₂ has 8 oxygen? Actually PO₄ is 4 oxygens × 2 = 8, yes. But SO₄ is 4×3=12 → total O=20? That can’t be right because products will have different.

Wait — let’s do proper balancing.

Set coefficients:

a Al₂(SO₄)₃ + b Ca₃(PO₄)₂ → c AlPO₄ + d CaSO₄

Balance Al: 2a = c
Balance S: 3a = d
Balance Ca: 3b = d
Balance P: 2b = c
Balance O: too messy — use above.

From Al: c = 2a
From P: c = 2b → so 2a = 2b → a = b
From S: d = 3a
From Ca: d = 3b → since a=b, consistent.

So set a=1, then b=1, c=2, d=3

Equation:
Al₂(SO₄)₃ + Ca₃(PO₄)₂ → 2AlPO₄ + 3CaSO₄

Check atoms:

Left:
Al=2, S=3, O=12 (from SO₄) + 8 (from PO₄) = 20? Wait — actually:

Al₂(SO₄)₃: Al=2, S=3, O=12
Ca₃(PO₄)₂: Ca=3, P=2, O=8
Total left: Al=2, S=3, Ca=3, P=2, O=20

Right:
2AlPO₄: Al=2, P=2, O=8
3CaSO₄: Ca=3, S=3, O=12
Total right: Al=2, P=2, Ca=3, S=3, O=20 ✔

Perfect.

Type: Two compounds swapping ions → DOUBLE

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## Problem 8:
__ FeBr₂ + __ K₂CO₃ → __ FeCO₃ + __ KBr

Left: Fe=1, Br=2, K=2, C=1, O=3
Right: Fe=1, C=1, O=3, K=1, Br=1 → need to fix K and Br

Try:
FeBr₂ + K₂CO₃ → FeCO₃ + 2KBr
Check:
Left: Fe=1, Br=2, K=2, C=1, O=3
Right: Fe=1, C=1, O=3, K=2, Br=2 ✔

Type: Swap → DOUBLE

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## Problem 9:
__ Cu + __ Ag₂SO₄ → __ Ag + __ CuSO₄

Left: Cu=1, Ag=2, S=1, O=4
Right: Ag=1, Cu=1, S=1, O=4 → need 2 Ag on right

Try:
Cu + Ag₂SO₄ → 2Ag + CuSO₄
Check:
Left: Cu=1, Ag=2, S=1, O=4
Right: Ag=2, Cu=1, S=1, O=4 ✔

Type: Copper replaces silver → SINGLE

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## Problem 10:
__ CuS + __ KCl → __ CuCl₂ + __ K₂S

Left: Cu=1, S=1, K=1, Cl=1
Right: Cu=1, Cl=2, K=2, S=1 → need to fix K and Cl

Try:
CuS + 2KCl → CuCl₂ + K₂S
Check:
Left: Cu=1, S=1, K=2, Cl=2
Right: Cu=1, Cl=2, K=2, S=1 ✔

Type: Swap → DOUBLE

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Final Answer:

1) 1 Cl₂ + 2 KI → 2 KCl + 1 I₂ → SINGLE
2) 1 K₂CO₃ + 1 BaCl₂ → 2 KCl + 1 BaCO₃ → DOUBLE
3) 2 Na + 1 MgCl₂ → 2 NaCl + 1 Mg → SINGLE
4) 2 Al + 3 CuCl₂ → 2 AlCl₃ + 3 Cu → SINGLE
5) 2 Al + 3 Pb(NO₃)₂ → 2 Al(NO₃)₃ + 3 Pb → SINGLE
6) 2 AgNO₃ + 1 CaBr₂ → 2 AgBr + 1 Ca(NO₃)₂ → DOUBLE
7) 1 Al₂(SO₄)₃ + 1 Ca₃(PO)₂ → 2 AlPO₄ + 3 CaSO₄ → DOUBLE
8) 1 FeBr₂ + 1 K₂CO₃ → 1 FeCO₃ + 2 KBr → DOUBLE
9) 1 Cu + 1 Ag₂SO₄ → 2 Ag + 1 CuSO₄ → SINGLE
10) 1 CuS + 2 KCl → 1 CuCl₂ + 1 K₂S → DOUBLE