Chemistry worksheet focusing on balancing chemical reactions and identifying their types, including examples like combustion, single replacement, and double replacement reactions.

Let’s go step by step through each problem. We’ll identify the reaction type for Part A, and for Part B we’ll balance the equation first, then identify the type.

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Part A: Identify each type of chemical reactions

1) C₉H₂₀ + 14 O₂ → 9 CO₂ + 10 H₂O
This is a combustion reaction — a hydrocarbon (C₉H₂₀) reacts with oxygen to produce carbon dioxide and water.

2) SrCl₂ + 2 AgNO₃ → Sr(NO₃)₂ + 2 AgCl
Two compounds swap partners: Sr pairs with NO₃, Ag pairs with Cl. This is a double replacement reaction.

3) 2 LiBr + F₂ → 2 LiF + Br₂
Fluorine replaces bromine in the compound. One element kicks out another from a compound. This is a single replacement reaction.

4) 2 NaF → Na + F₂
One compound breaks down into two elements. This is a decomposition reaction.

5) ZnBr₂ + F₂ → ZnF₂ + Br₂
Fluorine replaces bromine again. Another single replacement reaction.

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Part B: Balance the chemical reactions and identify each type

1) CuCl₂ + H₂S → CuS + HCl
Left: Cu=1, Cl=2, H=2, S=1
Right: Cu=1, S=1, H=1, Cl=1 → Not balanced!
We need 2 HCl on right to match 2 Cl and 2 H.
Balanced: CuCl₂ + H₂S → CuS + 2 HCl
Type: Two compounds swap parts → double replacement

2) Ca + AgCl → CaCl₂ + Ag
Left: Ca=1, Ag=1, Cl=1
Right: Ca=1, Cl=2, Ag=1 → Cl not balanced
Need 2 AgCl on left to get 2 Cl, which means 2 Ag on right.
Balanced: Ca + 2 AgCl → CaCl₂ + 2 Ag
Type: Element replaces another in compound → single replacement

3) C₄H₈ + O₂ → CO₂ + H₂O
Left: C=4, H=8, O=?
Right: C=1, H=2, O=3 (from CO₂ and H₂O) → Need to balance
Start with C: put 4 in front of CO₂ → now C=4
H: 8 on left → need 4 H₂O on right (since each has 2 H) → H=8
Now O: Right side = 4×2 (from CO₂) + 4×1 (from H₂O) = 8 + 4 = 12 O atoms
So left needs 6 O₂ (since each O₂ has 2 O) → 6×2=12
Balanced: C₄H₈ + 6 O₂ → 4 CO₂ + 4 H₂O
Type: Hydrocarbon + oxygen → CO₂ + H₂O → combustion

4) PbSO₄ + AgNO₃ → Ag₂SO₄ + Pb(NO₃)₂
Left: Pb=1, S=1, O=4+3=7? Wait — let’s count properly.
PbSO₄: Pb=1, S=1, O=4
AgNO₃: Ag=1, N=1, O=3
Right: Ag₂SO₄: Ag=2, S=1, O=4
Pb(NO₃)₂: Pb=1, N=2, O=6
So total right: Ag=2, S=1, O=4+6=10, Pb=1, N=2
Left: To get 2 Ag, need 2 AgNO₃ → gives 2 Ag, 2 N, 6 O
Plus PbSO₄: Pb=1, S=1, O=4 → total O = 6+4=10 → matches!
Balanced: PbSO₄ + 2 AgNO₃ → Ag₂SO₄ + Pb(NO₃)₂
Type: Swap of ions → double replacement

5) Li + N₂ → Li₃N
Left: Li=1, N=2
Right: Li=3, N=1 → Not balanced
Need even number of N on right → try 2 Li₃N → Li=6, N=2
Then left needs 6 Li and 1 N₂
Balanced: 6 Li + N₂ → 2 Li₃N
Type: Two elements combine to form one compound → synthesis (also called combination)

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Final Answer:

Part A:
1) Combustion
2) Double Replacement
3) Single Replacement
4) Decomposition
5) Single Replacement

Part B:
1) CuCl₂ + H₂S → CuS + 2 HCl — Double Replacement
2) Ca + 2 AgCl → CaCl₂ + 2 Ag — Single Replacement
3) C₄H₈ + 6 O₂ → 4 CO₂ + 4 H₂O — Combustion
4) PbSO₄ + 2 AgNO₃ → Ag₂SO₄ + Pb(NO₃)₂ — Double Replacement
5) 6 Li + N₂ → 2 Li₃N — Synthesis