Question

For each of the following chemical reactions
Balance the reactions (make sure to write out your ‘inventory’ for each side of the reaction)
Classify each type of reaction: single replacement, double replacement, synthesis, decomposition, combustion

1) _____ P4 + _____ O2 → _____ P2O3





Type of reaction:_________________

2) _____ RbNO3 + _____ BeF2→ _____ Be(NO3)2 + _____ RbF






Type of reaction:_________________

3) _____ Na3PO4 + _____ KOH→ _____ NaOH + _____ K3PO4






Type of reaction:_________________

4) _____ C2H4 + _____ O2→ _____ CO2 + _____H2O







Type of reaction:_________________

5) _____V2I10 + ____ HF → ____ VF5 + ____ HI




Type of reaction:_________________

6) ____ MgF2 + ____ Li2CO3 → ____ MgCO3 + ____ LiF






Type of reaction:_________________

7) ____ AgNO3 + ____ Cu → ____ Cu(NO3)2 + ____ Ag







Type of reaction:_________________

8) ____ CF4 + ____ Br2 → ____ CBr4 + ____ F2






Type of reaction:_________________






9) ____ HCN + ____ CuSO4 → ____ H2SO4 + ____ Cu(CN)2






Type of reaction:_________________

10) ____ GaF3 + ____ Cs → ____ CsF + ____ Ga






Type of reaction:_________________

11) ____ BaS + ____ PtF2 → ____ BaF2 + ____ PtS







Type of reaction:_________________

Answer 1

1) Balanced equation:
1 P4 + 3 O2 → 2 P2O3

Inventory:
- Left: P = 4, O = 6
- Right: P = 4 (2 × 2), O = 6 (2 × 3)

Type of reaction: synthesis (combination)

2) Balanced equation:
2 RbNO3 + 1 BeF2 → 1 Be(NO3)2 + 2 RbF

Inventory:
- Left: Rb = 2, N = 2, O = 6, Be = 1, F = 2
- Right: Be = 1, N = 2, O = 6, Rb = 2, F = 2

Type of reaction: double replacement (metathesis)

3) Balanced equation:
1 Na3PO4 + 3 KOH → 3 NaOH + 1 K3PO4

Inventory:
- Left: Na = 3, P = 1, O = 4 + 3 = 7, K = 3, H = 3
- Right: Na = 3, O = 3 + 4 = 7, H = 3, K = 3, P = 1

Type of reaction: double replacement

4) Balanced equation:
1 C2H4 + 3 O2 → 2 CO2 + 2 H2O

Inventory:
- Left: C = 2, H = 4, O = 6
- Right: C = 2, O = 4 + 2 = 6, H = 4

Type of reaction: combustion

5) Balanced equation:
1 V2I10 + 10 HF → 2 VF5 + 10 HI

Inventory:
- Left: V = 2, I = 10, H = 10, F = 10
- Right: V = 2, F = 10 (2 × 5), H = 10, I = 10

Type of reaction: double replacement (halide exchange)

6) Balanced equation:
1 MgF2 + 1 Li2CO3 → 1 MgCO3 + 2 LiF

Inventory:
- Left: Mg = 1, F = 2, Li = 2, C = 1, O = 3
- Right: Mg = 1, C = 1, O = 3, Li = 2, F = 2

Type of reaction: double replacement

7) Balanced equation:
2 AgNO3 + 1 Cu → 1 Cu(NO3)2 + 2 Ag

Inventory:
- Left: Ag = 2, N = 2, O = 6, Cu = 1
- Right: Cu = 1, N = 2, O = 6, Ag = 2

Type of reaction: single replacement (metal displacement)

8) Balanced equation:
1 CF4 + 2 Br2 → 1 CBr4 + 2 F2

Inventory:
- Left: C = 1, F = 4, Br = 4
- Right: C = 1, Br = 4 (1 × 4), F = 4 (2 × 2)

Type of reaction: single replacement (halogen displacement)

9) Balanced equation:
2 HCN + 1 CuSO4 → 1 H2SO4 + 1 Cu(CN)2

Inventory:
- Left: H = 2, C = 2, N = 2, Cu = 1, S = 1, O = 4
- Right: H = 2, S = 1, O = 4, Cu = 1, C = 2, N = 2

Type of reaction: double replacement

10) Balanced equation:
1 GaF3 + 3 Cs → 3 CsF + 1 Ga

Inventory:
- Left: Ga = 1, F = 3, Cs = 3
- Right: Cs = 3, F = 3, Ga = 1

Type of reaction: single replacement (metal displacement)

11) Balanced equation:
1 BaS + 1 PtF2 → 1 BaF2 + 1 PtS

Inventory:
- Left: Ba = 1, S = 1, Pt = 1, F = 2
- Right: Ba = 1, F = 2, Pt = 1, S = 1

Type of reaction: double replacement