Types of Chemical Reaction and Predicting Products Worksheet | PDF ... - Free Printable

Let’s go through each reaction one by one. For each, we’ll:

1. Balance the chemical equation (make sure atoms are equal on both sides).
2. Identify the type of reaction based on what’s happening.

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1. Na₃PO₄(aq) + KOH(aq) → NaOH(aq) + K₃PO₄(s)

Wait — this looks wrong. Potassium phosphate (K₃PO₄) is usually soluble, not solid. But let’s balance as written.

Left: Na=3, P=1, O=4+1=5? Wait — better to count per compound.

Actually, let’s write ions or just balance atoms.

Na₃PO₄ + 3KOH → 3NaOH + K₃PO₄

Check:
Na: 3 = 3 ✔️
P: 1 = 1 ✔️
O: 4 + 3 = 7 vs 3 + 4 = 7 ✔️
H: 3 = 3 ✔️
K: 3 = 3 ✔️

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

But note: K₃PO₄ is typically aqueous, not solid. Maybe typo? Anyway, reaction type: two compounds swapping partners → double-displacement

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2. MgCl₂(s) + Li₂CO₃(s) → MgCO₃(s) + LiCl(aq)

Balance:

MgCl₂ + Li₂CO₃ → MgCO₃ + 2LiCl

Check:
Mg: 1=1 ✔️
Cl: 2=2 ✔️
Li: 2=2 ✔️
C: 1=1 ✔️
O: 3=3 ✔️

Balanced: 1 MgCl₂ + 1 Li₂CO₃ → 1 MgCO₃ + 2 LiCl

Type: two ionic compounds exchange ions → double-displacement (also precipitation since MgCO₃ is solid)

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3. C₆H₁₂(l) + O₂(g) → CO₂(g) + H₂O(l)

This is combustion of a hydrocarbon.

General form: CₓHᵧ + O₂ → CO₂ + H₂O

Balance C: 6 on left → 6CO₂
Balance H: 12 H → 6H₂O
Now O: right side: 6×2 + 6×1 = 12 + 6 = 18 O atoms → need 9 O₂ molecules

So: 1 C₆H₁₂ + 9 O₂ → 6 CO₂ + 6 H₂O

Type: combustion

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4. Pb(s) + FeSO₄(aq) → PbSO₄(s) + Fe(s)

Check activity series: Lead is less reactive than iron? Actually, no — lead is *less* active than iron? Wait, standard activity series: K > Na > Ca > Mg > Al > Zn > Fe > Pb > H > Cu...

Iron is more reactive than lead, so iron should displace lead, not the other way. But here lead displaces iron? That wouldn’t happen spontaneously.

But for balancing purposes:

Pb + FeSO₄ → PbSO₄ + Fe

Already balanced: 1 Pb, 1 Fe, 1 S, 4 O on each side.

Type: single element replaces another in compound → single-displacement

*(Note: This reaction may not occur in reality due to reactivity, but for worksheet purposes, we classify it as single-displacement.)*

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5. CaCO₃(s) → CaO(s) + CO₂(g)

Already balanced: 1 Ca, 1 C, 3 O on each side.

Type: one compound breaks into two → decomposition

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6. P₄(s) + O₂(g) → P₂O₃(s)

Balance P: left has 4, right has 2 → put 2 before P₂O₃ → now P=4, O=6

So O₂ needs to give 6 O → 3 O₂

Thus: 1 P₄ + 3 O₂ → 2 P₂O₃

Type: two elements combine → synthesis

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7. RbNO₃(aq) + BeF₂(aq) → Be(NO₃)₂(aq) + RbF(aq)

Balance:

RbNO₃ + BeF₂ → Be(NO₃)₂ + RbF

Need 2 RbNO₃ to get 2 NO₃ for Be(NO₃)₂

Then 2 RbF on right

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

Check: Rb=2, N=2, O=6, Be=1, F=2 on both sides ✔️

Type: ions swap → double-displacement

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8. AgNO₃(aq) + Cu(s) → Cu(NO₃)₂(aq) + Ag(s)

Copper displaces silver? Yes, copper is more reactive than silver.

Balance:

AgNO₃ + Cu → Cu(NO₃)₂ + Ag

Need 2 AgNO₃ to provide 2 NO₃ for Cu(NO₃)₂

Then 2 Ag on right

So: 2 AgNO₃ + 1 Cu → 1 Cu(NO₃)₂ + 2 Ag

Type: single element displaces another → single-displacement

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9. C₃H₆O(l) + O₂(g) → CO₂(g) + H₂O(l)

Combustion of organic compound.

Balance C: 3 → 3CO₂
Balance H: 6 H → 3H₂O
Now O: right side: 3×2 + 3×1 = 6 + 3 = 9 O atoms

Left: C₃H₆O has 1 O, so O₂ must supply 8 O → 4 O₂

Check: left O: 1 + 8 = 9 ✔️

So: 1 C₃H₆O + 4 O₂ → 3 CO₂ + 3 H₂O

Type: combustion

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10. C₅H₅(l) + Fe(s) → Fe(C₅H₅)₂(s)

This is forming ferrocene! Iron with two cyclopentadienyl rings.

Balance:

Need 2 C₅H₅ to make Fe(C₅H₅)₂

So: 2 C₅H₅ + 1 Fe → 1 Fe(C₅H₅)₂

Type: two substances combine to form one → synthesis

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11. SeCl₆(s) + O₂(g) → SeO₂(s) + Cl₂(g)

Balance Se: 1=1 ✔️
Cl: 6 on left → need 3 Cl₂ on right (since Cl₂ has 2 Cl)
O: right has 2 O (from SeO₂), so need 1 O₂ on left

Check: SeCl₆ + O₂ → SeO₂ + 3Cl₂

Atoms: Se=1, Cl=6, O=2 on both sides ✔️

Type: ? One compound reacts with element to give different compounds. Not classic categories.

SeCl₆ loses Cl and gains O — could be seen as redox or displacement.

But looking at pattern: SeCl₆ + O₂ → SeO₂ + Cl₂ — oxygen is replacing chlorine? Not exactly.

Actually, this might be classified as single-displacement if we think of O displacing Cl, but it's not standard.

Alternatively, some might call it redox, but the worksheet lists specific types.

Looking at products: SeO₂ and Cl₂ — perhaps it’s a decomposition triggered by O₂? Not really.

Another view: SeCl₆ is being oxidized? Actually, Se goes from +6 in SeCl₆ to +4 in SeO₂ — reduced? Confusing.

Perhaps best fit: single-displacement — oxygen displaces chlorine? But O₂ is diatomic.

Wait — maybe it’s not a standard type. But let’s see: reactants are compound + element → products are compound + element.

That fits single-displacement definition loosely.

Some sources might call it “other” but since worksheet includes only those types, and it’s not synthesis, decomposition, etc., I’ll go with single-displacement.

*(Note: In reality, this reaction may not be straightforward, but for classification, we’ll use single-displacement.)*

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12. MgI₂(aq) + Mn(SO₃)₂(s) → MgSO₃(s) + MnI₄(aq)

First, check if Mn(SO₃)₂ is correct — manganese(IV) sulfite? Unusual, but okay.

Balance:

MgI₂ + Mn(SO₃)₂ → MgSO₃ + MnI₄

Need 2 MgSO₃ to balance SO₃? No — Mn(SO₃)₂ has 2 SO₃ groups.

Right: MnI₄ has 4 I, left has 2 I from MgI₂ → need 2 MgI₂

Then 2 Mg on left → need 2 MgSO₃ on right

But Mn(SO₃)₂ has 2 SO₃, and 2 MgSO₃ needs 2 SO₃ — good.

So: 2 MgI₂ + 1 Mn(SO₃)₂ → 2 MgSO₃ + 1 MnI₄

Check: Mg=2, I=4, Mn=1, S=2, O=6 on both sides ✔️

Type: ions swap → double-displacement

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13. NO₂(g) → O₂(g) + N₂(g)

Balance:

Left: N=2, O=2? Wait, NO₂ has 1N and 2O per molecule.

To get N₂, need 2 NO₂ → gives 2N and 4O

Right: N₂ + O₂ — but O₂ has 2O, so need 2 O₂ to use 4O

So: 2 NO₂ → 1 N₂ + 2 O₂

Check: N=2, O=4 on both sides ✔️

Type: one compound breaks into two elements → decomposition

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Now, compiling all answers:

1. Balanced: 1, 3, 3, 1 → Type: double-displacement
2. Balanced: 1, 1, 1, 2 → Type: double-displacement
3. Balanced: 1, 9, 6, 6 → Type: combustion
4. Balanced: 1, 1, 1, 1 → Type: single-displacement
5. Balanced: 1, 1, 1 → Type: decomposition
6. Balanced: 1, 3, 2 → Type: synthesis
7. Balanced: 2, 1, 1, 2 → Type: double-displacement
8. Balanced: 2, 1, 1, 2 → Type: single-displacement
9. Balanced: 1, 4, 3, 3 → Type: combustion
10. Balanced: 2, 1, 1 → Type: synthesis
11. Balanced: 1, 1, 1, 3 → Type: single-displacement *(or possibly other, but we'll use this)*
12. Balanced: 2, 1, 2, 1 → Type: double-displacement
13. Balanced: 2, 1, 1 → Type: decomposition

For #11, let me double-check. Some might argue it's not single-displacement because O₂ isn't displacing Cl directly. But in many curricula, any reaction where an element reacts with a compound to produce another element and compound is single-displacement. So I'll stick with that.

Final Answer:

1. 1, 3, 3, 1; double-displacement
2. 1, 1, 1, 2; double-displacement
3. 1, 9, 6, 6; combustion
4. 1, 1, 1, 1; single-displacement
5. 1, 1, 1; decomposition
6. 1, 3, 2; synthesis
7. 2, 1, 1, 2; double-displacement
8. 2, 1, 1, 2; single-displacement
9. 1, 4, 3, 3; combustion
10. 2, 1, 1; synthesis
11. 1, 1, 1, 3; single-displacement
12. 2, 1, 2, 1; double-displacement
13. 2, 1, 1; decomposition