Solved Chemistry: Predict the outcome of the following | Chegg.com - Free Printable

Let’s go through each reaction one by one. We’ll figure out what products form (if any), and then balance the equation.

Remember:
- For double replacement reactions (two compounds swapping partners), we assume both are aqueous. A reaction only happens if a precipitate, gas, or water forms.
- For combustion reactions (hydrocarbon + O₂), products are always CO₂ and H₂O.
- If no reaction occurs, write “no reaction”.

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1. Ag₂SO₄ + NaNO₃ →

Both are soluble salts. Possible products: AgNO₃ and Na₂SO₄ — both are also soluble. No precipitate, gas, or water → no reaction

→ no reaction

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2. AlCl₃ + (NH₄)₃PO₄ →

Double replacement: Al³⁺ swaps with NH₄⁺ → AlPO₄ and NH₄Cl

AlPO₄ is insoluble (precipitate). So reaction occurs.

Unbalanced:
AlCl₃ + (NH₄)₃PO₄ → AlPO₄ + NH₄Cl

Balance:
Left: Al=1, Cl=3, N=3, H=12, P=1, O=4
Right: Al=1, P=1, O=4, N=1, H=4, Cl=1 → need 3 NH₄Cl

So:
AlCl₃ + (NH₄)₃PO₄ → AlPO₄ + 3NH₄Cl

Check atoms:
Al:1=1, Cl:3=3, N:3=3, H:12=12, P:1=1, O:4=4 → balanced!

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3. C₃H₆ + O₂ →

Combustion of hydrocarbon → CO₂ + H₂O

Unbalanced:
C₃H₆ + O₂ → CO₂ + H₂O

Balance C: 3 on left → 3CO₂
Balance H: 6 on left → 3H₂O (since each has 2 H)
Now O: right = 3×2 + 3×1 = 9 → so O₂ must be 9/2 → multiply all by 2:

2C₃H₆ + 9O₂ → 6CO₂ + 6H₂O

Check:
C: 6=6, H:12=12, O:18=12+6=18 → good!

→ 2C₃H₆ + 9O₂ → 6CO₂ + 6H₂O

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4. Cu(NO₃)₂ + NH₄CN →

Double replacement: Cu²⁺ swaps with NH₄⁺ → Cu(CN)₂ and NH₄NO₃

Cu(CN)₂ is insoluble (cyanides of transition metals often precipitate). Reaction occurs.

Unbalanced:
Cu(NO₃)₂ + NH₄CN → Cu(CN)₂ + NH₄NO₃

Need 2 CN on right → 2 NH₄CN on left → gives 2 NH₄⁺ → need 2 NH₄NO₃

So:
Cu(NO₃)₂ + 2NH₄CN → Cu(CN)₂ + 2NH₄NO₃

Check:
Cu:1=1, N:2+2=4 vs 2+2=4? Wait:
Left: Cu=1, N from NO₃:2, N from CN:2 → total N=4; H:8, C:2, O:6
Right: Cu=1, C:2, N:2 (from CN) + 2 (from NO₃ in 2NH₄NO₃) =4, H:8, O:6 → yes!

Balanced.

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5. Pb(NO₃)₂ + CuOH →

Wait — CuOH? That’s not standard. Copper(I) hydroxide is unstable, but let’s assume it exists for this problem.

Double replacement: Pb²⁺ swaps with Cu⁺ → Pb(OH)₂ and CuNO₃

But charges: Pb²⁺ needs two OH⁻ → Pb(OH)₂
Cu⁺ needs one NO₃⁻ → CuNO₃

Unbalanced:
Pb(NO₃)₂ + CuOH → Pb(OH)₂ + CuNO₃

Need 2 OH on right → 2 CuOH on left → gives 2 Cu⁺ → need 2 CuNO₃

Also, left has 2 NO₃, right has 2 NO₃ → ok.

So:
Pb(NO₃)₂ + 2CuOH → Pb(OH)₂ + 2CuNO₃

Check:
Pb:1=1, N:2=2, O:6+2=8 vs 2+6=8?
Left: O from NO₃:6, from OH:2 → total 8
Right: Pb(OH)₂ has 2O, 2CuNO₃ has 6O → total 8 → good.
H:2=2, Cu:2=2 → balanced.

Note: In reality, CuOH might decompose, but per worksheet instructions, we proceed.

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6. AgNO₃ + MgCl₂ →

Double replacement: Ag⁺ swaps with Mg²⁺ → AgCl and Mg(NO₃)₂

AgCl is insoluble (classic precipitate). Reaction occurs.

Unbalanced:
AgNO₃ + MgCl₂ → AgCl + Mg(NO₃)₂

Need 2 AgCl to match 2 Cl → so 2 AgNO₃ → gives 2 NO₃ → matches Mg(NO₃)₂

So:
2AgNO₃ + MgCl₂ → 2AgCl + Mg(NO₃)₂

Check: Ag:2=2, N:2=2, O:6=6, Mg:1=1, Cl:2=2 → good.

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7. C₄H₈ + O₂ →

Combustion → CO₂ + H₂O

Unbalanced:
C₄H₈ + O₂ → CO₂ + H₂O

C:4 → 4CO₂
H:8 → 4H₂O
O: right = 8 + 4 = 12 → so O₂ = 6

→ C₄H₈ + 6O₂ → 4CO₂ + 4H₂O

Check: C:4=4, H:8=8, O:12=8+4=12 → good.

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8. NaF + CaBr₂ →

Double replacement: Na⁺ swaps with Ca²⁺ → NaBr and CaF₂

CaF₂ is insoluble (fluoride of calcium precipitates). Reaction occurs.

Unbalanced:
NaF + CaBr₂ → NaBr + CaF₂

Need 2 F on right → 2 NaF on left → gives 2 Na⁺ → need 2 NaBr

CaBr₂ already has 2 Br → matches 2 NaBr

So:
2NaF + CaBr₂ → 2NaBr + CaF₂

Check: Na:2=2, F:2=2, Ca:1=1, Br:2=2 → good.

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9. AlCl₃ + Ag₂SO₄ →

Double replacement: Al³⁺ swaps with Ag⁺ → Al₂(SO₄)₃ and AgCl

AgCl is insoluble → reaction occurs.

Unbalanced:
AlCl₃ + Ag₂SO₄ → Al₂(SO₄)₃ + AgCl

Need 2 Al on right → 2 AlCl₃ on left → gives 6 Cl → need 6 AgCl
Need 3 SO₄ on right → 3 Ag₂SO₄ on left → gives 6 Ag → matches 6 AgCl

So:
2AlCl₃ + 3Ag₂SO₄ → Al₂(SO₄)₃ + 6AgCl

Check: Al:2=2, Cl:6=6, Ag:6=6, S:3=3, O:12=12 → good.

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10. LiOH + Fe(NO₃)₃ →

Double replacement: Li⁺ swaps with Fe³⁺ → LiNO₃ and Fe(OH)₃

Fe(OH)₃ is insoluble → reaction occurs.

Unbalanced:
LiOH + Fe(NO₃)₃ → LiNO₃ + Fe(OH)₃

Need 3 OH on right → 3 LiOH on left → gives 3 Li⁺ → need 3 LiNO₃
Fe(NO₃)₃ has 3 NO₃ → matches 3 LiNO₃

So:
3LiOH + Fe(NO₃)₃ → 3LiNO₃ + Fe(OH)₃

Check: Li:3=3, O:3+9=12 vs 9+3=12?
Left: LiOH:3O, Fe(NO₃)₃:9O → total 12
Right: 3LiNO₃:9O, Fe(OH)₃:3O → total 12 → good.
H:3=3, Fe:1=1, N:3=3 → balanced.

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11. NaOH + H₃PO₄ →

Acid-base reaction → salt + water

Products: Na₃PO₄ and H₂O (since H₃PO₄ has 3 H⁺)

Unbalanced:
NaOH + H₃PO₄ → Na₃PO₄ + H₂O

Need 3 Na on right → 3 NaOH on left → gives 3 OH → combines with 3 H⁺ to make 3 H₂O

So:
3NaOH + H₃PO₄ → Na₃PO₄ + 3H₂O

Check: Na:3=3, O:3+4=7 vs 4+3=7?
Left: NaOH:3O, H₃PO₄:4O → 7
Right: Na₃PO₄:4O, 3H₂O:3O → 7 → good.
H:3+3=6 vs 6 → good. P:1=1 → balanced.

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12. KOH + MgCl₂ →

Double replacement: K⁺ swaps with Mg²⁺ → KCl and Mg(OH)₂

Mg(OH)₂ is slightly soluble but often considered precipitate in such worksheets → reaction occurs.

Unbalanced:
KOH + MgCl₂ → KCl + Mg(OH)₂

Need 2 OH on right → 2 KOH on left → gives 2 K⁺ → need 2 KCl
MgCl₂ has 2 Cl → matches 2 KCl

So:
2KOH + MgCl₂ → 2KCl + Mg(OH)₂

Check: K:2=2, O:2=2, H:2=2, Mg:1=1, Cl:2=2 → good.

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13. Li₂O + FeCl₃ →

This is not double replacement in aqueous sense — Li₂O is basic oxide, reacts with acid or salt?

Actually, Li₂O will react with water first, but since we’re assuming direct reaction...

Better to think: Li₂O + FeCl₃ → ? Possibly metathesis: Li⁺ swaps with Fe³⁺ → LiCl and Fe₂O₃? But that’s not typical.

Alternatively, in aqueous solution, Li₂O becomes 2LiOH, then reacts with FeCl₃ → same as #10 but with Li.

Assume: Li₂O + H₂O → 2LiOH (but water not given) — perhaps skip?

Wait — maybe they expect:
Li₂O + FeCl₃ → no reaction? Or force it?

Actually, metal oxides can react with salts if heated, but in aqueous context, likely no direct reaction unless specified.

But let’s try balancing as if it works:

Suppose: Li₂O + FeCl₃ → LiCl + Fe₂O₃? Not balanced.

To get Fe₂O₃, need 2 FeCl₃ → 6 Cl → need 6 LiCl → need 3 Li₂O → gives 3 O → but Fe₂O₃ needs 3 O → ok.

So:
3Li₂O + 2FeCl₃ → 6LiCl + Fe₂O₃

Is Fe₂O₃ a product? It’s rust, solid. Maybe acceptable.

In many curricula, this is accepted.

So:
3Li₂O + 2FeCl₃ → 6LiCl + Fe₂O₃

Check: Li:6=6, O:3=3, Fe:2=2, Cl:6=6 → balanced.

We’ll go with that.

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14. C₂H₂ + O₂ →

Combustion → CO₂ + H₂O

Unbalanced:
C₂H₂ + O₂ → CO₂ + H₂O

C:2 → 2CO₂
H:2 → 1H₂O
O: right = 4 + 1 = 5 → O₂ = 5/2 → multiply by 2:

2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O

Check: C:4=4, H:4=4, O:10=8+2=10 → good.

→ 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O

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15. BaS + NH₄Cl →

Double replacement: Ba²⁺ swaps with NH₄ → BaCl₂ and (NH₄)₂S

Both are soluble? BaCl₂ is soluble, (NH₄)₂S is soluble → no precipitate, gas, or water → no reaction

Wait — (NH₄)₂S can decompose to NH₃ and H₂S gases? But typically in these worksheets, if both products are soluble, no reaction.

However, ammonium sulfide is unstable and releases H₂S gas — so sometimes considered a reaction.

But strictly by solubility rules: both products soluble → no reaction.

I think for consistency, we say no reaction unless gas evolution is expected.

Upon second thought: (NH₄)₂S ⇌ 2NH₃ + H₂S — so gas is produced → reaction occurs.

So:
BaS + 2NH₄Cl → BaCl₂ + (NH₄)₂S → which decomposes to 2NH₃ + H₂S

But the worksheet says "predict products" — probably expect the initial double replacement, then note if it decomposes.

To keep it simple, many texts would write:

BaS + 2NH₄Cl → BaCl₂ + (NH₄)₂S

And since (NH₄)₂S is unstable, but for balancing, we leave it.

But actually, in practice, you’d see bubbling due to H₂S.

Given ambiguity, I’ll follow common textbook approach: if both products are soluble and no obvious gas/water, no reaction. But here, (NH₄)₂S does produce gas.

Let me check online or standard rule: Ammonium salts with sulfides do release H₂S.

So reaction occurs.

Unbalanced:
BaS + NH₄Cl → BaCl₂ + (NH₄)₂S

Need 2 NH₄ on right → 2 NH₄Cl on left → gives 2 Cl → matches BaCl₂

So:
BaS + 2NH₄Cl → BaCl₂ + (NH₄)₂S

And (NH₄)₂S may further decompose, but for this level, this is fine.

Some might argue no reaction, but I think it's better to include it.

Final decision: reaction occurs, products as above.

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Now compiling all answers:

1. no reaction
2. AlCl₃ + (NH₄)₃PO₄ → AlPO₄ + 3NH₄Cl
3. 2C₃H₆ + 9O₂ → 6CO₂ + 6H₂O
4. Cu(NO₃)₂ + 2NH₄CN → Cu(CN)₂ + 2NH₄NO₃
5. Pb(NO₃)₂ + 2CuOH → Pb(OH)₂ + 2CuNO₃
6. 2AgNO₃ + MgCl₂ → 2AgCl + Mg(NO₃)₂
7. C₄H₈ + 6O₂ → 4CO₂ + 4H₂O
8. 2NaF + CaBr₂ → 2NaBr + CaF₂
9. 2AlCl₃ + 3Ag₂SO₄ → Al₂(SO₄) + 6AgCl
10. 3LiOH + Fe(NO₃)₃ → 3LiNO₃ + Fe(OH)₃
11. 3NaOH + H₃PO₄ → Na₃PO₄ + 3H₂O
12. 2KOH + MgCl₂ → 2KCl + Mg(OH)₂
13. 3Li₂O + 2FeCl₃ → 6LiCl + Fe₂O₃
14. 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
15. BaS + 2NH₄Cl → BaCl₂ + (NH₄)₂S

For #15, if your teacher considers (NH₄)₂S as decomposing, they might want gases, but I think this is acceptable.

Final Answer:

1. no reaction
2. AlCl₃ + (NH₄)₃PO₄ → AlPO₄ + 3NH₄Cl
3. 2C₃H₆ + 9O₂ → 6CO₂ + 6H₂O
4. Cu(NO₃)₂ + 2NH₄CN → Cu(CN)₂ + 2NH₄NO₃
5. Pb(NO₃)₂ + 2CuOH → Pb(OH)₂ + 2CuNO₃
6. 2AgNO₃ + MgCl₂ → 2AgCl + Mg(NO₃)₂
7. C₄H + 6O₂ → 4CO₂ + 4H₂O
8. 2NaF + CaBr₂ → 2NaBr + CaF₂
9. 2AlCl₃ + 3Ag₂SO₄ → Al₂(SO₄)₃ + 6AgCl
10. 3LiOH + Fe(NO₃)₃ → 3LiNO₃ + Fe(OH)₃
11. 3NaOH + H₃PO₄ → Na₃PO₄ + 3H₂O
12. 2KOH + MgCl₂ → 2KCl + Mg(OH)₂
13. 3Li₂O + 2FeCl₃ → 6LiCl + Fe₂O₃
14. 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
15. BaS + 2NH₄Cl → BaCl₂ + (NH₄)₂S