Reaction Products Worksheet for predicting and balancing chemical reactions.

Let’s solve each reaction step by step. We’ll predict the products, balance the equation, and check if any product is a precipitate (solid that forms in water). Remember: all reactions happen in water.

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1) Ca(OH)₂ + HF →

This is an acid-base reaction. Calcium hydroxide (base) reacts with hydrofluoric acid (acid) to form salt and water.

Products: CaF₂ and H₂O

Unbalanced:
Ca(OH)₂ + HF → CaF₂ + H₂O

Balance F: need 2 HF on left
Ca(OH)₂ + 2HF → CaF₂ + H₂O

Now balance H and O: Left has 2H from OH and 2H from HF = 4H; right has 2H in H₂O → so we need 2 H₂O

Final balanced:
Ca(OH)₂ + 2HF → CaF₂ + 2H₂O

Check solubility: CaF₂ is insoluble → it will precipitate! ✔

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2) Pb(NO₃)₂ + K₂CrO₄ →

Double displacement. Swap ions: Pb²⁺ pairs with CrO₄²⁻, K⁺ pairs with NO₃⁻

Products: PbCrO₄ and KNO₃

Unbalanced:
Pb(NO₃)₂ + K₂CrO₄ → PbCrO₄ + KNO₃

Balance K and NO₃: need 2 KNO₃ on right

Balanced:
Pb(NO₃)₂ + K₂CrO₄ → PbCrO₄ + 2KNO₃

Solubility: PbCrO₄ is insoluble → precipitate! ✔
KNO₃ is soluble → stays in solution.

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3) NaC₂H₃O₂ + H₂SO₄ →

Sodium acetate + sulfuric acid → double displacement or acid-base?

Actually, this can produce acetic acid (weak acid) and sodium sulfate.

Products: HC₂H₃O₂ (acetic acid) and Na₂SO₄

Unbalanced:
NaC₂H₃O₂ + H₂SO₄ → HC₂H₃O₂ + Na₂SO₄

Need 2 Na on right → so 2 NaC₂H₃O₂ on left

Also, H₂SO₄ provides 2H⁺ → makes 2 HC₂H₃O₂

Balanced:
2NaC₂H₃O₂ + H₂SO₄ → 2HC₂H₃O₂ + Na₂SO₄

Solubility: All products are soluble! Acetic acid is weak but still dissolves. Na₂SO₄ is soluble. No precipitate. ✘

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4) Cu(OH)₂ + H₃PO₄ →

Acid-base reaction. Copper(II) hydroxide + phosphoric acid → copper phosphate + water

Products: Cu₃(PO)₂ and H₂O

Unbalanced:
Cu(OH)₂ + H₃PO₄ → Cu₃(PO₄)₂ + H₂O

Balance Cu: need 3 Cu(OH)₂
Balance PO₄: need 2 H₃PO₄

So:
3Cu(OH)₂ + 2H₃PO₄ → Cu₃(PO₄)₂ + H₂O

Now count H and O:

Left: H from 3×2=6 OH → 6H, plus 2×3=6H from acid → total 12H
Right: currently only 2H in one H₂O → need 6 H₂O to get 12H

Check O: Left: 3×2=6 O from OH, 2×4=8 O from PO₄ → 14 O? Wait — better to recount atoms properly.

Actually, let’s write full atom count:

Left:
Cu: 3
O from OH: 3×2 = 6
H from OH: 3×2 = 6
From 2H₃PO₄: H: 6, P: 2, O: 8
Total left: Cu=3, P=2, H=12, O=14

Right: Cu₃(PO₄)₂ → Cu=3, P=2, O=8
Plus 6H₂O → H=12, O=6 → total O=14 ✔

Balanced:
3Cu(OH)₂ + 2H₃PO₄ → Cu₃(PO₄)₂ + 6H₂O

Solubility: Cu₃(PO₄)₂ is insoluble → precipitate! ✔

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5) AgNO₃ + Na₂CO₃ →

Double displacement: Ag⁺ with CO₃²⁻, Na⁺ with NO₃⁻

Products: Ag₂CO₃ and NaNO₃

Unbalanced:
AgNO₃ + Na₂CO₃ → Ag₂CO₃ + NaNO₃

Balance Ag: need 2 AgNO₃
Balance Na: need 2 NaNO₃

Balanced:
2AgNO₃ + Na₂CO₃ → Ag₂CO₃ + 2NaNO₃

Solubility: Ag₂CO₃ is insoluble → precipitate! ✔
NaNO₃ is soluble.

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6) Zn + H₂CO₃ →

Zinc metal + carbonic acid → single displacement? But carbonic acid is weak and unstable.

Actually, metals react with acids to produce hydrogen gas and salt.

But H₂CO₃ decomposes easily to CO₂ and H₂O. So likely:

Zn + H₂CO₃ → ZnCO₃ + H₂

But ZnCO₃ may not be stable? Let’s think.

Standard reaction: Metal + Acid → Salt + Hydrogen Gas

Here, acid is H₂CO₃ → salt would be ZnCO₃

Unbalanced:
Zn + H₂CO₃ → ZnCO₃ + H₂

Already balanced! Atoms: Zn=1, H=2, C=1, O=3 on both sides.

Balanced:
Zn + H₂CO₃ → ZnCO₃ + H₂

Solubility: ZnCO₃ is insoluble → precipitate! ✔
H₂ is gas → bubbles out.

Note: In reality, H₂CO₃ breaks down quickly, but for worksheet purposes, we treat it as written.

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7) Pb(OH)₂ + Hg₂S →

Wait — Hg₂S? That’s mercury(I) sulfide. But Pb(OH)₂ is lead(II) hydroxide.

Is this a double displacement? Let’s see possible swap:

Pb²⁺ with S²⁻ → PbS
Hg₂²⁺ with OH⁻ → Hg₂(OH)₂? But mercury(I) hydroxide is unstable and decomposes.

Actually, Hg₂S is already very insoluble. And Pb(OH)₂ is also mostly insoluble.

But let’s try to write products:

Possible: PbS + Hg₂(OH)₂

But Hg₂(OH)₂ decomposes to Hg + HgO + H₂O — too messy.

Alternatively, maybe no reaction? But worksheet says “assume all reactions take place”.

Perhaps it’s meant to be HgS? But it says Hg₂S.

Looking at common reactions — perhaps this is a trick.

Wait — Hg₂S is not standard. Mercury(I) compounds are rare and often disproportionate.

But for school level, maybe they expect:

Pb(OH)₂ + Hg₂S → PbS + Hg₂(OH)₂

Then balance:

Already 1:1:1:1

But Hg₂(OH)₂ is not stable — it turns into Hg + HgO + H₂O.

However, since the problem says “assume all reactions take place in water” and doesn’t mention decomposition, we’ll go with direct swap.

Balanced:
Pb(OH)₂ + Hg₂S → PbS + Hg₂(OH)₂

Solubility: PbS is extremely insoluble → precipitate! ✔
Hg₂(OH)₂ — actually, mercury(I) hydroxide decomposes immediately, but if we consider it as formed, it might also be insoluble. However, typically we say PbS precipitates.

In many textbooks, this reaction isn't common, but based on ion exchange, PbS should form.

To be safe, we'll note PbS precipitates.

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Now, compiling final answers for each:

1) Ca(OH)₂ + 2HF → CaF₂↓ + 2H₂O
2) Pb(NO₃)₂ + K₂CrO₄ → PbCrO₄↓ + 2KNO₃
3) 2NaC₂H₃O₂ + H₂SO₄ → 2HC₂H₃O₂ + Na₂SO₄ (no ppt)
4) 3Cu(OH)₂ + 2H₃PO₄ → Cu₃(PO₄)₂↓ + 6H₂O
5) 2AgNO₃ + Na₂CO₃ → Ag₂CO₃↓ + 2NaNO₃
6) Zn + H₂CO₃ → ZnCO₃↓ + H₂↑
7) Pb(OH)₂ + Hg₂S → PbS↓ + Hg₂(OH)₂ (though Hg₂(OH)₂ decomposes)

For #7, some sources might say no reaction, but per instructions, we assume reaction occurs.

Final Answer:

1) Ca(OH)₂ + 2HF → CaF₂ + 2H₂O
2) Pb(NO₃)₂ + K₂CrO₄ → PbCrO₄ + 2KNO₃
3) 2NaC₂H₃O₂ + H₂SO₄ → 2HC₂H₃O₂ + Na₂SO₄
4) 3Cu(OH)₂ + 2H₃PO₄ → Cu(PO₄)₂ + 6H₂O
5) 2AgNO₃ + Na₂CO₃ → Ag₂CO₃ + 2NaNO₃
6) Zn + H₂CO₃ → ZnCO₃ + H₂
7) Pb(OH)₂ + Hg₂S → PbS + Hg₂(OH)₂

(Note: Precipitates are CaF₂, PbCrO₄, Cu₃(PO₄)₂, Ag₂CO₃, ZnCO₃, PbS — but since question didn’t ask to mark them, just to write balanced equations, we provide those.)