Neutralization Reactions Worksheet for chemistry practice.

Let’s solve each neutralization reaction step by step. Remember: in a neutralization reaction, an acid reacts with a base to produce a salt and water. We need to balance the atoms on both sides — especially H⁺ from the acid and OH⁻ from the base — so that they form H₂O correctly.

We’ll go one row at a time.

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Row 1:
HCl(aq) + Ca(OH)₂(aq) → CaCl₂(s) + H₂O(l)

- HCl has 1 H⁺, Ca(OH)₂ has 2 OH⁻ → need 2 HCl to provide 2 H⁺ to make 2 H₂O.
- So: 2 HCl + Ca(OH)₂ → CaCl₂ + 2 H₂O

Check:
Left: H=2+2=4, Cl=2, Ca=1, O=2
Right: Ca=1, Cl=2, H=4, O=2 → Balanced ✔

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Row 2:
HCN(aq) + Ba(OH)₂(aq) → Ba(CN)₂(aq) + H₂O(l)

- HCN gives 1 H⁺, Ba(OH)₂ gives 2 OH⁻ → need 2 HCN.
- So: 2 HCN + Ba(OH)₂ → Ba(CN)₂ + 2 H₂O

Check:
Left: H=2+2=4, C=2, N=2, Ba=1, O=2
Right: Ba=1, C=2, N=2, H=4, O=2 → Balanced ✔

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Row 3:
HC₂H₃O₂(aq) + Zn(OH)₂(aq) → Zn(C₂H₃O₂)₂(aq) + H₂O(l)

- Acetic acid (HC₂H₃O₂) is monoprotic → 1 H⁺ per molecule.
- Zn(OH)₂ has 2 OH⁻ → need 2 acetic acid molecules.
- So: 2 HC₂H₃O₂ + Zn(OH)₂ → Zn(C₂H₃O₂)₂ + 2 H₂O

Check:
Left: H=2×4 + 2 = 8+2=10? Wait — let’s count carefully.

Actually, HC₂H₃O₂ has 4 H atoms total? No — only 1 acidic H. But for balancing, we care about how many H⁺ react with OH⁻.

Each HC₂H₃O₂ provides 1 H⁺ → needs 2 to match Zn(OH)₂’s 2 OH⁻ → makes 2 H₂O.

Atoms:

Left:
C: 2×2 = 4
H: 2×4 (from 2 HC₂H₃O₂) + 2 (from Zn(OH)₂) = 8 + 2 = 10? Wait — no! In HC₂H₃O₂, formula is CH₃COOH → actually 4 H atoms per molecule? Let’s write it as C₂H₄O₂ to avoid confusion.

Better to think:
Acid: HC₂H₃O₂ → when it donates H⁺, becomes C₂H₃O₂
Base: Zn(OH)₂ → Zn²⁺ and 2 OH⁻

So product is Zn(C₂H₃O₂)₂ → which means 2 acetate ions.

Thus, we need 2 HC₂H₃O₂ to give 2 H⁺ and 2 C₂H₃O₂.

Then 2 H + 2 OH⁻ → 2 H₂O.

So equation:
2 HC₂H₃O₂ + Zn(OH)₂ → Zn(C₂H₃O₂)₂ + 2 H₂O

Now check atom counts:

Left:
H: from 2 HC₂H₃O₂ → each has 4 H? Actually, molecular formula of acetic acid is C₂H₄O₂, but written as HC₂H₃O₂ meaning one H is acidic. Total H atoms per molecule: 4? Let’s use standard counting.

Standard way:
HC₂H₃O₂ = C₂H₄O₂ (since H + C₂H₃O₂ = C₂H₄O₂)

So 2 HC₂H₃O₂ = 2 C₂H₄O₂ → C₄H₈O₄
Zn(OH)₂ = Zn H₂ O₂
Total left: C₄H₁₀O₆ Zn

Right:
Zn(C₂H₃O₂)₂ = Zn C₄ H₆ O₄
2 H₂O = H₄ O₂
Total right: Zn C₄ H₁₀ O₆ → matches ✔

Balanced.

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Row 4:
HClO₄(aq) + LiOH(aq) → LiClO₄(aq) + H₂O(l)

- Both are monoprotic/monobasic → 1:1 ratio.
- Already balanced: HClO₄ + LiOH → LiClO₄ + H₂O

Check:
Left: H=1+1=2, Cl=1, O=4+1=5, Li=1
Right: Li=1, Cl=1, O=4+1=5, H=2 → Balanced ✔

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Row 5:
HF(aq) + Sr(OH)₂(aq) → SrF₂(s) + H₂O(l)

- HF gives 1 H⁺, Sr(OH)₂ gives 2 OH⁻ → need 2 HF.
- So: 2 HF + Sr(OH)₂ → SrF₂ + 2 H₂O

Check:
Left: H=2+2=4, F=2, Sr=1, O=2
Right: Sr=1, F=2, H=4, O=2 → Balanced ✔

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Row 6:
H₂S(aq) + CsOH(aq) → Cs₂S(aq) + H₂O(l)

- H₂S can donate 2 H⁺ (diprotic), CsOH gives 1 OH⁻ → need 2 CsOH.
- So: H₂S + 2 CsOH → Cs₂S + 2 H₂O

Check:
Left: H=2+2=4, S=1, Cs=2, O=2
Right: Cs=2, S=1, H=4, O=2 → Balanced ✔

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Row 7:
H₃PO₄(aq) + Ba(OH)₂(aq) → Ba₃(PO₄)₂(s) + H₂O(l)

- H₃PO₄ has 3 H⁺, Ba(OH)₂ has 2 OH⁻ → find LCM of 3 and 2 = 6.
- So need 2 H₃PO₄ (gives 6 H⁺) and 3 Ba(OH)₂ (gives 6 OH⁻) → makes 6 H₂O.
- Product: Ba₃(PO₄)₂ requires 3 Ba²⁺ and 2 PO₄³⁻ → which matches 2 H₃PO₄ giving 2 PO₄³⁻ after losing 6 H⁺.

Equation:
2 H₃PO₄ + 3 Ba(OH)₂ → Ba₃(PO₄)₂ + 6 H₂O

Check atoms:

Left:
H: 2×3 + 3×2 = 6 + 6 = 12
P: 2
O: 2×4 + 3×2 = 8 + 6 = 14? Wait — H₃PO₄ has 4 O, Ba(OH)₂ has 2 O.

Better:

2 H₃PO₄ = H₆ P₂ O₈
3 Ba(OH)₂ = Ba₃ H₆ O₆
Total left: H₁₂ P₂ O₁₄ Ba₃

Right:
Ba₃(PO₄)₂ = Ba₃ P₂ O₈
6 H₂O = H₁₂ O₆
Total right: Ba₃ P₂ O₁₄ H₁₂ → matches ✔

Balanced.

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Row 8:
HNO₃(aq) + Hg(OH)₂(aq) → Hg(NO₃)₂(aq) + H₂O(l)

- HNO₃ is monoprotic, Hg(OH)₂ has 2 OH⁻ → need 2 HNO₃.
- So: 2 HNO₃ + Hg(OH)₂ → Hg(NO₃)₂ + 2 H₂O

Check:
Left: H=2+2=4, N=2, O=6+2=8, Hg=1
Right: Hg=1, N=2, O=6+2=8, H=4 → Balanced ✔

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Row 9:
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

- Classic 1:1 reaction → already balanced: HCl + NaOH → NaCl + H₂O

Check:
Left: H=1+1=2, Cl=1, Na=1, O=1
Right: Na=1, Cl=1, H=2, O=1 → Balanced ✔

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Final Answer:

1. 2 HCl(aq) + Ca(OH)₂(aq) → CaCl₂(s) + 2 H₂O(l)
2. 2 HCN(aq) + Ba(OH)₂(aq) → Ba(CN)₂(aq) + 2 H₂O(l)
3. 2 HC₂H₃O₂(aq) + Zn(OH)₂(aq) → Zn(C₂H₃O₂)₂(aq) + 2 H₂O(l)
4. HClO₄(aq) + LiOH(aq) → LiClO₄(aq) + H₂O(l)
5. 2 HF(aq) + Sr(OH)₂(aq) → SrF₂(s) + 2 H₂O(l)
6. H₂S(aq) + 2 CsOH(aq) → Cs₂S(aq) + 2 H₂O(l)
7. 2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) → Ba₃(PO₄)₂(s) + 6 H₂O(l)
8. 2 HNO₃(aq) + Hg(OH)₂(aq) → Hg(NO₃)₂(aq) + 2 H₂O(l)
9. HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)