Neutralization Reactions Worksheet for Chemistry II, featuring examples and problems on balancing neutralization reactions.

Let's solve the Neutralization Reactions Worksheet step by step, following the rules:

> Acid + Base → Salt + Water
> H⁺ from acid and OH⁻ from base combine to form H₂O.
> The leftover ions form the salt.
> In writing salts: Write the metal first, then the anion.

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🔬 Example 2 Recap:

$$
2\ \text{HF} + 1\ \text{Ca(OH)}_2 \rightarrow 2\ \text{H}_2\text{O} + \text{CaF}_2
$$
- 2 H⁺ from HF react with 2 OH⁻ from Ca(OH)₂ to make 2 H₂O.
- Leftover: 1 Ca²⁺ and 2 F⁻ → CaF₂ (calcium fluoride).

Now let’s solve each problem.

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## ✔ Problem 1: Complete & Balance

1. $ \text{H}_2\text{SO}_4 + \text{Sr(OH)}_2 \rightarrow \text{H}_2\text{O} + \text{SrSO}_4 $

- H₂SO₄ provides 2 H⁺, Sr(OH)₂ provides 2 OH⁻ → makes 2 H₂O.
- Leftover: Sr²⁺ and SO₄²⁻ → SrSO₄ (already given).
- Balanced as is:

$$
\boxed{1}\ \text{H}_2\text{SO}_4 + \boxed{1}\ \text{Sr(OH)}_2 \rightarrow \boxed{2}\ \text{H}_2\text{O} + \boxed{1}\ \text{SrSO}_4
$$

✔ Check: H: 2+2=4 → 4 in 2 H₂O; O, S, Sr balanced.

---

2. $ \text{Ca(OH)}_2 + \text{H}_3\text{PO}_4 \rightarrow \text{H}_2\text{O} + \text{Ca}_3(\text{PO}_4)_2 $

- Phosphate ion PO₄³⁻ requires 3 Ca²⁺ → so need 3 Ca(OH)₂.
- H₃PO₄ has 3 H⁺ → needs 3 OH⁻ per molecule.
- So for 1 H₃PO₄: need 3 OH⁻ → but each Ca(OH)₂ gives 2 OH⁻ → need 3/2 = 1.5 Ca(OH)₂? Not whole number.

Better: Use lowest common multiple.

- To get 6 OH⁻ (from 3 Ca(OH)₂), we need 6 H⁺ → 2 H₃PO₄ (since each gives 3 H⁺).
- So:
- 3 Ca(OH)₂ → 6 OH⁻
- 2 H₃PO₄ → 6 H⁺ → forms 6 H₂O
- Leftover: 3 Ca²⁺ and 2 PO₄³⁻ → Ca₃(PO₄)₂

So:

$$
\boxed{3}\ \text{Ca(OH)}_2 + \boxed{2}\ \text{H}_3\text{PO}_4 \rightarrow \boxed{6}\ \text{H}_2\text{O} + \boxed{1}\ \text{Ca}_3(\text{PO}_4)_2
$$

✔ Check:
- Ca: 3 = 3
- P: 2 = 2
- O, H: left: 3×2 + 2×4 = 6+8=14 O from bases, plus 8 from acids → total O: 22? Better count atoms later if needed. But it balances.

---

3. $ \text{HBr} + \text{Ba(OH)}_2 \rightarrow \text{H}_2\text{O} + \_\_\_\_\_\_\_\_ $

- Ba(OH)₂ has 2 OH⁻ → needs 2 H⁺ → so need 2 HBr.
- 2 H⁺ + 2 OH⁻ → 2 H₂O
- Leftover: Ba²⁺ and 2 Br⁻ → BaBr₂

$$
\boxed{2}\ \text{HBr} + \boxed{1}\ \text{Ba(OH)}_2 \rightarrow \boxed{2}\ \text{H}_2\text{O} + \boxed{1}\ \text{BaBr}_2
$$

✔ Check: H: 2+2=4 → 4 in 2 H₂O; Br: 2 = 2; Ba: 1 = 1.

---

4. $ \text{Zn(OH)}_2 + \text{HNO}_3 \rightarrow \_\_\_\_\_\_\_\_ + \text{Zn(NO}_3)_2 $

- Zn(OH)₂ has 2 OH⁻ → needs 2 H⁺ → so 2 HNO₃
- 2 H⁺ + 2 OH⁻ → 2 H₂O
- Leftover: Zn²⁺ and 2 NO₃⁻ → Zn(NO₃)₂ (given)

$$
\boxed{1}\ \text{Zn(OH)}_2 + \boxed{2}\ \text{HNO}_3 \rightarrow \boxed{2}\ \text{H}_2\text{O} + \boxed{1}\ \text{Zn(NO}_3)_2
$$

✔ Check: Zn:1, N:2, O, H all balance.

---

5. $ \text{Al(OH)}_3 + \text{HCl} \rightarrow \_\_\_\_\_\_\_\_ + \_\_\_\_\_\_\_\_ $

- Al(OH)₃ has 3 OH⁻ → needs 3 H⁺ → 3 HCl
- 3 H⁺ + 3 OH⁻ → 3 H₂O
- Leftover: Al³⁺ and 3 Cl⁻ → AlCl₃

$$
\boxed{1}\ \text{Al(OH)}_3 + \boxed{3}\ \text{HCl} \rightarrow \boxed{3}\ \text{H}_2\text{O} + \boxed{1}\ \text{AlCl}_3
$$

✔ Check: Al:1, Cl:3, H:3+3=6 → 6 in 3 H₂O; O:3 → 3 in water.

---

## ✔ Problem 2: Complete & Balance

a) $ \_\_\_\_\_\_ \text{CsOH} + \text{H}_2\text{SO}_3 \rightarrow \_\_\_\_\_\_ \text{H}_2\text{O} + \text{Cs}_2\text{CO}_3 $

Wait — H₂SO₃ is sulfurous acid, not sulfuric. But product is Cs₂CO₃ → carbonate!

That doesn't match! Carbonate comes from H₂CO₃, not H₂SO₃.

This is likely a typo. Probably meant:

> $ \text{CsOH} + \text{H}_2\text{CO}_3 \rightarrow \text{H}_2\text{O} + \text{Cs}_2\text{CO}_3 $

But since it says H₂SO₃ and Cs₂CO₃, that can’t be right.

Wait — maybe it's a mix-up.

Let’s re-read:

> a) ___ CsOH + H₂SO₃ → ___ H₂O + Cs₂CO₃

But H₂SO₃ → sulfite ion SO₃²⁻, not CO₃²⁻.

So unless it's a mistake, this equation is invalid.

But perhaps it's supposed to be H₂CO₃ instead of H₂SO₃?

Or maybe it's H₂SO₃ and Cs₂SO₃?

Let’s assume typo: It should be:

> $ \text{CsOH} + \text{H}_2\text{SO}_3 \rightarrow \text{H}_2\text{O} + \text{Cs}_2\text{SO}_3 $

Then:

- H₂SO₃ has 2 H⁺ → needs 2 OH⁻ → 2 CsOH
- 2 H⁺ + 2 OH⁻ → 2 H₂O
- Leftover: 2 Cs⁺ and SO₃²⁻ → Cs₂SO₃

So:

$$
\boxed{2}\ \text{CsOH} + \boxed{1}\ \text{H}_2\text{SO}_3 \rightarrow \boxed{2}\ \text{H}_2\text{O} + \boxed{1}\ \text{Cs}_2\text{SO}_3
$$

But the worksheet says Cs₂CO₃ — which implies carbonate, not sulfite.

So either:
- It's a typo, or
- We're supposed to write the correct salt based on acid.

But H₂SO₃ cannot form Cs₂CO₃.

So likely H₂CO₃ was intended.

Assuming H₂CO₃ instead of H₂SO₃:

$$
\boxed{2}\ \text{CsOH} + \boxed{1}\ \text{H}_2\text{CO}_3 \rightarrow \boxed{2}\ \text{H}_2\text{O} + \boxed{1}\ \text{Cs}_2\text{CO}_3
$$

✔ This makes sense.

But since the problem says H₂SO₃, and asks for Cs₂CO₃, it's inconsistent.

Wait — maybe it's H₂CO₃ and Cs₂CO₃? Let's go with that.

Alternatively, perhaps the acid is H₂CO₃, and it's written as H₂SO₃ by mistake.

We’ll proceed assuming H₂CO₃.

But let's check the original:

> a) ___ CsOH + H₂SO₃ → ___ H₂O + Cs₂CO₃

No. That’s wrong.

Maybe it's H₂CO₃ and Cs₂CO₃ — but it says H₂SO₃.

I think it's a typo. Likely meant:

> a) ___ CsOH + H₂CO₃ → ___ H₂O + Cs₂CO₃

So:

$$
\boxed{2}\ \text{CsOH} + \boxed{1}\ \text{H}_2\text{CO}_3 \rightarrow \boxed{2}\ \text{H}_2\text{O} + \boxed{1}\ \text{Cs}_2\text{CO}_3
$$

We'll go with that.

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b) $ \_\_\_\_\_\_ \text{HF} + \text{Mg(OH)}_2 \rightarrow \_\_\_\_\_\_ \text{H}_2\text{O} + \_\_\_\_\_\_\_\_ $

- Mg(OH)₂ has 2 OH⁻ → needs 2 H⁺ → 2 HF
- 2 H⁺ + 2 OH⁻ → 2 H₂O
- Leftover: Mg²⁺ and 2 F⁻ → MgF₂

$$
\boxed{2}\ \text{HF} + \boxed{1}\ \text{Mg(OH)}_2 \rightarrow \boxed{2}\ \text{H}_2\text{O} + \boxed{1}\ \text{MgF}_2
$$

✔ Check: Mg:1, F:2, H:2+2=4 → 4 in 2 H₂O.

---

c) $ \_\_\_\_\_\_ \text{HNO}_3 + \text{Al(OH)}_3 \rightarrow \_\_\_\_\_\_ \text{H}_2\text{O} + \text{Al(NO}_3)_3 $

- Al(OH)₃ has 3 OH⁻ → needs 3 H⁺ → 3 HNO₃
- 3 H⁺ + 3 OH⁻ → 3 H₂O
- Leftover: Al³⁺ and 3 NO₃⁻ → Al(NO₃)₃ (given)

$$
\boxed{3}\ \text{HNO}_3 + \boxed{1}\ \text{Al(OH)}_3 \rightarrow \boxed{3}\ \text{H}_2\text{O} + \boxed{1}\ \text{Al(NO}_3)_3
$$

✔ Balanced.

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d) $ \text{HCl} + \text{KOH} \rightarrow \_\_\_\_\_\_ + \_\_\_\_\_\_\_\_ $

- HCl + KOH → H₂O + KCl
- 1 H⁺ + 1 OH⁻ → 1 H₂O
- Leftover: K⁺ and Cl⁻ → KCl

$$
\boxed{1}\ \text{HCl} + \boxed{1}\ \text{KOH} \rightarrow \boxed{1}\ \text{H}_2\text{O} + \boxed{1}\ \text{KCl}
$$

✔ Simple neutralization.

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e) $ \text{HBrO}_3 + \text{LiOH} \rightarrow \text{H}_2\text{O} + \_\_\_\_\_\_\_\_ $

- HBrO₃ has 1 H⁺ → needs 1 OH⁻ → 1 LiOH
- H⁺ + OH⁻ → H₂O
- Leftover: Li⁺ and BrO₃⁻ → LiBrO₃

$$
\boxed{1}\ \text{HBrO}_3 + \boxed{1}\ \text{LiOH} \rightarrow \boxed{1}\ \text{H}_2\text{O} + \boxed{1}\ \text{LiBrO}_3
$$

✔ Balanced.

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## ✔ Final Answers:

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Problem 1:

1. $ \boxed{1} \text{H}_2\text{SO}_4 + \boxed{1} \text{Sr(OH)}_2 \rightarrow \boxed{2} \text{H}_2\text{O} + \boxed{1} \text{SrSO}_4 $

2. $ \boxed{3} \text{Ca(OH)}_2 + \boxed{2} \text{H}_3\text{PO}_4 \rightarrow \boxed{6} \text{H}_2\text{O} + \boxed{1} \text{Ca}_3(\text{PO}_4)_2 $

3. $ \boxed{2} \text{HBr} + \boxed{1} \text{Ba(OH)}_2 \rightarrow \boxed{2} \text{H}_2\text{O} + \boxed{1} \text{BaBr}_2 $

4. $ \boxed{1} \text{Zn(OH)}_2 + \boxed{2} \text{HNO}_3 \rightarrow \boxed{2} \text{H}_2\text{O} + \boxed{1} \text{Zn(NO}_3)_2 $

5. $ \boxed{1} \text{Al(OH)}_3 + \boxed{3} \text{HCl} \rightarrow \boxed{3} \text{H}_2\text{O} + \boxed{1} \text{AlCl}_3 $

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Problem 2:

a) $ \boxed{2} \text{CsOH} + \text{H}_2\text{CO}_3 \rightarrow \boxed{2} \text{H}_2\text{O} + \boxed{1} \text{Cs}_2\text{CO}_3 $
*(Note: Likely typo — H₂SO₃ should be H₂CO₃)*

b) $ \boxed{2} \text{HF} + \boxed{1} \text{Mg(OH)}_2 \rightarrow \boxed{2} \text{H}_2\text{O} + \boxed{1} \text{MgF}_2 $

c) $ \boxed{3} \text{HNO}_3 + \boxed{1} \text{Al(OH)}_3 \rightarrow \boxed{3} \text{H}_2\text{O} + \boxed{1} \text{Al(NO}_3)_3 $

d) $ \boxed{1} \text{HCl} + \boxed{1} \text{KOH} \rightarrow \boxed{1} \text{H}_2\text{O} + \boxed{1} \text{KCl} $

e) $ \boxed{1} \text{HBrO}_3 + \boxed{1} \text{LiOH} \rightarrow \boxed{1} \text{H}_2\text{O} + \boxed{1} \text{LiBrO}_3 $

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📝 Summary Tips:

- Always balance H⁺ and OH⁻ to form H₂O.
- Then combine leftover ions to form salt.
- Write metal first in salt (e.g., NaCl, not ClNa).
- Use PT (Periodic Table) to confirm charges (e.g., Ca²⁺, F⁻ → CaF₂).

Let me know if you'd like a printable version or explanations for any step!