Worksheet for writing chemical formulas and balancing chemical equations.

Let's solve each part of the problem step by step.

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1. Write the chemical formula of the following substances:

We will determine the correct chemical formulas based on the names and known ionic charges.

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a) Nitric acid
- Nitric acid is a common strong acid composed of hydrogen (H⁺) and nitrate ion (NO₃⁻).
- Formula: HNO₃

✔ Ans: HNO₃

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b) Sulfuric acid
- Sulfuric acid contains hydrogen (H⁺) and sulfate ion (SO₄²⁻).
- Two H⁺ ions are needed to balance SO₄²⁻.
- Formula: H₂SO₄

✔ Ans: H₂SO₄

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c) Magnesium hydroxide
- Magnesium ion: Mg²⁺
- Hydroxide ion: OH⁻
- Need two OH⁻ to balance Mg²⁺ → Mg(OH)₂
- Note: Parentheses are used because there are two OH groups.

✔ Ans: Mg(OH)₂

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d) Ammonium chloride
- Ammonium ion: NH₄⁺
- Chloride ion: Cl⁻
- One-to-one ratio: NH₄Cl

✔ Ans: NH₄Cl

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e) Sodium hydrogen carbonate
- Sodium: Na⁺
- Hydrogen carbonate (bicarbonate): HCO₃⁻
- Combine: NaHCO₃

✔ Ans: NaHCO₃

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f) Copper (I) oxide
- Copper (I) means Cu⁺
- Oxide: O²⁻
- To balance: need two Cu⁺ for one O²⁻ → Cu₂O

✔ Ans: Cu₂O

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g) Zinc oxide
- Zinc: Zn²⁺
- Oxide: O²⁻
- Ratio 1:1 → ZnO

✔ Ans: ZnO

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h) Iron (III) sulfate
- Iron (III): Fe³⁺
- Sulfate: SO₄²⁻
- Find least common multiple:
- Charge: Fe³⁺ and SO₄²⁻ → need 2 Fe³⁺ (total +6) and 3 SO₄²⁻ (total -6)
→ Fe₂(SO₄)₃

✔ Ans: Fe₂(SO₄)₃

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i) Silver sulfide
- Silver: Ag⁺
- Sulfide: S²⁻
- Need two Ag⁺ per S²⁻ → Ag₂S

✔ Ans: Ag₂S

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j) Potassium nitride
- Potassium: K⁺
- Nitride: N³⁻
- Need three K⁺ per N³⁻ → K₃N

✔ Ans: K₃N

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k) Lead (IV) nitrate
- Lead (IV): Pb⁴⁺
- Nitrate: NO₃⁻
- Need four NO₃⁻ to balance Pb⁴⁺ → Pb(NO₃)₄

✔ Ans: Pb(NO₃)₄

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l) Diboron trioxide
- "Di" = 2 boron, "tri" = 3 oxygen
- Formula: B₂O₃

✔ Ans: B₂O₃

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m) Hydrogen gas
- Elemental hydrogen is diatomic → H₂

✔ Ans: H₂

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n) Bromine gas
- Bromine is also diatomic in elemental form → Br₂

✔ Ans: Br₂

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o) Ammonia
- Ammonia is NH₃ (common compound)

✔ Ans: NH₃

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✔ Summary of Answers for Part 1:

| Substance | Formula |
|---------|--------|
| a) Nitric acid | HNO₃ |
| b) Sulfuric acid | H₂SO₄ |
| c) Magnesium hydroxide | Mg(OH)₂ |
| d) Ammonium chloride | NH₄Cl |
| e) Sodium hydrogen carbonate | NaHCO₃ |
| f) Copper (I) oxide | Cu₂O |
| g) Zinc oxide | ZnO |
| h) Iron (III) sulfate | Fe₂(SO₄)₃ |
| i) Silver sulfide | Ag₂S |
| j) Potassium nitride | K₃N |
| k) Lead (IV) nitrate | Pb(NO₃)₄ |
| l) Diboron trioxide | B₂O₃ |
| m) Hydrogen gas | H₂ |
| n) Bromine gas | Br₂ |
| o) Ammonia | NH₃ |

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2. Balance the following chemical equations:

We will balance each equation using the method of adjusting coefficients.

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a) ___ SnO₂ + ___ H₂ → ___ Sn + ___ H₂O

Unbalanced:
SnO₂ + H₂ → Sn + H₂O

- Sn: 1 on both sides → OK
- O: 2 on left, 1 on right → need 2 H₂O
- H: 2 on left, but 4 on right (if 2 H₂O) → need 2 H₂

Try:
SnO₂ + 2H₂ → Sn + 2H₂O

Check:
- Sn: 1 = 1 ✔
- O: 2 = 2 ✔
- H: 4 = 4 ✔

✔ Balanced: SnO₂ + 2H₂ → Sn + 2H₂O

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b) ___ KOH + ___ H₃PO₄ → ___ K₃PO₄ + ___ H₂O

Unbalanced:
KOH + H₃PO₄ → K₃PO₄ + H₂O

- K: 1 on left, 3 on right → need 3 KOH
- P: 1 on both sides → OK
- H: from 3 KOH (3H) + H₃PO₄ (3H) = 6H total on left
- On right: H₂O has 2H per molecule → need 3 H₂O for 6H
- O: check later

Try:
3KOH + H₃PO₄ → K₃PO₄ + 3H₂O

Check atoms:
- K: 3 = 3 ✔
- O: 3 (from KOH) + 4 (from H₃PO₄) = 7; right: 4 (K₃PO₄) + 3 (H₂O) = 7 ✔
- H: 3 (KOH) + 3 (H₃PO₄) = 6; right: 3×2 = 6 ✔
- P: 1 = 1 ✔

✔ Balanced: 3KOH + H₃PO₄ → K₃PO₄ + 3H₂O

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c) ___ KNO₃ + ___ H₂CO₃ → ___ K₂CO₃ + ___ HNO₃

Unbalanced:
KNO₃ + H₂CO₃ → K₂CO₃ + HNO₃

- K: 1 on left, 2 on right → need 2 KNO₃
- CO₃: 1 on both sides → OK
- H: 2 on left, 1 on right → need 2 HNO₃
- N: 2 on left (from 2 KNO₃), 2 on right (2 HNO₃) → OK

Try:
2KNO₃ + H₂CO₃ → K₂CO₃ + 2HNO₃

Check:
- K: 2 = 2 ✔
- N: 2 = 2 ✔
- C: 1 = 1 ✔
- O: Left: 6 (2KNO₃) + 3 (H₂CO₃) = 9; Right: 3 (K₂CO₃) + 6 (2HNO₃) = 9 ✔
- H: 2 = 2 ✔

✔ Balanced: 2KNO₃ + H₂CO₃ → K₂CO₃ + 2HNO₃

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d) ___ Na₃PO₄ + ___ HCl → ___ NaCl + ___ H₃PO₄

Unbalanced:
Na₃PO₄ + HCl → NaCl + H₃PO₄

- Na: 3 on left, 1 on right → need 3 NaCl
- Cl: 3 on right → need 3 HCl
- PO₄: 1 on both sides → OK
- H: 3 on right (H₃PO₄), 3 on left (3 HCl) → OK

Try:
Na₃PO₄ + 3HCl → 3NaCl + H₃PO₄

Check:
- Na: 3 = 3 ✔
- P: 1 = 1 ✔
- O: 4 = 4 ✔
- H: 3 = 3 ✔
- Cl: 3 = 3 ✔

✔ Balanced: Na₃PO₄ + 3HCl → 3NaCl + H₃PO₄

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✔ Final Answers for Part 2:

| Equation | Balanced Equation |
|--------|-------------------|
| a) | SnO₂ + 2H₂ → Sn + 2H₂O |
| b) | 3KOH + H₃PO₄ → K₃PO₄ + 3H₂O |
| c) | 2KNO₃ + H₂CO₃ → K₂CO₃ + 2HNO₃ |
| d) | Na₃PO₄ + 3HCl → 3NaCl + H₃PO₄ |

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✔ Final Answer Summary:

#### 1. Chemical Formulas:
a) HNO₃
b) H₂SO₄
c) Mg(OH)₂
d) NH₄Cl
e) NaHCO₃
f) Cu₂O
g) ZnO
h) Fe₂(SO₄)₃
i) Ag₂S
j) K₃N
k) Pb(NO₃)₄
l) B₂O₃
m) H₂
n) Br₂
o) NH₃

#### 2. Balanced Equations:
a) SnO₂ + 2H₂ → Sn + 2H₂O
b) 3KOH + H₃PO₄ → K₃PO₄ + 3H₂O
c) 2KNO₃ + H₂CO₃ → K₂CO₃ + 2HNO₃
d) Na₃PO₄ + 3HCl → 3NaCl + H₃PO₄

Let me know if you'd like these written out neatly for submission!