Worksheet for writing balanced chemical equations from word equations.

Explanation:
Let's solve each problem step by step. We need to write balanced chemical equations from the word equations. First, we identify the correct chemical formulas for each compound, then write the unbalanced equation, and finally balance it.

1. zinc chromate + iron III carbonate → zinc carbonate + iron III chromate
- Zinc chromate: ZnCrO₄ (Zn²⁺, CrO₄²⁻)
- Iron III carbonate: Fe₂(CO₃)₃ (Fe³⁺, CO₃²⁻)
- Zinc carbonate: ZnCO₃
- Iron III chromate: Fe₂(CrO₄)₃
Unbalanced: ZnCrO₄ + Fe₂(CO₃)₃ → ZnCO₃ + Fe₂(CrO₄)₃
Now balance:
We need 3 ZnCrO₄ to give 3 CrO₄, and 2 Fe₂(CrO₄)₃ needs 6 CrO₄ → so 6 ZnCrO₄
But wait, let's check charges. Actually, chromate is CrO₄²⁻, so zinc chromate is ZnCrO₄ (Zn²⁺, CrO₄²⁻) — that’s correct.
Iron III chromate: Fe³⁺ and CrO₄²⁻ → Fe₂(CrO₄)₃
Zinc carbonate: ZnCO₃
Iron III carbonate: Fe₂(CO₃)₃
So unbalanced:
ZnCrO₄ + Fe₂(CO₃)₃ → ZnCO₃ + Fe₂(CrO₄)₃
Now balance:
Left: Zn=1, Cr=1, O=4, Fe=2, C=3, O=9 → total O=13
Right: Zn=1, C=1, O=3, Fe=2, Cr=3, O=12 → total O=15
We need to balance Cr and CO₃.
Let’s try 3 ZnCrO₄ → 3 Zn, 3 Cr
Then Fe₂(CrO₄)₃ has 2 Fe, 3 Cr
So 3 ZnCrO₄ + Fe₂(CO₃)₃ → 3 ZnCO₃ + Fe₂(CrO₄)₃
Now check atoms:
Left: Zn=3, Cr=3, O=12 (from 3×4=12), Fe=2, C=3, O=9 → total O=21
Right: Zn=3, C=3, O=9 (from 3×3=9), Fe=2, Cr=3, O=12 → total O=21
Oxygen: left: 3×4=12 from ZnCrO₄, 3×3=9 from CO₃ → 21
Right: 3×3=9 from ZnCO₃, 3×4=12 from CrO₄ → 21
Yes, balanced.
So: 3 ZnCrO₄ + Fe₂(CO₃)₃ → 3 ZnCO₃ + Fe₂(CrO₄)₃

2. barium hypochlorite + iron III carbonate → iron III hypochlorite + barium carbonate
- Barium hypochlorite: Ba(ClO)₂ (Ba²⁺, ClO⁻)
- Iron III carbonate: Fe₂(CO₃)₃
- Iron III hypochlorite: Fe(ClO)₃
- Barium carbonate: BaCO₃
Unbalanced: Ba(ClO)₂ + Fe₂(CO₃)₃ → Fe(ClO)₃ + BaCO₃
Balance:
We need 3 Ba(ClO)₂ to give 3 Ba, 6 ClO
But Fe(ClO)₃ has 1 Fe, 3 ClO → so 2 Fe(ClO)₃ needs 6 ClO
So 3 Ba(ClO)₂ → 3 Ba, 6 ClO
2 Fe(ClO)₃ → 2 Fe, 6 ClO
Now Fe₂(CO₃)₃ has 2 Fe, 3 CO₃
BaCO₃ has 1 Ba, 1 CO₃ → so 3 BaCO₃ → 3 Ba, 3 CO₃
So: 3 Ba(ClO)₂ + Fe₂(CO₃)₃ → 2 Fe(ClO)₃ + 3 BaCO₃
Check:
Left: Ba=3, ClO=6, Fe=2, C=3, O=9
Right: Fe=2, ClO=6, Ba=3, C=3, O=9
Balanced.

3. cobalt II chlorate → cobalt II chloride + oxygen
- Cobalt II chlorate: Co(ClO₃)₂
- Cobalt II chloride: CoCl₂
- Oxygen: O₂
Unbalanced: Co(ClO₃)₂ → CoCl₂ + O₂
Chlorate decomposes: ClO₃⁻ → Cl⁻ + O₂
So Co(ClO₃)₂ → CoCl₂ + 3 O₂
Check: Left: Co=1, Cl=2, O=6
Right: Co=1, Cl=2, O=6
Balanced.

4. copper I hydroxide + nickel II sulfate → nickel II hydroxide + cuprous sulfate
- Copper I hydroxide: CuOH (Cu⁺, OH⁻)
- Nickel II sulfate: NiSO₄
- Nickel II hydroxide: Ni(OH)₂
- Cuprous sulfate: Cu₂SO₄ (Cu⁺, SO₄²⁻)
Unbalanced: CuOH + NiSO₄ → Ni(OH)₂ + Cu₂SO₄
Balance:
Left: Cu=1, O=1, H=1, Ni=1, S=1, O=4
Right: Ni=1, O=2, H=2, Cu=2, S=1, O=4
So need 2 CuOH → 2 Cu, 2 O, 2 H
Right: Ni(OH)₂ has 2 H, 2 O
So 2 CuOH + NiSO₄ → Ni(OH)₂ + Cu₂SO₄
Check:
Left: Cu=2, O=2+4=6, H=2, Ni=1, S=1
Right: Ni=1, O=2+4=6, H=2, Cu=2, S=1
Balanced.

5. sulfuric acid + strontium fluoride → strontium sulfate + hydrogen fluoride
- Sulfuric acid: H₂SO₄
- Strontium fluoride: SrF₂
- Strontium sulfate: SrSO₄
- Hydrogen fluoride: HF
Unbalanced: H₂SO₄ + SrF₂ → SrSO₄ + HF
Balance:
Left: H=2, S=1, O=4, Sr=1, F=2
Right: Sr=1, S=1, O=4, H=1, F=1
So need 2 HF → 2 H, 2 F
So: H₂SO₄ + SrF₂ → SrSO₄ + 2 HF
Balanced.

6. sodium tartrate + mercury II nitrate → sodium nitrate + mercury II tartrate
- Sodium tartrate: Na₂C₄H₄O₆ (tartrate is C₄H₄O₆²⁻)
- Mercury II nitrate: Hg(NO₃)₂
- Sodium nitrate: NaNO₃
- Mercury II tartrate: HgC₄H₄O₆
Unbalanced: Na₂C₄H₄O₆ + Hg(NO₃)₂ → NaNO₃ + HgC₄H₄O₆
Balance:
Left: Na=2, C₄H₄O₆=1, Hg=1, N=2, O=6
Right: Na=1, N=1, O=3, Hg=1, C₄H₄O₆=1
So need 2 NaNO₃ → 2 Na, 2 N, 6 O
So: Na₂C₄H₄O₆ + Hg(NO₃)₂ → 2 NaNO₃ + HgC₄H₄O₆
Balanced.

7. cadmium + tin IV sulfite → tin + cadmium sulfite
- Cadmium: Cd
- Tin IV sulfite: Sn(SO₃)₂ (Sn⁴⁺, SO₃²⁻)
- Tin: Sn
- Cadmium sulfite: CdSO₃
Unbalanced: Cd + Sn(SO₃)₂ → Sn + CdSO₃
Balance:
Left: Cd=1, Sn=1, S=2, O=6
Right: Sn=1, Cd=1, S=1, O=3
So need 2 CdSO₃ → 2 Cd, 2 S, 6 O
But Cd is only 1 on left → so 2 Cd
So: 2 Cd + Sn(SO₃)₂ → Sn + 2 CdSO₃
Balanced.

8. hydrogen + nitrogen → ammonia
- Hydrogen: H₂
- Nitrogen: N₂
- Ammonia: NH₃
Unbalanced: H₂ + N₂ → NH₃
Balance:
Left: H=2, N=2
Right: N=1, H=3
So need 2 NH₃ → 2 N, 6 H
So need 3 H₂ → 6 H
So: 3 H₂ + N₂ → 2 NH₃
Balanced.

9. lead IV carbonate → lead IV oxide + carbon dioxide
- Lead IV carbonate: Pb(CO₃)₂ (Pb⁴⁺, CO₃²⁻)
- Lead IV oxide: PbO₂
- Carbon dioxide: CO₂
Unbalanced: Pb(CO₃)₂ → PbO₂ + CO₂
Balance:
Left: Pb=1, C=2, O=6
Right: Pb=1, O=2, C=1, O=2 → total O=4, C=1
So need 2 CO₂ → 2 C, 4 O
So: Pb(CO₃)₂ → PbO₂ + 2 CO₂
Check: Left: Pb=1, C=2, O=6
Right: Pb=1, O=2, C=2, O=4 → total O=6
Balanced.

10. chromium III hydroxide → chromium III oxide + water
- Chromium III hydroxide: Cr(OH)₃
- Chromium III oxide: Cr₂O₃
- Water: H₂O
Unbalanced: Cr(OH)₃ → Cr₂O₃ + H₂O
Balance:
Left: Cr=1, O=3, H=3
Right: Cr=2, O=3, H=2, O=1 → total O=4, H=2
So need 2 Cr(OH)₃ → 2 Cr, 6 O, 6 H
Right: Cr₂O₃ → 2 Cr, 3 O
So need 3 H₂O → 6 H, 3 O
Total right: Cr=2, O=3+3=6, H=6
So: 2 Cr(OH)₃ → Cr₂O₃ + 3 H₂O
Balanced.

Final Answer:
1. 3 ZnCrO₄ + Fe₂(CO₃)₃ → 3 ZnCO₃ + Fe₂(CrO₄)₃
2. 3 Ba(ClO)₂ + Fe₂(CO₃)₃ → 2 Fe(ClO)₃ + 3 BaCO₃
3. Co(ClO₃)₂ → CoCl₂ + 3 O₂
4. 2 CuOH + NiSO₄ → Ni(OH)₂ + Cu₂SO₄
5. H₂SO₄ + SrF₂ → SrSO₄ + 2 HF
6. Na₂C₄H₄O₆ + Hg(NO₃)₂ → 2 NaNO₃ + HgC₄H₄O₆
7. 2 Cd + Sn(SO₃)₂ → Sn + 2 CdSO₃
8. 3 H₂ + N₂ → 2 NH₃
9. Pb(CO₃)₂ → PbO₂ + 2 CO₂
10. 2 Cr(OH)₃ → Cr₂O₃ + 3 H₂O