Let’s go through each problem one by one. We’ll write the correct chemical formulas first, then balance the equations so that the number of atoms of each element is the same on both sides.
---
1. phosphorus (P₄) + oxygen → diphosphorus pentoxide
- Phosphorus is given as P₄.
- Oxygen is O₂ (diatomic).
- Diphosphorus pentoxide = P₂O₅ (but we need to balance with P₄).
Unbalanced:
P₄ + O₂ → P₂O₅
Balance P: 4 P on left → need 2 P₂O₅ on right (gives 4 P and 10 O)
Now O: 10 O on right → need 5 O₂ on left (5 × 2 = 10)
✔ Balanced:
P₄ + 5O₂ → 2P₂O₅
---
2. potassium hydroxide → potassium oxide + water
- Potassium hydroxide = KOH
- Potassium oxide = K₂O (K⁺ and O²⁻)
- Water = H₂O
Unbalanced:
KOH → K₂O + H₂O
Balance K: 2 K on right → need 2 KOH on left
Now H: 2 H on left → matches H₂O on right
O: 2 O on left (from 2 KOH), 1 in K₂O + 1 in H₂O = 2 → good
✔ Balanced:
2KOH → K₂O + H₂O
---
3. aluminum + copper(II) nitrate → aluminum nitrate + copper
- Aluminum = Al
- Copper(II) nitrate = Cu(NO₃)₂
- Aluminum nitrate = Al(NO₃)₃ (Al³⁺, NO₃⁻)
- Copper = Cu
Unbalanced:
Al + Cu(NO₃)₂ → Al(NO₃)₃ + Cu
Balance NO₃: LCM of 2 and 3 is 6 → use 3 Cu(NO₃)₂ and 2 Al(NO₃)₃
That gives 6 NO₃ on each side.
Now Cu: 3 on left → 3 Cu on right
Al: 2 on right → 2 Al on left
✔ Balanced:
2Al + 3Cu(NO₃)₂ → 2Al(NO₃)₃ + 3Cu
---
4. pentane (C₅H₁₂) + oxygen → carbon dioxide + water
Combustion reaction.
Unbalanced:
C₅H₁₂ + O₂ → CO₂ + H₂O
Balance C: 5 on left → 5 CO₂ on right
Balance H: 12 on left → 6 H₂O on right (6×2=12 H)
Now O: Right side = 5×2 + 6×1 = 10 + 6 = 16 O → need 8 O₂ on left (8×2=16)
✔ Balanced:
C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O
---
5. ammonium dichromate → chromium(III) oxide + nitrogen + water
- Ammonium dichromate = (NH₄)₂Cr₂O₇
- Chromium(III) oxide = Cr₂O₃
- Nitrogen = N₂
- Water = H₂O
Unbalanced:
(NH₄)₂Cr₂O₇ → Cr₂O₃ + N₂ + H₂O
Check atoms:
Left: N=2, H=8, Cr=2, O=7
Right: Cr₂O₃ → Cr=2, O=3; N₂ → N=2; H₂O → H=2, O=1 per molecule
We have 8 H on left → need 4 H₂O on right (4×2=8 H, 4 O)
Total O on right: Cr₂O₃ (3) + 4 H₂O (4) = 7 → matches left!
N: 2 on left → N₂ on right → good
Cr: 2 on both sides → good
✔ Balanced:
(NH₄)₂Cr₂O₇ → Cr₂O₃ + N₂ + 4H₂O
---
6. barium chloride + sodium hydroxide → sodium chloride + barium hydroxide
- Barium chloride = BaCl₂
- Sodium hydroxide = NaOH
- Sodium chloride = NaCl
- Barium hydroxide = Ba(OH)₂
Unbalanced:
BaCl₂ + NaOH → NaCl + Ba(OH)₂
Balance Cl: 2 on left → 2 NaCl on right
Balance Na: 2 on right → 2 NaOH on left
Balance OH: 2 on left → Ba(OH)₂ has 2 OH → good
Ba: 1 on each side → good
✔ Balanced:
BaCl₂ + 2NaOH → 2NaCl + Ba(OH)₂
---
7. nickel chlorate → nickel chloride + oxygen
- Nickel chlorate = Ni(ClO₃)₂ (Ni²⁺, ClO₃⁻)
- Nickel chloride = NiCl₂
- Oxygen = O₂
Unbalanced:
Ni(ClO₃)₂ → NiCl₂ + O₂
Left: Ni=1, Cl=2, O=6
Right: Ni=1, Cl=2, O=2 (in O₂) → need 3 O₂ to get 6 O
✔ Balanced:
Ni(ClO₃)₂ → NiCl₂ + 3O₂
---
8. aluminum + hydrochloric acid → aluminum chloride + hydrogen
- Aluminum = Al
- Hydrochloric acid = HCl
- Aluminum chloride = AlCl₃ (Al³⁺, Cl⁻)
- Hydrogen = H₂
Unbalanced:
Al + HCl → AlCl₃ + H₂
Balance Cl: 3 on right → 3 HCl on left
But H: 3 on left → not even for H₂ (needs even number)
Try 2 Al → 2 AlCl₃ → needs 6 Cl → 6 HCl
Then H: 6 on left → 3 H₂ on right
Check:
Left: Al=2, H=6, Cl=6
Right: Al=2, Cl=6, H=6 → perfect
✔ Balanced:
2Al + 6HCl → 2AlCl₃ + 3H₂
---
Final Answer:
1. P₄ + 5O₂ → 2P₂O₅
2. 2KOH → K₂O + H₂O
3. 2Al + 3Cu(NO₃)₂ → 2Al(NO₃)₃ + 3Cu
4. C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O
5. (NH₄)₂Cr₂O₇ → Cr₂O₃ + N₂ + 4H₂O
6. BaCl₂ + 2NaOH → 2NaCl + Ba(OH)₂
7. Ni(ClO₃)₂ → NiCl₂ + 3O₂
8. 2Al + 6HCl → 2AlCl₃ + 3H₂
Parent Tip: Review the logic above to help your child master the concept of chemical word equations worksheet.
