Free Printable Balancing and Classifying Chemical Equations Worksheets - Free Printable
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Step-by-step solution for: Free Printable Balancing and Classifying Chemical Equations Worksheets
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Show Answer Key & Explanations
Step-by-step solution for: Free Printable Balancing and Classifying Chemical Equations Worksheets
Let’s go through each reaction one by one and balance them. Remember: in a balanced chemical equation, the number of atoms of each element must be the same on both sides (left and right).
We’ll count atoms for each element and adjust coefficients (the numbers in front) until everything matches.
---
i. HBr + NaOH → NaBr + H₂O
Left: H=1+1=2, Br=1, Na=1, O=1
Right: Na=1, Br=1, H=2, O=1 → Already balanced!
✔ Answer: 1 HBr + 1 NaOH → 1 NaBr + 1 H₂O
---
ii. H₂SO₄ + KOH → K₂SO₄ + H₂O
Left: H=2+1=3? Wait — let’s write with variables first.
Try:
H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
Check:
Left: H=2+2=4, S=1, O=4+2=6, K=2
Right: K=2, S=1, O=4+2=6, H=4 → Balanced!
✔ Answer: 1 H₂SO₄ + 2 KOH → 1 K₂SO₄ + 2 H₂O
---
iii. HCl + Ca(OH)₂ → CaCl₂ + H₂O
Ca(OH)₂ has 2 OH groups → needs 2 H⁺ to make 2 H₂O.
So:
2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O
Check:
Left: H=2+2=4, Cl=2, Ca=1, O=2
Right: Ca=1, Cl=2, H=4, O=2 → Balanced!
✔ Answer: 2 HCl + 1 Ca(OH)₂ → 1 CaCl₂ + 2 H₂O
---
iv. Fe(OH)₃ + H₂SO₄ → Fe₂(SO₄)₃ + H₂O
Fe₂(SO₄)₃ means 2 Fe and 3 SO₄ groups.
So we need 2 Fe(OH)₃ and 3 H₂SO₄.
Then water: 2×3 = 6 OH from base, 3×2 = 6 H from acid → makes 6 H₂O.
Equation:
2Fe(OH)₃ + 3H₂SO₄ → Fe₂(SO₄)₃ + 6H₂O
Check:
Left: Fe=2, O=6+12=18, H=6+6=12, S=3
Right: Fe=2, S=3, O=12+6=18, H=12 → Balanced!
✔ Answer: 2 Fe(OH)₃ + 3 H₂SO₄ → 1 Fe₂(SO₄)₃ + 6 H₂O
---
v. H₂SO₄ + B(OH)₃ → B₂(SO₄)₃ + H₂O
B₂(SO₄)₃ → 2 B, 3 SO₄
So need 2 B(OH)₃ and 3 H₂SO₄
Water: 2×3 = 6 OH, 3×2 = 6 H → 6 H₂O
Equation:
3H₂SO₄ + 2B(OH)₃ → B₂(SO₄)₃ + 6H₂O
Check:
Left: H=6+6=12, S=3, O=12+6=18, B=2
Right: B=2, S=3, O=12+6=18, H=12 → Balanced!
✔ Answer: 3 H₂SO₄ + 2 B(OH)₃ → 1 B₂(SO₄)₃ + 6 H₂O
---
vi. Pb(OH)₂ + HCl → PbCl₂ + H₂O
Pb(OH)₂ has 2 OH → needs 2 HCl → makes 2 H₂O
Equation:
Pb(OH)₂ + 2HCl → PbCl₂ + 2H₂O
Check:
Left: Pb=1, O=2, H=2+2=4, Cl=2
Right: Pb=1, Cl=2, H=4, O=2 → Balanced!
✔ Answer: 1 Pb(OH)₂ + 2 HCl → 1 PbCl₂ + 2 H₂O
---
vii. H₂SO₄ + NH₄OH → (NH₄)₂SO₄ + H₂O
(NH₄)₂SO₄ needs 2 NH₄⁺ → so 2 NH₄OH
Then: H₂SO₄ + 2NH₄OH → (NH₄)₂SO₄ + 2H₂O
Check:
Left: H=2+8=10? Wait — better break down:
H₂SO₄: H=2, S=1, O=4
2NH₄OH: N=2, H=8+2=10? Actually NH₄OH is NH₄⁺ and OH⁻ → formula unit has N, 5H, O
Wait — standard way: NH₄OH as written has N, 5H, O per molecule? Let's treat it as is.
Actually, better to think ionically, but for balancing:
Assume NH₄OH = NH₅O? No — actually, it’s often written as aqueous ammonia, but here we take it as given.
Standard balancing:
To get (NH₄)₂SO₄, you need 2 NH₄OH and 1 H₂SO₄ → produces 2 H₂O
Equation:
H₂SO₄ + 2NH₄OH → (NH₄)₂SO₄ + 2H₂O
Atoms:
Left: H=2 + 2*(5)=12? Wait — NH₄OH: N=1, H=5, O=1? That’s not standard.
Actually, NH₄OH is ammonium hydroxide — it’s NH₃(aq) + H₂O, but in formulas, we treat it as having 1 N, 5 H, 1 O? Let me check atom count properly.
Better approach: Think of it as providing OH⁻.
Each NH₄OH provides 1 OH⁻. H₂SO₄ provides 2 H⁺ → needs 2 OH⁻ → so 2 NH₄OH.
Products: (NH₄)₂SO₄ and 2 H₂O.
Now count atoms:
Left:
H₂SO₄: H=2, S=1, O=4
2NH₄OH: N=2, H=2*5=10? But NH₄OH is typically considered as N, 5H, O — yes.
Total left: N=2, H=2+10=12, S=1, O=4+2=6
Right:
(NH₄)₂SO₄: N=2, H=8, S=1, O=4
2H₂O: H=4, O=2 → total H=12, O=6 → matches!
✔ Answer: 1 H₂SO₄ + 2 NH₄OH → 1 (NH₄)₂SO₄ + 2 H₂O
---
viii. H₂CO₃ + CsOH → Cs₂CO₃ + H₂O
Cs₂CO₃ needs 2 Cs → so 2 CsOH
H₂CO₃ + 2CsOH → Cs₂CO₃ + 2H₂O
Check:
Left: H=2+2=4, C=1, O=3+2=5, Cs=2
Right: Cs=2, C=1, O=3+2=5, H=4 → Balanced!
✔ Answer: 1 H₂CO₃ + 2 CsOH → 1 Cs₂CO₃ + 2 H₂O
---
ix. HF + Mg(OH)₂ → MgF₂ + H₂O
Mg(OH)₂ has 2 OH → needs 2 HF → makes 2 H₂O
Equation:
2HF + Mg(OH)₂ → MgF₂ + 2H₂O
Check:
Left: H=2+2=4, F=2, Mg=1, O=2
Right: Mg=1, F=2, H=4, O=2 → Balanced!
✔ Answer: 2 HF + 1 Mg(OH)₂ → 1 MgF₂ + 2 H₂O
---
x. HNO₃ + Al(OH)₃ → Al(NO₃)₃ + H₂O
Al(NO₃)₃ has 3 NO₃ → needs 3 HNO₃
Al(OH)₃ has 3 OH → makes 3 H₂O
Equation:
3HNO₃ + Al(OH)₃ → Al(NO₃)₃ + 3H₂O
Check:
Left: H=3+3=6, N=3, O=9+3=12, Al=1
Right: Al=1, N=3, O=9+3=12, H=6 → Balanced!
✔ Answer: 3 HNO₃ + 1 Al(OH)₃ → 1 Al(NO₃)₃ + 3 H₂O
---
xi. HNO₃ + Zn(OH)₂ → Zn(NO₃)₂ + H₂O
Zn(NO₃)₂ has 2 NO₃ → needs 2 HNO₃
Zn(OH)₂ has 2 OH → makes 2 H₂O
Equation:
2HNO₃ + Zn(OH)₂ → Zn(NO₃)₂ + 2H₂O
Check:
Left: H=2+2=4, N=2, O=6+2=8, Zn=1
Right: Zn=1, N=2, O=6+2=8, H=4 → Balanced!
✔ Answer: 2 HNO₃ + 1 Zn(OH)₂ → 1 Zn(NO₃)₂ + 2 H₂O
---
xii. H₃PO₄ + Ca(OH)₂ → Ca₃(PO₄)₂ + H₂O
Ca₃(PO₄)₂ has 3 Ca and 2 PO₄
So need 2 H₃PO₄ and 3 Ca(OH)₂
Water: 2×3 = 6 H from acid, 3×2 = 6 OH from base → 6 H₂O
Equation:
2H₃PO₄ + 3Ca(OH)₂ → Ca₃(PO₄)₂ + 6H₂O
Check:
Left: H=6+6=12, P=2, O=8+6=14, Ca=3
Right: Ca=3, P=2, O=8+6=14, H=12 → Balanced!
✔ Answer: 2 H₃PO₄ + 3 Ca(OH)₂ → 1 Ca₃(PO₄)₂ + 6 H₂O
---
xiii. HClO₃ + Al(OH)₃ → Al(ClO₃)₃ + H₂O
Al(ClO₃)₃ has 3 ClO₃ → needs 3 HClO₃
Al(OH)₃ has 3 OH → makes 3 H₂O
Equation:
3HClO₃ + Al(OH)₃ → Al(ClO₃)₃ + 3H₂O
Check:
Left: H=3+3=6, Cl=3, O=9+3=12, Al=1
Right: Al=1, Cl=3, O=9+3=12, H=6 → Balanced!
✔ Answer: 3 HClO₃ + 1 Al(OH)₃ → 1 Al(ClO₃)₃ + 3 H₂O
---
xiv. HF + Ba(OH)₂ → BaF₂ + H₂O
Ba(OH)₂ has 2 OH → needs 2 HF → makes 2 H₂O
Equation:
2HF + Ba(OH)₂ → BaF₂ + 2H₂O
Check:
Left: H=2+2=4, F=2, Ba=1, O=2
Right: Ba=1, F=2, H=4, O=2 → Balanced!
✔ Answer: 2 HF + 1 Ba(OH)₂ → 1 BaF₂ + 2 H₂O
---
xv. HCl + Al(OH)₃ → AlCl₃ + H₂O
AlCl₃ has 3 Cl → needs 3 HCl
Al(OH)₃ has 3 OH → makes 3 H₂O
Equation:
3HCl + Al(OH)₃ → AlCl₃ + 3H₂O
Check:
Left: H=3+3=6, Cl=3, Al=1, O=3
Right: Al=1, Cl=3, H=6, O=3 → Balanced!
✔ Answer: 3 HCl + 1 Al(OH)₃ → 1 AlCl₃ + 3 H₂O
---
Final Answer:
i. 1 HBr + 1 NaOH → 1 NaBr + 1 H₂O
ii. 1 H₂SO₄ + 2 KOH → 1 K₂SO₄ + 2 H₂O
iii. 2 HCl + 1 Ca(OH)₂ → 1 CaCl₂ + 2 H₂O
iv. 2 Fe(OH)₃ + 3 H₂SO₄ → 1 Fe₂(SO₄)₃ + 6 H₂O
v. 3 H₂SO₄ + 2 B(OH)₃ → 1 B₂(SO₄)₃ + 6 H₂O
vi. 1 Pb(OH)₂ + 2 HCl → 1 PbCl₂ + 2 H₂O
vii. 1 H₂SO₄ + 2 NH₄OH → 1 (NH₄)₂SO₄ + 2 H₂O
viii. 1 H₂CO₃ + 2 CsOH → 1 Cs₂CO₃ + 2 H₂O
ix. 2 HF + 1 Mg(OH)₂ → 1 MgF₂ + 2 H₂O
x. 3 HNO₃ + 1 Al(OH)₃ → 1 Al(NO₃)₃ + 3 H₂O
xi. 2 HNO₃ + 1 Zn(OH)₂ → 1 Zn(NO₃)₂ + 2 H₂O
xii. 2 H₃PO₄ + 3 Ca(OH)₂ → 1 Ca₃(PO)₂ + 6 H₂O
xiii. 3 HClO₃ + 1 Al(OH)₃ → 1 Al(ClO₃)₃ + 3 H₂O
xiv. 2 HF + 1 Ba(OH)₂ → 1 BaF₂ + 2 H₂O
xv. 3 HCl + 1 Al(OH)₃ → 1 AlCl₃ + 3 H₂O
We’ll count atoms for each element and adjust coefficients (the numbers in front) until everything matches.
---
i. HBr + NaOH → NaBr + H₂O
Left: H=1+1=2, Br=1, Na=1, O=1
Right: Na=1, Br=1, H=2, O=1 → Already balanced!
✔ Answer: 1 HBr + 1 NaOH → 1 NaBr + 1 H₂O
---
ii. H₂SO₄ + KOH → K₂SO₄ + H₂O
Left: H=2+1=3? Wait — let’s write with variables first.
Try:
H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
Check:
Left: H=2+2=4, S=1, O=4+2=6, K=2
Right: K=2, S=1, O=4+2=6, H=4 → Balanced!
✔ Answer: 1 H₂SO₄ + 2 KOH → 1 K₂SO₄ + 2 H₂O
---
iii. HCl + Ca(OH)₂ → CaCl₂ + H₂O
Ca(OH)₂ has 2 OH groups → needs 2 H⁺ to make 2 H₂O.
So:
2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O
Check:
Left: H=2+2=4, Cl=2, Ca=1, O=2
Right: Ca=1, Cl=2, H=4, O=2 → Balanced!
✔ Answer: 2 HCl + 1 Ca(OH)₂ → 1 CaCl₂ + 2 H₂O
---
iv. Fe(OH)₃ + H₂SO₄ → Fe₂(SO₄)₃ + H₂O
Fe₂(SO₄)₃ means 2 Fe and 3 SO₄ groups.
So we need 2 Fe(OH)₃ and 3 H₂SO₄.
Then water: 2×3 = 6 OH from base, 3×2 = 6 H from acid → makes 6 H₂O.
Equation:
2Fe(OH)₃ + 3H₂SO₄ → Fe₂(SO₄)₃ + 6H₂O
Check:
Left: Fe=2, O=6+12=18, H=6+6=12, S=3
Right: Fe=2, S=3, O=12+6=18, H=12 → Balanced!
✔ Answer: 2 Fe(OH)₃ + 3 H₂SO₄ → 1 Fe₂(SO₄)₃ + 6 H₂O
---
v. H₂SO₄ + B(OH)₃ → B₂(SO₄)₃ + H₂O
B₂(SO₄)₃ → 2 B, 3 SO₄
So need 2 B(OH)₃ and 3 H₂SO₄
Water: 2×3 = 6 OH, 3×2 = 6 H → 6 H₂O
Equation:
3H₂SO₄ + 2B(OH)₃ → B₂(SO₄)₃ + 6H₂O
Check:
Left: H=6+6=12, S=3, O=12+6=18, B=2
Right: B=2, S=3, O=12+6=18, H=12 → Balanced!
✔ Answer: 3 H₂SO₄ + 2 B(OH)₃ → 1 B₂(SO₄)₃ + 6 H₂O
---
vi. Pb(OH)₂ + HCl → PbCl₂ + H₂O
Pb(OH)₂ has 2 OH → needs 2 HCl → makes 2 H₂O
Equation:
Pb(OH)₂ + 2HCl → PbCl₂ + 2H₂O
Check:
Left: Pb=1, O=2, H=2+2=4, Cl=2
Right: Pb=1, Cl=2, H=4, O=2 → Balanced!
✔ Answer: 1 Pb(OH)₂ + 2 HCl → 1 PbCl₂ + 2 H₂O
---
vii. H₂SO₄ + NH₄OH → (NH₄)₂SO₄ + H₂O
(NH₄)₂SO₄ needs 2 NH₄⁺ → so 2 NH₄OH
Then: H₂SO₄ + 2NH₄OH → (NH₄)₂SO₄ + 2H₂O
Check:
Left: H=2+8=10? Wait — better break down:
H₂SO₄: H=2, S=1, O=4
2NH₄OH: N=2, H=8+2=10? Actually NH₄OH is NH₄⁺ and OH⁻ → formula unit has N, 5H, O
Wait — standard way: NH₄OH as written has N, 5H, O per molecule? Let's treat it as is.
Actually, better to think ionically, but for balancing:
Assume NH₄OH = NH₅O? No — actually, it’s often written as aqueous ammonia, but here we take it as given.
Standard balancing:
To get (NH₄)₂SO₄, you need 2 NH₄OH and 1 H₂SO₄ → produces 2 H₂O
Equation:
H₂SO₄ + 2NH₄OH → (NH₄)₂SO₄ + 2H₂O
Atoms:
Left: H=2 + 2*(5)=12? Wait — NH₄OH: N=1, H=5, O=1? That’s not standard.
Actually, NH₄OH is ammonium hydroxide — it’s NH₃(aq) + H₂O, but in formulas, we treat it as having 1 N, 5 H, 1 O? Let me check atom count properly.
Better approach: Think of it as providing OH⁻.
Each NH₄OH provides 1 OH⁻. H₂SO₄ provides 2 H⁺ → needs 2 OH⁻ → so 2 NH₄OH.
Products: (NH₄)₂SO₄ and 2 H₂O.
Now count atoms:
Left:
H₂SO₄: H=2, S=1, O=4
2NH₄OH: N=2, H=2*5=10? But NH₄OH is typically considered as N, 5H, O — yes.
Total left: N=2, H=2+10=12, S=1, O=4+2=6
Right:
(NH₄)₂SO₄: N=2, H=8, S=1, O=4
2H₂O: H=4, O=2 → total H=12, O=6 → matches!
✔ Answer: 1 H₂SO₄ + 2 NH₄OH → 1 (NH₄)₂SO₄ + 2 H₂O
---
viii. H₂CO₃ + CsOH → Cs₂CO₃ + H₂O
Cs₂CO₃ needs 2 Cs → so 2 CsOH
H₂CO₃ + 2CsOH → Cs₂CO₃ + 2H₂O
Check:
Left: H=2+2=4, C=1, O=3+2=5, Cs=2
Right: Cs=2, C=1, O=3+2=5, H=4 → Balanced!
✔ Answer: 1 H₂CO₃ + 2 CsOH → 1 Cs₂CO₃ + 2 H₂O
---
ix. HF + Mg(OH)₂ → MgF₂ + H₂O
Mg(OH)₂ has 2 OH → needs 2 HF → makes 2 H₂O
Equation:
2HF + Mg(OH)₂ → MgF₂ + 2H₂O
Check:
Left: H=2+2=4, F=2, Mg=1, O=2
Right: Mg=1, F=2, H=4, O=2 → Balanced!
✔ Answer: 2 HF + 1 Mg(OH)₂ → 1 MgF₂ + 2 H₂O
---
x. HNO₃ + Al(OH)₃ → Al(NO₃)₃ + H₂O
Al(NO₃)₃ has 3 NO₃ → needs 3 HNO₃
Al(OH)₃ has 3 OH → makes 3 H₂O
Equation:
3HNO₃ + Al(OH)₃ → Al(NO₃)₃ + 3H₂O
Check:
Left: H=3+3=6, N=3, O=9+3=12, Al=1
Right: Al=1, N=3, O=9+3=12, H=6 → Balanced!
✔ Answer: 3 HNO₃ + 1 Al(OH)₃ → 1 Al(NO₃)₃ + 3 H₂O
---
xi. HNO₃ + Zn(OH)₂ → Zn(NO₃)₂ + H₂O
Zn(NO₃)₂ has 2 NO₃ → needs 2 HNO₃
Zn(OH)₂ has 2 OH → makes 2 H₂O
Equation:
2HNO₃ + Zn(OH)₂ → Zn(NO₃)₂ + 2H₂O
Check:
Left: H=2+2=4, N=2, O=6+2=8, Zn=1
Right: Zn=1, N=2, O=6+2=8, H=4 → Balanced!
✔ Answer: 2 HNO₃ + 1 Zn(OH)₂ → 1 Zn(NO₃)₂ + 2 H₂O
---
xii. H₃PO₄ + Ca(OH)₂ → Ca₃(PO₄)₂ + H₂O
Ca₃(PO₄)₂ has 3 Ca and 2 PO₄
So need 2 H₃PO₄ and 3 Ca(OH)₂
Water: 2×3 = 6 H from acid, 3×2 = 6 OH from base → 6 H₂O
Equation:
2H₃PO₄ + 3Ca(OH)₂ → Ca₃(PO₄)₂ + 6H₂O
Check:
Left: H=6+6=12, P=2, O=8+6=14, Ca=3
Right: Ca=3, P=2, O=8+6=14, H=12 → Balanced!
✔ Answer: 2 H₃PO₄ + 3 Ca(OH)₂ → 1 Ca₃(PO₄)₂ + 6 H₂O
---
xiii. HClO₃ + Al(OH)₃ → Al(ClO₃)₃ + H₂O
Al(ClO₃)₃ has 3 ClO₃ → needs 3 HClO₃
Al(OH)₃ has 3 OH → makes 3 H₂O
Equation:
3HClO₃ + Al(OH)₃ → Al(ClO₃)₃ + 3H₂O
Check:
Left: H=3+3=6, Cl=3, O=9+3=12, Al=1
Right: Al=1, Cl=3, O=9+3=12, H=6 → Balanced!
✔ Answer: 3 HClO₃ + 1 Al(OH)₃ → 1 Al(ClO₃)₃ + 3 H₂O
---
xiv. HF + Ba(OH)₂ → BaF₂ + H₂O
Ba(OH)₂ has 2 OH → needs 2 HF → makes 2 H₂O
Equation:
2HF + Ba(OH)₂ → BaF₂ + 2H₂O
Check:
Left: H=2+2=4, F=2, Ba=1, O=2
Right: Ba=1, F=2, H=4, O=2 → Balanced!
✔ Answer: 2 HF + 1 Ba(OH)₂ → 1 BaF₂ + 2 H₂O
---
xv. HCl + Al(OH)₃ → AlCl₃ + H₂O
AlCl₃ has 3 Cl → needs 3 HCl
Al(OH)₃ has 3 OH → makes 3 H₂O
Equation:
3HCl + Al(OH)₃ → AlCl₃ + 3H₂O
Check:
Left: H=3+3=6, Cl=3, Al=1, O=3
Right: Al=1, Cl=3, H=6, O=3 → Balanced!
✔ Answer: 3 HCl + 1 Al(OH)₃ → 1 AlCl₃ + 3 H₂O
---
Final Answer:
i. 1 HBr + 1 NaOH → 1 NaBr + 1 H₂O
ii. 1 H₂SO₄ + 2 KOH → 1 K₂SO₄ + 2 H₂O
iii. 2 HCl + 1 Ca(OH)₂ → 1 CaCl₂ + 2 H₂O
iv. 2 Fe(OH)₃ + 3 H₂SO₄ → 1 Fe₂(SO₄)₃ + 6 H₂O
v. 3 H₂SO₄ + 2 B(OH)₃ → 1 B₂(SO₄)₃ + 6 H₂O
vi. 1 Pb(OH)₂ + 2 HCl → 1 PbCl₂ + 2 H₂O
vii. 1 H₂SO₄ + 2 NH₄OH → 1 (NH₄)₂SO₄ + 2 H₂O
viii. 1 H₂CO₃ + 2 CsOH → 1 Cs₂CO₃ + 2 H₂O
ix. 2 HF + 1 Mg(OH)₂ → 1 MgF₂ + 2 H₂O
x. 3 HNO₃ + 1 Al(OH)₃ → 1 Al(NO₃)₃ + 3 H₂O
xi. 2 HNO₃ + 1 Zn(OH)₂ → 1 Zn(NO₃)₂ + 2 H₂O
xii. 2 H₃PO₄ + 3 Ca(OH)₂ → 1 Ca₃(PO)₂ + 6 H₂O
xiii. 3 HClO₃ + 1 Al(OH)₃ → 1 Al(ClO₃)₃ + 3 H₂O
xiv. 2 HF + 1 Ba(OH)₂ → 1 BaF₂ + 2 H₂O
xv. 3 HCl + 1 Al(OH)₃ → 1 AlCl₃ + 3 H₂O
Parent Tip: Review the logic above to help your child master the concept of balancing chemical equations worksheet.
