49 Balancing Chemical Equations Worksheets [with Answers] - Free Printable
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Step-by-step solution for: 49 Balancing Chemical Equations Worksheets [with Answers]
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Show Answer Key & Explanations
Step-by-step solution for: 49 Balancing Chemical Equations Worksheets [with Answers]
Let’s go through each equation one by one and balance them. Balancing means making sure the number of atoms of each element is the same on both sides of the arrow.
We’ll use coefficients (numbers in front) to balance — never change the subscripts!
---
1) Na₃PO₄ + KOH → NaOH + K₃PO₄
Left: Na=3, P=1, O=4+1=5, K=1, H=1
Right: Na=1, O=1+4=5, H=1, K=3, P=1
→ Need 3 NaOH and 3 KOH to match Na and K
✔ 1 Na₃PO₄ + 3 KOH → 3 NaOH + 1 K₃PO₄
---
2) MgF₂ + Li₂CO₃ → MgCO₃ + LiF
Left: Mg=1, F=2, Li=2, C=1, O=3
Right: Mg=1, C=1, O=3, Li=1, F=1
→ Need 2 LiF to match Li and F
✔ 1 MgF₂ + 1 Li₂CO₃ → 1 MgCO₃ + 2 LiF
---
3) P₄ + O₂ → P₂O₃
Left: P=4, O=2
Right: P=2, O=3
→ Make P equal: 2 P₂O₃ has 4P and 6O → need 3 O₂
✔ 1 P₄ + 3 O₂ → 2 P₂O₃
---
4) RbNO₃ + BeF₂ → Be(NO₃)₂ + RbF
Left: Rb=1, N=1, O=3, Be=1, F=2
Right: Be=1, N=2, O=6, Rb=1, F=1
→ Need 2 RbNO₃ and 2 RbF
✔ 2 RbNO₃ + 1 BeF₂ → 1 Be(NO₃)₂ + 2 RbF
---
5) AgNO₃ + Cu → Cu(NO₃)₂ + Ag
Left: Ag=1, N=1, O=3, Cu=1
Right: Cu=1, N=2, O=6, Ag=1
→ Need 2 AgNO₃ and 2 Ag
✔ 2 AgNO₃ + 1 Cu → 1 Cu(NO₃)₂ + 2 Ag
---
6) CF₄ + Br₂ → CBr₄ + F₂
Left: C=1, F=4, Br=2
Right: C=1, Br=4, F=2
→ Need 2 Br₂ and 2 F₂
✔ 1 CF₄ + 2 Br₂ → 1 CBr₄ + 2 F₂
---
7) HCN + CuSO₄ → H₂SO₄ + Cu(CN)₂
Left: H=1, C=1, N=1, Cu=1, S=1, O=4
Right: H=2, S=1, O=4, Cu=1, C=2, N=2
→ Need 2 HCN to get 2C, 2N, 2H
✔ 2 HCN + 1 CuSO₄ → 1 H₂SO₄ + 1 Cu(CN)₂
---
8) GaF₃ + Cs → CsF + Ga
Left: Ga=1, F=3, Cs=1
Right: Cs=1, F=1, Ga=1
→ Need 3 CsF and 3 Cs
✔ 1 GaF₃ + 3 Cs → 3 CsF + 1 Ga
---
9) BaS + PtF₂ → BaF₂ + PtS
Left: Ba=1, S=1, Pt=1, F=2
Right: Ba=1, F=2, Pt=1, S=1
→ Already balanced!
✔ 1 BaS + 1 PtF₂ → 1 BaF₂ + 1 PtS
---
10) N₂ + H₂ → NH₃
Left: N=2, H=2
Right: N=1, H=3
→ Need 2 NH₃ (gives 2N, 6H) → so 3 H₂
✔ 1 N₂ + 3 H₂ → 2 NH₃
---
11) NaF + Br₂ → NaBr + F₂
Left: Na=1, F=1, Br=2
Right: Na=1, Br=1, F=2
→ Need 2 NaF and 2 NaBr
✔ 2 NaF + 1 Br₂ → 2 NaBr + 1 F₂
---
12) Pb(OH)₂ + HCl → H₂O + PbCl₂
Left: Pb=1, O=2, H=2+1=3, Cl=1
Right: H=2, O=1, Pb=1, Cl=2
→ Need 2 HCl and 2 H₂O
✔ 1 Pb(OH)₂ + 2 HCl → 2 H₂O + 1 PbCl₂
---
13) AlBr₃ + K₂SO₄ → KBr + Al₂(SO₄)₃
Left: Al=1, Br=3, K=2, S=1, O=4
Right: K=1, Br=1, Al=2, S=3, O=12
→ Need 2 AlBr₃, 3 K₂SO₄, 6 KBr
✔ 2 AlBr₃ + 3 K₂SO₄ → 6 KBr + 1 Al₂(SO₄)₃
---
14) CH₄ + O₂ → CO₂ + H₂O
Left: C=1, H=4, O=2
Right: C=1, O=2+1=3, H=2
→ Need 2 H₂O (makes 4H, 2O from water + 2O from CO₂ = 4O total) → so 2 O₂
✔ 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
---
15) Na₃PO₄ + CaCl₂ → NaCl + Ca₃(PO₄)₂
Left: Na=3, P=1, O=4, Ca=1, Cl=2
Right: Na=1, Cl=1, Ca=3, P=2, O=8
→ Need 2 Na₃PO₄, 3 CaCl₂, 6 NaCl
✔ 2 Na₃PO₄ + 3 CaCl₂ → 6 NaCl + 1 Ca₃(PO₄)₂
---
16) K + Cl₂ → KCl
Left: K=1, Cl=2
Right: K=1, Cl=1
→ Need 2 KCl and 2 K
✔ 2 K + 1 Cl₂ → 2 KCl
---
17) Al + HCl → H₂ + AlCl₃
Left: Al=1, H=1, Cl=1
Right: H=2, Al=1, Cl=3
→ Need 6 HCl, 3 H₂, 2 AlCl₃? Wait — let’s do step by step:
Al + HCl → H₂ + AlCl₃
To balance Cl: need 3 HCl for 1 AlCl₃ → but then H=3, which doesn’t make H₂ easily.
Better: 2 Al + 6 HCl → 3 H₂ + 2 AlCl₃
✔ 2 Al + 6 HCl → 3 H₂ + 2 AlCl₃
---
18) N₂ + F₂ → NF₃
Left: N=2, F=2
Right: N=1, F=3
→ Need 2 NF₃ (2N, 6F) → so 3 F₂
✔ 1 N₂ + 3 F₂ → 2 NF₃
---
19) SO₂ + Li₂Se → SSe₂ + Li₂O
Left: S=1, O=2, Li=2, Se=1
Right: S=1, Se=2, Li=2, O=1
→ Need 2 Li₂O and 2 Li₂Se? Let’s see:
Try: SO₂ + 2 Li₂Se → SSe₂ + 2 Li₂O
Check: Left: S=1, O=2, Li=4, Se=2
Right: S=1, Se=2, Li=4, O=2 → YES!
✔ 1 SO₂ + 2 Li₂Se → 1 SSe₂ + 2 Li₂O
---
20) NH₃ + H₂SO₄ → (NH₄)₂SO₄
Left: N=1, H=3+2=5, S=1, O=4
Right: N=2, H=8, S=1, O=4
→ Need 2 NH₃ to get 2N and 6H → plus H₂SO₄ gives 8H total
✔ 2 NH₃ + 1 H₂SO₄ → 1 (NH₄)₂SO₄
---
Final Answer:
1) 1, 3, 3, 1
2) 1, 1, 1, 2
3) 1, 3, 2
4) 2, 1, 1, 2
5) 2, 1, 1, 2
6) 1, 2, 1, 2
7) 2, 1, 1, 1
8) 1, 3, 3, 1
9) 1, 1, 1, 1
10) 1, 3, 2
11) 2, 1, 2, 1
12) 1, 2, 2, 1
13) 2, 3, 6, 1
14) 1, 2, 1, 2
15) 2, 3, 6, 1
16) 2, 1, 2
17) 2, 6, 3, 2
18) 1, 3, 2
19) 1, 2, 1, 2
20) 2, 1, 1
We’ll use coefficients (numbers in front) to balance — never change the subscripts!
---
1) Na₃PO₄ + KOH → NaOH + K₃PO₄
Left: Na=3, P=1, O=4+1=5, K=1, H=1
Right: Na=1, O=1+4=5, H=1, K=3, P=1
→ Need 3 NaOH and 3 KOH to match Na and K
✔ 1 Na₃PO₄ + 3 KOH → 3 NaOH + 1 K₃PO₄
---
2) MgF₂ + Li₂CO₃ → MgCO₃ + LiF
Left: Mg=1, F=2, Li=2, C=1, O=3
Right: Mg=1, C=1, O=3, Li=1, F=1
→ Need 2 LiF to match Li and F
✔ 1 MgF₂ + 1 Li₂CO₃ → 1 MgCO₃ + 2 LiF
---
3) P₄ + O₂ → P₂O₃
Left: P=4, O=2
Right: P=2, O=3
→ Make P equal: 2 P₂O₃ has 4P and 6O → need 3 O₂
✔ 1 P₄ + 3 O₂ → 2 P₂O₃
---
4) RbNO₃ + BeF₂ → Be(NO₃)₂ + RbF
Left: Rb=1, N=1, O=3, Be=1, F=2
Right: Be=1, N=2, O=6, Rb=1, F=1
→ Need 2 RbNO₃ and 2 RbF
✔ 2 RbNO₃ + 1 BeF₂ → 1 Be(NO₃)₂ + 2 RbF
---
5) AgNO₃ + Cu → Cu(NO₃)₂ + Ag
Left: Ag=1, N=1, O=3, Cu=1
Right: Cu=1, N=2, O=6, Ag=1
→ Need 2 AgNO₃ and 2 Ag
✔ 2 AgNO₃ + 1 Cu → 1 Cu(NO₃)₂ + 2 Ag
---
6) CF₄ + Br₂ → CBr₄ + F₂
Left: C=1, F=4, Br=2
Right: C=1, Br=4, F=2
→ Need 2 Br₂ and 2 F₂
✔ 1 CF₄ + 2 Br₂ → 1 CBr₄ + 2 F₂
---
7) HCN + CuSO₄ → H₂SO₄ + Cu(CN)₂
Left: H=1, C=1, N=1, Cu=1, S=1, O=4
Right: H=2, S=1, O=4, Cu=1, C=2, N=2
→ Need 2 HCN to get 2C, 2N, 2H
✔ 2 HCN + 1 CuSO₄ → 1 H₂SO₄ + 1 Cu(CN)₂
---
8) GaF₃ + Cs → CsF + Ga
Left: Ga=1, F=3, Cs=1
Right: Cs=1, F=1, Ga=1
→ Need 3 CsF and 3 Cs
✔ 1 GaF₃ + 3 Cs → 3 CsF + 1 Ga
---
9) BaS + PtF₂ → BaF₂ + PtS
Left: Ba=1, S=1, Pt=1, F=2
Right: Ba=1, F=2, Pt=1, S=1
→ Already balanced!
✔ 1 BaS + 1 PtF₂ → 1 BaF₂ + 1 PtS
---
10) N₂ + H₂ → NH₃
Left: N=2, H=2
Right: N=1, H=3
→ Need 2 NH₃ (gives 2N, 6H) → so 3 H₂
✔ 1 N₂ + 3 H₂ → 2 NH₃
---
11) NaF + Br₂ → NaBr + F₂
Left: Na=1, F=1, Br=2
Right: Na=1, Br=1, F=2
→ Need 2 NaF and 2 NaBr
✔ 2 NaF + 1 Br₂ → 2 NaBr + 1 F₂
---
12) Pb(OH)₂ + HCl → H₂O + PbCl₂
Left: Pb=1, O=2, H=2+1=3, Cl=1
Right: H=2, O=1, Pb=1, Cl=2
→ Need 2 HCl and 2 H₂O
✔ 1 Pb(OH)₂ + 2 HCl → 2 H₂O + 1 PbCl₂
---
13) AlBr₃ + K₂SO₄ → KBr + Al₂(SO₄)₃
Left: Al=1, Br=3, K=2, S=1, O=4
Right: K=1, Br=1, Al=2, S=3, O=12
→ Need 2 AlBr₃, 3 K₂SO₄, 6 KBr
✔ 2 AlBr₃ + 3 K₂SO₄ → 6 KBr + 1 Al₂(SO₄)₃
---
14) CH₄ + O₂ → CO₂ + H₂O
Left: C=1, H=4, O=2
Right: C=1, O=2+1=3, H=2
→ Need 2 H₂O (makes 4H, 2O from water + 2O from CO₂ = 4O total) → so 2 O₂
✔ 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
---
15) Na₃PO₄ + CaCl₂ → NaCl + Ca₃(PO₄)₂
Left: Na=3, P=1, O=4, Ca=1, Cl=2
Right: Na=1, Cl=1, Ca=3, P=2, O=8
→ Need 2 Na₃PO₄, 3 CaCl₂, 6 NaCl
✔ 2 Na₃PO₄ + 3 CaCl₂ → 6 NaCl + 1 Ca₃(PO₄)₂
---
16) K + Cl₂ → KCl
Left: K=1, Cl=2
Right: K=1, Cl=1
→ Need 2 KCl and 2 K
✔ 2 K + 1 Cl₂ → 2 KCl
---
17) Al + HCl → H₂ + AlCl₃
Left: Al=1, H=1, Cl=1
Right: H=2, Al=1, Cl=3
→ Need 6 HCl, 3 H₂, 2 AlCl₃? Wait — let’s do step by step:
Al + HCl → H₂ + AlCl₃
To balance Cl: need 3 HCl for 1 AlCl₃ → but then H=3, which doesn’t make H₂ easily.
Better: 2 Al + 6 HCl → 3 H₂ + 2 AlCl₃
✔ 2 Al + 6 HCl → 3 H₂ + 2 AlCl₃
---
18) N₂ + F₂ → NF₃
Left: N=2, F=2
Right: N=1, F=3
→ Need 2 NF₃ (2N, 6F) → so 3 F₂
✔ 1 N₂ + 3 F₂ → 2 NF₃
---
19) SO₂ + Li₂Se → SSe₂ + Li₂O
Left: S=1, O=2, Li=2, Se=1
Right: S=1, Se=2, Li=2, O=1
→ Need 2 Li₂O and 2 Li₂Se? Let’s see:
Try: SO₂ + 2 Li₂Se → SSe₂ + 2 Li₂O
Check: Left: S=1, O=2, Li=4, Se=2
Right: S=1, Se=2, Li=4, O=2 → YES!
✔ 1 SO₂ + 2 Li₂Se → 1 SSe₂ + 2 Li₂O
---
20) NH₃ + H₂SO₄ → (NH₄)₂SO₄
Left: N=1, H=3+2=5, S=1, O=4
Right: N=2, H=8, S=1, O=4
→ Need 2 NH₃ to get 2N and 6H → plus H₂SO₄ gives 8H total
✔ 2 NH₃ + 1 H₂SO₄ → 1 (NH₄)₂SO₄
---
Final Answer:
1) 1, 3, 3, 1
2) 1, 1, 1, 2
3) 1, 3, 2
4) 2, 1, 1, 2
5) 2, 1, 1, 2
6) 1, 2, 1, 2
7) 2, 1, 1, 1
8) 1, 3, 3, 1
9) 1, 1, 1, 1
10) 1, 3, 2
11) 2, 1, 2, 1
12) 1, 2, 2, 1
13) 2, 3, 6, 1
14) 1, 2, 1, 2
15) 2, 3, 6, 1
16) 2, 1, 2
17) 2, 6, 3, 2
18) 1, 3, 2
19) 1, 2, 1, 2
20) 2, 1, 1
Parent Tip: Review the logic above to help your child master the concept of chemical equations to balance worksheet.
![49 Balancing Chemical Equations Worksheets [with Answers]](/images2/chemical-equations-to-balance-worksheet/chemical-equations-to-balance-worksheet-11.jpg)
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