49 Balancing Chemical Equations Worksheets [with Answers] - Free Printable
![49 Balancing Chemical Equations Worksheets [with Answers]](/images2/how-to-balance-chemical-equations-worksheet/how-to-balance-chemical-equations-worksheet-0.jpg)
Educational worksheet: 49 Balancing Chemical Equations Worksheets [with Answers]. Download and print for classroom or home learning activities.
JPG
1932×2500
196.5 KB
Free · Personal Use
Quality Assured by Worksheets Library Team
Reviewed for educational accuracy and age-appropriateness
ID: #1513709
⭐
Show Answer Key & Explanations
Step-by-step solution for: 49 Balancing Chemical Equations Worksheets [with Answers]
▼
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 make that happen — 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 on right
✔ 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: put 2 in front of P₂O₃ → now P=4, O=6
→ So need 3 O₂ (since 3×2=6 oxygen atoms)
✔ 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 (to get 4 Br) and 2 F₂ (to get 4 F)
✔ 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 on left
✔ 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 → so 3 Cs on left
✔ 1 GaF₃ + 3 Cs → 3 CsF + 1 Ga
---
9) BaS + PtF₂ → BaF₂ + PtS
Already balanced! All elements have 1 atom on each side.
✔ 1 BaS + 1 PtF₂ → 1 BaF₂ + 1 PtS
---
10) N₂ + H₂ → NH₃
Left: N=2, H=2
Right: N=1, H=3
→ Put 2 in front of NH₃ → N=2, H=6
→ Then need 3 H₂ (3×2=6 H)
✔ 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 → then H=2+2=4? Wait — let’s count again.
Pb(OH)₂ has 2 OH → 2 O and 2 H
Plus HCl → say 2 HCl → adds 2 H and 2 Cl
Total left: Pb=1, O=2, H=4, Cl=2
Right: PbCl₂ → Pb=1, Cl=2; H₂O → if we put 2 H₂O → H=4, O=2
✔ 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 → Al=2, Br=6
→ Need 3 K₂SO₄ → K=6, S=3, O=12
→ Right: 6 KBr and 1 Al₂(SO₄)₃
✔ 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
→ Put 2 H₂O → H=4, O=2+2=4? Wait: CO₂ has 2 O, 2 H₂O has 2 O → total 4 O
→ So need 2 O₂ (2×2=4 O)
✔ 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 → Na=6, P=2, O=8
→ Need 3 CaCl₂ → Ca=3, Cl=6
→ Right: 6 NaCl and 1 Ca₃(PO)₂
✔ 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 → so 2 K on left
✔ 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 → H=6, Cl=6
→ Right: 3 H₂ (H=6), 2 AlCl₃ (Al=2, Cl=6) → wait, Al mismatch
Better:
Put 2 Al → need 2 AlCl₃ → Cl=6 → so 6 HCl → H=6 → 3 H₂
✔ 2 Al + 6 HCl → 3 H₂ + 2 AlCl₃
---
18) N₂ + F₂ → NF₃
Left: N=2, F=2
Right: N=1, F=3
→ Put 2 NF₃ → N=2, F=6
→ Need 3 F₂ (3×2=6 F)
✔ 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 → O=2, Li=4 → so need 2 Li₂Se → Se=2, Li=4
→ Right: 1 SSe₂ and 2 Li₂O
✔ 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 → N=2, H=6 → plus H₂SO₄ → H=6+2=8 → perfect
✔ 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 make that happen — 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 on right
✔ 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: put 2 in front of P₂O₃ → now P=4, O=6
→ So need 3 O₂ (since 3×2=6 oxygen atoms)
✔ 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 (to get 4 Br) and 2 F₂ (to get 4 F)
✔ 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 on left
✔ 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 → so 3 Cs on left
✔ 1 GaF₃ + 3 Cs → 3 CsF + 1 Ga
---
9) BaS + PtF₂ → BaF₂ + PtS
Already balanced! All elements have 1 atom on each side.
✔ 1 BaS + 1 PtF₂ → 1 BaF₂ + 1 PtS
---
10) N₂ + H₂ → NH₃
Left: N=2, H=2
Right: N=1, H=3
→ Put 2 in front of NH₃ → N=2, H=6
→ Then need 3 H₂ (3×2=6 H)
✔ 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 → then H=2+2=4? Wait — let’s count again.
Pb(OH)₂ has 2 OH → 2 O and 2 H
Plus HCl → say 2 HCl → adds 2 H and 2 Cl
Total left: Pb=1, O=2, H=4, Cl=2
Right: PbCl₂ → Pb=1, Cl=2; H₂O → if we put 2 H₂O → H=4, O=2
✔ 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 → Al=2, Br=6
→ Need 3 K₂SO₄ → K=6, S=3, O=12
→ Right: 6 KBr and 1 Al₂(SO₄)₃
✔ 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
→ Put 2 H₂O → H=4, O=2+2=4? Wait: CO₂ has 2 O, 2 H₂O has 2 O → total 4 O
→ So need 2 O₂ (2×2=4 O)
✔ 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 → Na=6, P=2, O=8
→ Need 3 CaCl₂ → Ca=3, Cl=6
→ Right: 6 NaCl and 1 Ca₃(PO)₂
✔ 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 → so 2 K on left
✔ 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 → H=6, Cl=6
→ Right: 3 H₂ (H=6), 2 AlCl₃ (Al=2, Cl=6) → wait, Al mismatch
Better:
Put 2 Al → need 2 AlCl₃ → Cl=6 → so 6 HCl → H=6 → 3 H₂
✔ 2 Al + 6 HCl → 3 H₂ + 2 AlCl₃
---
18) N₂ + F₂ → NF₃
Left: N=2, F=2
Right: N=1, F=3
→ Put 2 NF₃ → N=2, F=6
→ Need 3 F₂ (3×2=6 F)
✔ 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 → O=2, Li=4 → so need 2 Li₂Se → Se=2, Li=4
→ Right: 1 SSe₂ and 2 Li₂O
✔ 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 → N=2, H=6 → plus H₂SO₄ → H=6+2=8 → perfect
✔ 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 how to balance chemical equations worksheet.
