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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Step-by-step solution for: 49 Balancing Chemical Equations Worksheets [with Answers]
Let's solve each of these chemical equations by balancing them. The goal is to ensure the number of atoms of each element is the same on both sides of the equation.
---
- Left: Li = 3, P = 1, O = 4 + 1 = 5 (from PO₄ and OH), Na = 1, H = 1
- Right: Li = 1, O = 1 + 4 = 5, H = 1, Na = 3, P = 1
We need 3 LiOH to balance Li and H:
→ 1 Li₃PO₄ + 3 NaOH → 3 LiOH + 1 Na₃PO₄
✔ Balanced.
Answer:
1 Li₃PO₄ + 3 NaOH → 3 LiOH + 1 Na₃PO₄
---
- 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 balance Li and F:
→ MgF₂ + Li₂CO₃ → MgCO₃ + 2 LiF
✔ Balanced.
Answer:
1 MgF₂ + 1 Li₂CO₃ → 1 MgCO₃ + 2 LiF
---
- Left: P = 4, O = 2
- Right: P = 2, O = 3
We want 2 P₂O₃ → 4 P, 6 O → so we need 3 O₂
→ P₄ + 3 O₂ → 2 P₂O₃
✔ Balanced.
Answer:
1 P₄ + 3 O₂ → 2 P₂O₃
---
- Left: Rb = 1, N = 1, O = 3, Mg = 1, F = 2
- Right: Mg = 1, N = 2, O = 6, Rb = 1, F = 1
Need 2 RbNO₃ to get 2 NO₃⁻, and 2 RbF:
→ 2 RbNO₃ + MgF₂ → Mg(NO₃)₂ + 2 RbF
✔ Balanced.
Answer:
2 RbNO₃ + 1 MgF₂ → 1 Mg(NO₃)₂ + 2 RbF
---
- Left: Ag = 1, N = 1, O = 3, Cu = 1
- Right: Cu = 1, N = 2, O = 6, Ag = 1
Need 2 AgNO₃ to give 2 NO₃⁻, then 2 Ag on right:
→ 2 AgNO₃ + Cu → Cu(NO₃)₂ + 2 Ag
✔ Balanced.
Answer:
2 AgNO₃ + 1 Cu → 1 Cu(NO₃)₂ + 2 Ag
---
- 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:
→ CF₄ + 2 Br₂ → CBr₄ + 2 F₂
✔ Balanced.
Answer:
1 CF₄ + 2 Br₂ → 1 CBr₄ + 2 F₂
---
- 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 2 C, 2 N, 2 H → but H₂SO₄ needs 2 H, so okay.
→ 2 HCN + CuSO₄ → H₂SO₄ + Cu(CN)₂
✔ Balanced.
Answer:
2 HCN + 1 CuSO₄ → 1 H₂SO₄ + 1 Cu(CN)₂
---
- Left: Ga = 1, F = 3, Cs = 1
- Right: Cs = 1, F = 1, Ga = 1
Need 3 CsF → so 3 Cs and 3 F
→ GaF₃ + 3 Cs → 3 CsF + Ga
✔ Balanced.
Answer:
1 GaF₃ + 3 Cs → 3 CsF + 1 Ga
---
- Left: Sr = 1, S = 1, Pt = 1, F = 2
- Right: Sr = 1, F = 2, Pt = 1, S = 1
Already balanced.
Answer:
1 SrS + 1 PtF₂ → 1 SrF₂ + 1 PtS
---
- Left: N = 2, H = 2
- Right: N = 1, H = 3
Need 2 NH₃ → 2 N, 6 H → so 3 H₂
→ N₂ + 3 H₂ → 2 NH₃
✔ Balanced.
Answer:
1 N₂ + 3 H₂ → 2 NH₃
---
- Left: Li = 1, F = 1, Br = 2
- Right: Li = 1, Br = 1, F = 2
Need 2 LiF → 2 F, and 2 LiBr → 2 Li, 2 Br
→ 2 LiF + Br₂ → 2 LiBr + F₂
✔ Balanced.
Answer:
2 LiF + 1 Br₂ → 2 LiBr + 1 F₂
---
- Left: Pb = 1, O = 2, H = 2 + 1 = 3, Cl = 1
- Right: H = 2, O = 1, Pb = 1, Cl = 2
Need 2 HCl → 2 Cl, 2 H → then 2 H₂O? But H₂O has 2 H, 1 O.
Wait: Pb(OH)₂ has 2 OH groups → can react with 2 H⁺ → so need 2 HCl
→ Pb(OH)₂ + 2 HCl → 2 H₂O + PbCl₂
Check:
Left: Pb=1, O=2+2=4? No — wait:
Pb(OH)₂ → Pb, 2O, 2H
2 HCl → 2H, 2Cl
Total: Pb=1, O=2, H=4, Cl=2
Right: 2 H₂O → 4H, 2O; PbCl₂ → Pb, 2Cl
Yes!
✔ Balanced.
Answer:
1 Pb(OH)₂ + 2 HCl → 2 H₂O + 1 PbCl₂
---
- Left: Ga = 1, Br = 3, Na = 2, C = 1, O = 3
- Right: Na = 1, Br = 1, Ga = 2, C = 3, O = 9
Need 2 GaBr₃ → 2 Ga, 6 Br
Need 3 Na₂CO₃ → 6 Na, 3 C, 9 O
Then: 6 NaBr, and 1 Ga₂(CO₃)₃
→ 2 GaBr₃ + 3 Na₂CO₃ → 6 NaBr + 1 Ga₂(CO₃)₃
✔ Check:
Left: Ga=2, Br=6, Na=6, C=3, O=9
Right: Na=6, Br=6, Ga=2, C=3, O=9
Perfect.
Answer:
2 GaBr₃ + 3 Na₂CO₃ → 6 NaBr + 1 Ga₂(CO₃)₃
---
Classic combustion.
- CH₄ → C, 4H
- O₂ → 2O
- CO₂ → C, 2O
- H₂O → 2H, 1O
To balance H: need 2 H₂O → 4H → matches CH₄
So: CH₄ + O₂ → CO₂ + 2 H₂O
Now O: right = 2 (from CO₂) + 2 (from 2 H₂O) = 4 O → need 2 O₂
→ CH₄ + 2 O₂ → CO₂ + 2 H₂O
✔ Balanced.
Answer:
1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
---
- Left: Li = 3, P = 1, O = 4, Ca = 1, Cl = 2
- Right: Li = 1, Cl = 1, Ca = 3, P = 2, O = 8
Need 2 Li₃PO₄ → 6 Li, 2 P, 8 O
Need 3 CaCl₂ → 3 Ca, 6 Cl
Then: 6 LiCl, and 1 Ca₃(PO₄)₂
→ 2 Li₃PO₄ + 3 CaCl₂ → 6 LiCl + 1 Ca₃(PO₄)₂
✔ Check:
Left: Li=6, P=2, O=8, Ca=3, Cl=6
Right: Li=6, Cl=6, Ca=3, P=2, O=8
Perfect.
Answer:
2 Li₃PO₄ + 3 CaCl₂ → 6 LiCl + 1 Ca₃(PO₄)₂
---
- Left: Na = 1, Cl = 2
- Right: Na = 1, Cl = 1
Need 2 NaCl → 2 Na, 2 Cl → so 2 Na
→ 2 Na + Cl₂ → 2 NaCl
✔ Balanced.
Answer:
2 Na + 1 Cl₂ → 2 NaCl
---
- Left: Ga = 1, H = 1, Cl = 1
- Right: H = 2, Ga = 1, Cl = 3
Need 3 HCl → 3 H, 3 Cl → gives 3/2 H₂? But better to use whole numbers.
So: Ga + 3 HCl → (3/2) H₂ + GaCl₃ → multiply all by 2:
→ 2 Ga + 6 HCl → 3 H₂ + 2 GaCl₃
✔ Check:
Left: Ga=2, H=6, Cl=6
Right: H=6, Ga=2, Cl=6
Perfect.
Answer:
2 Ga + 6 HCl → 3 H₂ + 2 GaCl₃
---
- Left: N = 2, F = 2
- Right: N = 1, F = 3
Need 2 NF₃ → 2 N, 6 F → so 3 F₂
→ N₂ + 3 F₂ → 2 NF₃
✔ Balanced.
Answer:
1 N₂ + 3 F₂ → 2 NF₃
---
- Left: S = 1, O = 2, Li = 2, Se = 1
- Right: S = 1, Se = 2, Li = 2, O = 1
Need 2 SO₂ → 2 S, 4 O → but only 1 O in Li₂O → need 2 Li₂O?
But Li₂Se has 2 Li, 1 Se → so for 2 Se → need 2 Li₂Se → 4 Li
But 2 Li₂O → 4 Li, 2 O → good
So: 2 SO₂ + 2 Li₂Se → SSe₂ + 2 Li₂O? But SSe₂ has only one S, one Se?
Wait: SSe₂ → S = 1, Se = 2
But left: 2 SO₂ → 2 S, 4 O
2 Li₂Se → 4 Li, 2 Se
Right: SSe₂ → 1 S, 2 Se → not enough S
So need 2 SSe₂ → 2 S, 4 Se → so need 4 Se → 4 Li₂Se → 8 Li
And 2 Li₂O → 4 Li → no.
Wait: Let's try:
Want 2 SSe₂ → 2 S, 4 Se
So need 2 SO₂ → 2 S, 4 O
And 4 Li₂Se → 4 Li, 4 Se
Then: 2 Li₂O → 4 Li, 2 O → but we have 4 O from SO₂ → too much
No.
Alternative:
Try balancing:
Suppose: a SO₂ + b Li₂Se → c SSe₂ + d Li₂O
From S: a = c
From Se: b = 2c
From O: 2a = d
From Li: 2b = 2d → b = d
Now: b = 2c, d = b, and d = 2a → so b = 2a
But b = 2c and a = c → so b = 2a → yes.
So let a = 1 → c = 1, b = 2, d = 2
→ 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
✔ Balanced.
Answer:
1 SO₂ + 2 Li₂Se → 1 SSe₂ + 2 Li₂O
---
- Left: N = 1, H = 3 + 2 = 5, S = 1, O = 4
- Right: N = 2, H = 8, S = 1, O = 4
Need 2 NH₃ → 2 N, 6 H → plus H₂SO₄ → 2 H → total 8 H → matches (NH₄)₂SO₄
→ 2 NH₃ + H₂SO₄ → (NH₄)₂SO₄
✔ Balanced.
Answer:
2 NH₃ + 1 H₂SO₄ → 1 (NH₄)₂SO₄
---
## ✔ Final Answers Summary:
1) 1 Li₃PO₄ + 3 NaOH → 3 LiOH + 1 Na₃PO₄
2) 1 MgF₂ + 1 Li₂CO₃ → 1 MgCO₃ + 2 LiF
3) 1 P₄ + 3 O₂ → 2 P₂O₃
4) 2 RbNO₃ + 1 MgF₂ → 1 Mg(NO₃)₂ + 2 RbF
5) 2 AgNO₃ + 1 Cu → 1 Cu(NO₃)₂ + 2 Ag
6) 1 CF₄ + 2 Br₂ → 1 CBr₄ + 2 F₂
7) 2 HCN + 1 CuSO₄ → 1 H₂SO₄ + 1 Cu(CN)₂
8) 1 GaF₃ + 3 Cs → 3 CsF + 1 Ga
9) 1 SrS + 1 PtF₂ → 1 SrF₂ + 1 PtS
10) 1 N₂ + 3 H₂ → 2 NH₃
11) 2 LiF + 1 Br₂ → 2 LiBr + 1 F₂
12) 1 Pb(OH)₂ + 2 HCl → 2 H₂O + 1 PbCl₂
13) 2 GaBr₃ + 3 Na₂CO₃ → 6 NaBr + 1 Ga₂(CO₃)₃
14) 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
15) 2 Li₃PO₄ + 3 CaCl₂ → 6 LiCl + 1 Ca₃(PO₄)₂
16) 2 Na + 1 Cl₂ → 2 NaCl
17) 2 Ga + 6 HCl → 3 H₂ + 2 GaCl₃
18) 1 N₂ + 3 F₂ → 2 NF₃
19) 1 SO₂ + 2 Li₂Se → 1 SSe₂ + 2 Li₂O
20) 2 NH₃ + 1 H₂SO₄ → 1 (NH₄)₂SO₄
---
Let me know if you'd like this formatted as a printable PDF or need explanations for any specific step!
---
1) Li₃PO₄ + NaOH → LiOH + Na₃PO₄
- Left: Li = 3, P = 1, O = 4 + 1 = 5 (from PO₄ and OH), Na = 1, H = 1
- Right: Li = 1, O = 1 + 4 = 5, H = 1, Na = 3, P = 1
We need 3 LiOH to balance Li and H:
→ 1 Li₃PO₄ + 3 NaOH → 3 LiOH + 1 Na₃PO₄
✔ Balanced.
Answer:
1 Li₃PO₄ + 3 NaOH → 3 LiOH + 1 Na₃PO₄
---
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 balance Li and F:
→ MgF₂ + Li₂CO₃ → MgCO₃ + 2 LiF
✔ Balanced.
Answer:
1 MgF₂ + 1 Li₂CO₃ → 1 MgCO₃ + 2 LiF
---
3) P₄ + O₂ → P₂O₃
- Left: P = 4, O = 2
- Right: P = 2, O = 3
We want 2 P₂O₃ → 4 P, 6 O → so we need 3 O₂
→ P₄ + 3 O₂ → 2 P₂O₃
✔ Balanced.
Answer:
1 P₄ + 3 O₂ → 2 P₂O₃
---
4) RbNO₃ + MgF₂ → Mg(NO₃)₂ + RbF
- Left: Rb = 1, N = 1, O = 3, Mg = 1, F = 2
- Right: Mg = 1, N = 2, O = 6, Rb = 1, F = 1
Need 2 RbNO₃ to get 2 NO₃⁻, and 2 RbF:
→ 2 RbNO₃ + MgF₂ → Mg(NO₃)₂ + 2 RbF
✔ Balanced.
Answer:
2 RbNO₃ + 1 MgF₂ → 1 Mg(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 to give 2 NO₃⁻, then 2 Ag on right:
→ 2 AgNO₃ + Cu → Cu(NO₃)₂ + 2 Ag
✔ Balanced.
Answer:
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:
→ CF₄ + 2 Br₂ → CBr₄ + 2 F₂
✔ Balanced.
Answer:
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 2 C, 2 N, 2 H → but H₂SO₄ needs 2 H, so okay.
→ 2 HCN + CuSO₄ → H₂SO₄ + Cu(CN)₂
✔ Balanced.
Answer:
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 and 3 F
→ GaF₃ + 3 Cs → 3 CsF + Ga
✔ Balanced.
Answer:
1 GaF₃ + 3 Cs → 3 CsF + 1 Ga
---
9) SrS + PtF₂ → SrF₂ + PtS
- Left: Sr = 1, S = 1, Pt = 1, F = 2
- Right: Sr = 1, F = 2, Pt = 1, S = 1
Already balanced.
Answer:
1 SrS + 1 PtF₂ → 1 SrF₂ + 1 PtS
---
10) N₂ + H₂ → NH₃
- Left: N = 2, H = 2
- Right: N = 1, H = 3
Need 2 NH₃ → 2 N, 6 H → so 3 H₂
→ N₂ + 3 H₂ → 2 NH₃
✔ Balanced.
Answer:
1 N₂ + 3 H₂ → 2 NH₃
---
11) LiF + Br₂ → LiBr + F₂
- Left: Li = 1, F = 1, Br = 2
- Right: Li = 1, Br = 1, F = 2
Need 2 LiF → 2 F, and 2 LiBr → 2 Li, 2 Br
→ 2 LiF + Br₂ → 2 LiBr + F₂
✔ Balanced.
Answer:
2 LiF + 1 Br₂ → 2 LiBr + 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 → 2 Cl, 2 H → then 2 H₂O? But H₂O has 2 H, 1 O.
Wait: Pb(OH)₂ has 2 OH groups → can react with 2 H⁺ → so need 2 HCl
→ Pb(OH)₂ + 2 HCl → 2 H₂O + PbCl₂
Check:
Left: Pb=1, O=2+2=4? No — wait:
Pb(OH)₂ → Pb, 2O, 2H
2 HCl → 2H, 2Cl
Total: Pb=1, O=2, H=4, Cl=2
Right: 2 H₂O → 4H, 2O; PbCl₂ → Pb, 2Cl
Yes!
✔ Balanced.
Answer:
1 Pb(OH)₂ + 2 HCl → 2 H₂O + 1 PbCl₂
---
13) GaBr₃ + Na₂CO₃ → NaBr + Ga₂(CO₃)₃ (tough one)
- Left: Ga = 1, Br = 3, Na = 2, C = 1, O = 3
- Right: Na = 1, Br = 1, Ga = 2, C = 3, O = 9
Need 2 GaBr₃ → 2 Ga, 6 Br
Need 3 Na₂CO₃ → 6 Na, 3 C, 9 O
Then: 6 NaBr, and 1 Ga₂(CO₃)₃
→ 2 GaBr₃ + 3 Na₂CO₃ → 6 NaBr + 1 Ga₂(CO₃)₃
✔ Check:
Left: Ga=2, Br=6, Na=6, C=3, O=9
Right: Na=6, Br=6, Ga=2, C=3, O=9
Perfect.
Answer:
2 GaBr₃ + 3 Na₂CO₃ → 6 NaBr + 1 Ga₂(CO₃)₃
---
14) CH₄ + O₂ → CO₂ + H₂O
Classic combustion.
- CH₄ → C, 4H
- O₂ → 2O
- CO₂ → C, 2O
- H₂O → 2H, 1O
To balance H: need 2 H₂O → 4H → matches CH₄
So: CH₄ + O₂ → CO₂ + 2 H₂O
Now O: right = 2 (from CO₂) + 2 (from 2 H₂O) = 4 O → need 2 O₂
→ CH₄ + 2 O₂ → CO₂ + 2 H₂O
✔ Balanced.
Answer:
1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
---
15) Li₃PO₄ + CaCl₂ → LiCl + Ca₃(PO₄)₂ (also tough)
- Left: Li = 3, P = 1, O = 4, Ca = 1, Cl = 2
- Right: Li = 1, Cl = 1, Ca = 3, P = 2, O = 8
Need 2 Li₃PO₄ → 6 Li, 2 P, 8 O
Need 3 CaCl₂ → 3 Ca, 6 Cl
Then: 6 LiCl, and 1 Ca₃(PO₄)₂
→ 2 Li₃PO₄ + 3 CaCl₂ → 6 LiCl + 1 Ca₃(PO₄)₂
✔ Check:
Left: Li=6, P=2, O=8, Ca=3, Cl=6
Right: Li=6, Cl=6, Ca=3, P=2, O=8
Perfect.
Answer:
2 Li₃PO₄ + 3 CaCl₂ → 6 LiCl + 1 Ca₃(PO₄)₂
---
16) Na + Cl₂ → NaCl
- Left: Na = 1, Cl = 2
- Right: Na = 1, Cl = 1
Need 2 NaCl → 2 Na, 2 Cl → so 2 Na
→ 2 Na + Cl₂ → 2 NaCl
✔ Balanced.
Answer:
2 Na + 1 Cl₂ → 2 NaCl
---
17) Ga + HCl → H₂ + GaCl₃ (last tough one)
- Left: Ga = 1, H = 1, Cl = 1
- Right: H = 2, Ga = 1, Cl = 3
Need 3 HCl → 3 H, 3 Cl → gives 3/2 H₂? But better to use whole numbers.
So: Ga + 3 HCl → (3/2) H₂ + GaCl₃ → multiply all by 2:
→ 2 Ga + 6 HCl → 3 H₂ + 2 GaCl₃
✔ Check:
Left: Ga=2, H=6, Cl=6
Right: H=6, Ga=2, Cl=6
Perfect.
Answer:
2 Ga + 6 HCl → 3 H₂ + 2 GaCl₃
---
18) N₂ + F₂ → NF₃
- Left: N = 2, F = 2
- Right: N = 1, F = 3
Need 2 NF₃ → 2 N, 6 F → so 3 F₂
→ N₂ + 3 F₂ → 2 NF₃
✔ Balanced.
Answer:
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 SO₂ → 2 S, 4 O → but only 1 O in Li₂O → need 2 Li₂O?
But Li₂Se has 2 Li, 1 Se → so for 2 Se → need 2 Li₂Se → 4 Li
But 2 Li₂O → 4 Li, 2 O → good
So: 2 SO₂ + 2 Li₂Se → SSe₂ + 2 Li₂O? But SSe₂ has only one S, one Se?
Wait: SSe₂ → S = 1, Se = 2
But left: 2 SO₂ → 2 S, 4 O
2 Li₂Se → 4 Li, 2 Se
Right: SSe₂ → 1 S, 2 Se → not enough S
So need 2 SSe₂ → 2 S, 4 Se → so need 4 Se → 4 Li₂Se → 8 Li
And 2 Li₂O → 4 Li → no.
Wait: Let's try:
Want 2 SSe₂ → 2 S, 4 Se
So need 2 SO₂ → 2 S, 4 O
And 4 Li₂Se → 4 Li, 4 Se
Then: 2 Li₂O → 4 Li, 2 O → but we have 4 O from SO₂ → too much
No.
Alternative:
Try balancing:
Suppose: a SO₂ + b Li₂Se → c SSe₂ + d Li₂O
From S: a = c
From Se: b = 2c
From O: 2a = d
From Li: 2b = 2d → b = d
Now: b = 2c, d = b, and d = 2a → so b = 2a
But b = 2c and a = c → so b = 2a → yes.
So let a = 1 → c = 1, b = 2, d = 2
→ 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
✔ Balanced.
Answer:
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 → 2 N, 6 H → plus H₂SO₄ → 2 H → total 8 H → matches (NH₄)₂SO₄
→ 2 NH₃ + H₂SO₄ → (NH₄)₂SO₄
✔ Balanced.
Answer:
2 NH₃ + 1 H₂SO₄ → 1 (NH₄)₂SO₄
---
## ✔ Final Answers Summary:
1) 1 Li₃PO₄ + 3 NaOH → 3 LiOH + 1 Na₃PO₄
2) 1 MgF₂ + 1 Li₂CO₃ → 1 MgCO₃ + 2 LiF
3) 1 P₄ + 3 O₂ → 2 P₂O₃
4) 2 RbNO₃ + 1 MgF₂ → 1 Mg(NO₃)₂ + 2 RbF
5) 2 AgNO₃ + 1 Cu → 1 Cu(NO₃)₂ + 2 Ag
6) 1 CF₄ + 2 Br₂ → 1 CBr₄ + 2 F₂
7) 2 HCN + 1 CuSO₄ → 1 H₂SO₄ + 1 Cu(CN)₂
8) 1 GaF₃ + 3 Cs → 3 CsF + 1 Ga
9) 1 SrS + 1 PtF₂ → 1 SrF₂ + 1 PtS
10) 1 N₂ + 3 H₂ → 2 NH₃
11) 2 LiF + 1 Br₂ → 2 LiBr + 1 F₂
12) 1 Pb(OH)₂ + 2 HCl → 2 H₂O + 1 PbCl₂
13) 2 GaBr₃ + 3 Na₂CO₃ → 6 NaBr + 1 Ga₂(CO₃)₃
14) 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
15) 2 Li₃PO₄ + 3 CaCl₂ → 6 LiCl + 1 Ca₃(PO₄)₂
16) 2 Na + 1 Cl₂ → 2 NaCl
17) 2 Ga + 6 HCl → 3 H₂ + 2 GaCl₃
18) 1 N₂ + 3 F₂ → 2 NF₃
19) 1 SO₂ + 2 Li₂Se → 1 SSe₂ + 2 Li₂O
20) 2 NH₃ + 1 H₂SO₄ → 1 (NH₄)₂SO₄
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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-0.jpg)
