Balancing Equations Worksheet featuring 20 chemical reactions to practice balancing.
Balancing Equations Worksheet with 20 chemical equations to balance, including elements like Na, O, P, S, and compounds such as Na₂O, P₂O₅, and H₂SO₄.
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Show Answer Key & Explanations
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 start with #1 and work our way down to #20.
---
1) Na + O₂ → Na₂O
Left: Na=1, O=2
Right: Na=2, O=1
To fix oxygen: put 2 in front of Na₂O → now right has O=2, but Na=4
So put 4 in front of Na on left.
✔ Balanced: 4Na + O₂ → 2Na₂O
---
2) P + O₂ → P₂O₃
Left: P=1, O=2
Right: P=2, O=3
Make P match: put 2 in front of P → left P=2
Now O: right has 3, left has 2 → need common multiple? Let’s try 3 O₂ (gives 6 O) → then right needs 2 P₂O₃ (which gives 6 O and 4 P) → so left needs 4 P.
Wait — let’s do it step by step:
Try: 4P + 3O₂ → 2P₂O₃
Check: Left: P=4, O=6; Right: P=4, O=6 ✔
✔ Balanced: 4P + 3O₂ → 2P₂O₃
---
3) Sb₂S₃ + HCl → SbCl₃ + H₂S
Left: Sb=2, S=3, H=1, Cl=1
Right: Sb=1, Cl=3, H=2, S=1
Start with Sb: put 2 in front of SbCl₃ → now right: Sb=2, Cl=6
Then S: put 3 in front of H₂S → right: S=3, H=6
Now H and Cl on right: H=6, Cl=6 → so left needs 6 HCl
✔ Balanced: Sb₂S₃ + 6HCl → 2SbCl₃ + 3H₂S
---
4) 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 2 N → then H from NH₃ = 6, plus H₂SO₄ has 2 H → total H=8 → matches right!
✔ Balanced: 2NH₃ + H₂SO₄ → (NH₄)₂SO₄
---
5) CuO + HCl → CuCl₂ + H₂O
Left: Cu=1, O=1, H=1, Cl=1
Right: Cu=1, Cl=2, H=2, O=1
Need 2 HCl to get 2 Cl and 2 H → then H₂O uses 2 H and 1 O → perfect.
✔ Balanced: CuO + 2HCl → CuCl₂ + H₂O
---
6) AgNO₃ + H₂S → Ag₂S + HNO₃
Left: Ag=1, N=1, O=3, H=2, S=1
Right: Ag=2, S=1, H=1, N=1, O=3
Need 2 AgNO₃ → then right needs 2 HNO₃ → that gives H=2, which matches left H₂S.
✔ Balanced: 2AgNO₃ + H₂S → Ag₂S + 2HNO₃
---
7) Cu + S → Cu₂S
Left: Cu=1, S=1
Right: Cu=2, S=1
Put 2 in front of Cu.
✔ Balanced: 2Cu + S → Cu₂S
---
8) Al + H₃PO₄ → H₂ + AlPO₄
Left: Al=1, H=3, P=1, O=4
Right: H=2, Al=1, P=1, O=4
Hydrogen doesn’t match. Need even H on left? Try 2 H₃PO₄ → H=6 → then need 3 H₂ on right (since 3×2=6)
But then P=2 → so need 2 AlPO₄ → then Al=2 → so left needs 2 Al.
Check: 2Al + 2H₃PO₄ → 3H₂ + 2AlPO₄
Left: Al=2, H=6, P=2, O=8
Right: H=6, Al=2, P=2, O=8 ✔
✔ Balanced: 2Al + 2H₃PO₄ → 3H₂ + 2AlPO₄
---
9) NaNO₃ → NaNO₂ + O₂
Left: Na=1, N=1, O=3
Right: Na=1, N=1, O=2+2=4? Wait: NaNO₂ has O=2, O₂ has O=2 → total O=4 → too many.
Actually: NaNO₃ → NaNO₂ + ½O₂ → but we want whole numbers.
Multiply all by 2: 2NaNO₃ → 2NaNO₂ + O₂
Check: Left: Na=2, N=2, O=6
Right: Na=2, N=2, O=4 + 2 = 6 ✔
✔ Balanced: 2NaNO₃ → 2NaNO₂ + O₂
---
10) Mg(ClO₃)₂ → MgCl₂ + O₂
Left: Mg=1, Cl=2, O=6
Right: Mg=1, Cl=2, O=2 → need more O₂
Oxygen: 6 on left → need 3 O₂ on right (since 3×2=6)
✔ Balanced: Mg(ClO₃)₂ → MgCl₂ + 3O₂
---
11) H₂O₂ → H₂O + O₂
Left: H=2, O=2
Right: H=2, O=1+2=3 → not balanced
Try 2H₂O₂ → 2H₂O + O₂
Left: H=4, O=4
Right: H=4, O=2+2=4 ✔
✔ Balanced: 2H₂O₂ → 2H₂O + O₂
---
12) BaO₂ → BaO + O₂
Left: Ba=1, O=2
Right: Ba=1, O=1+2=3 → no
Try 2BaO₂ → 2BaO + O₂
Left: Ba=2, O=4
Right: Ba=2, O=2+2=4 ✔
✔ Balanced: 2BaO₂ → 2BaO + O₂
---
13) Pb(NO₃)₂ + KCl → PbCl₂ + KNO₃
Left: Pb=1, N=2, O=6, K=1, Cl=1
Right: Pb=1, Cl=2, K=1, N=1, O=3
Need 2 KCl → then Cl=2, K=2 → so right needs 2 KNO₃ → then N=2, O=6 → matches left.
✔ Balanced: Pb(NO₃)₂ + 2KCl → PbCl₂ + 2KNO₃
---
14) P + O₂ → P₂O₅
Left: P=1, O=2
Right: P=2, O=5
Find LCM for O: 2 and 5 → 10 → so 5 O₂ (10 O), 2 P₂O₅ (10 O and 4 P) → so left needs 4 P.
✔ Balanced: 4P + 5O₂ → 2P₂O₅
---
15) NH₄NO₂ → N₂ + H₂O
Left: N=2, H=4, O=2
Right: N=2, H=2, O=1
Need 2 H₂O → then H=4, O=2 → matches.
✔ Balanced: NH₄NO₂ → N₂ + 2H₂O
---
16) H₂ + N₂ → NH₃
Left: H=2, N=2
Right: N=1, H=3
LCM for H: 2 and 3 → 6 → so 3 H₂ (6 H), 2 NH₃ (6 H and 2 N) → left N₂ already has 2 N.
✔ Balanced: 3H₂ + N₂ → 2NH₃
---
17) Cl₂ + KBr → KCl + Br₂
Left: Cl=2, K=1, Br=1
Right: K=1, Cl=1, Br=2
Need 2 KBr → then Br=2, K=2 → so right needs 2 KCl → then Cl=2 → matches left Cl₂.
✔ Balanced: Cl₂ + 2KBr → 2KCl + Br₂
---
18) BaCl₂ + (NH₄)₂CO₃ → BaCO₃ + NH₄Cl
Left: Ba=1, Cl=2, N=2, H=8, C=1, O=3
Right: Ba=1, C=1, O=3, N=1, H=4, Cl=1
Need 2 NH₄Cl → then N=2, H=8, Cl=2 → matches left.
✔ Balanced: BaCl₂ + (NH₄)₂CO₃ → BaCO₃ + 2NH₄Cl
---
19) MgCO₃ + HCl → MgCl₂ + CO₂ + H₂O
Left: Mg=1, C=1, O=3, H=1, Cl=1
Right: Mg=1, Cl=2, C=1, O=2+1=3, H=2
Need 2 HCl → then H=2, Cl=2 → matches right.
✔ Balanced: MgCO₃ + 2HCl → MgCl₂ + CO₂ + H₂O
---
20) H₂CO₃ → H₂O + CO₂
Left: H=2, C=1, O=3
Right: H=2, O=1+2=3, C=1 → already balanced!
✔ Balanced: H₂CO₃ → H₂O + CO₂
---
Final Answer:
1) 4Na + O₂ → 2Na₂O
2) 4P + 3O₂ → 2P₂O₃
3) Sb₂S₃ + 6HCl → 2SbCl₃ + 3H₂S
4) 2NH₃ + H₂SO₄ → (NH₄)₂SO₄
5) CuO + 2HCl → CuCl₂ + H₂O
6) 2AgNO₃ + H₂S → Ag₂S + 2HNO₃
7) 2Cu + S → Cu₂S
8) 2Al + 2H₃PO₄ → 3H₂ + 2AlPO₄
9) 2NaNO₃ → 2NaNO₂ + O₂
10) Mg(ClO₃)₂ → MgCl₂ + 3O₂
11) 2H₂O₂ → 2H₂O + O₂
12) 2BaO₂ → 2BaO + O₂
13) Pb(NO₃)₂ + 2KCl → PbCl₂ + 2KNO₃
14) 4P + 5O₂ → 2P₂O₅
15) NH₄NO₂ → N₂ + 2H₂O
16) 3H₂ + N₂ → 2NH₃
17) Cl₂ + 2KBr → 2KCl + Br₂
18) BaCl₂ + (NH₄)₂CO₃ → BaCO₃ + 2NH₄Cl
19) MgCO₃ + 2HCl → MgCl₂ + CO₂ + H₂O
20) H₂CO₃ → H₂O + CO₂
We’ll start with #1 and work our way down to #20.
---
1) Na + O₂ → Na₂O
Left: Na=1, O=2
Right: Na=2, O=1
To fix oxygen: put 2 in front of Na₂O → now right has O=2, but Na=4
So put 4 in front of Na on left.
✔ Balanced: 4Na + O₂ → 2Na₂O
---
2) P + O₂ → P₂O₃
Left: P=1, O=2
Right: P=2, O=3
Make P match: put 2 in front of P → left P=2
Now O: right has 3, left has 2 → need common multiple? Let’s try 3 O₂ (gives 6 O) → then right needs 2 P₂O₃ (which gives 6 O and 4 P) → so left needs 4 P.
Wait — let’s do it step by step:
Try: 4P + 3O₂ → 2P₂O₃
Check: Left: P=4, O=6; Right: P=4, O=6 ✔
✔ Balanced: 4P + 3O₂ → 2P₂O₃
---
3) Sb₂S₃ + HCl → SbCl₃ + H₂S
Left: Sb=2, S=3, H=1, Cl=1
Right: Sb=1, Cl=3, H=2, S=1
Start with Sb: put 2 in front of SbCl₃ → now right: Sb=2, Cl=6
Then S: put 3 in front of H₂S → right: S=3, H=6
Now H and Cl on right: H=6, Cl=6 → so left needs 6 HCl
✔ Balanced: Sb₂S₃ + 6HCl → 2SbCl₃ + 3H₂S
---
4) 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 2 N → then H from NH₃ = 6, plus H₂SO₄ has 2 H → total H=8 → matches right!
✔ Balanced: 2NH₃ + H₂SO₄ → (NH₄)₂SO₄
---
5) CuO + HCl → CuCl₂ + H₂O
Left: Cu=1, O=1, H=1, Cl=1
Right: Cu=1, Cl=2, H=2, O=1
Need 2 HCl to get 2 Cl and 2 H → then H₂O uses 2 H and 1 O → perfect.
✔ Balanced: CuO + 2HCl → CuCl₂ + H₂O
---
6) AgNO₃ + H₂S → Ag₂S + HNO₃
Left: Ag=1, N=1, O=3, H=2, S=1
Right: Ag=2, S=1, H=1, N=1, O=3
Need 2 AgNO₃ → then right needs 2 HNO₃ → that gives H=2, which matches left H₂S.
✔ Balanced: 2AgNO₃ + H₂S → Ag₂S + 2HNO₃
---
7) Cu + S → Cu₂S
Left: Cu=1, S=1
Right: Cu=2, S=1
Put 2 in front of Cu.
✔ Balanced: 2Cu + S → Cu₂S
---
8) Al + H₃PO₄ → H₂ + AlPO₄
Left: Al=1, H=3, P=1, O=4
Right: H=2, Al=1, P=1, O=4
Hydrogen doesn’t match. Need even H on left? Try 2 H₃PO₄ → H=6 → then need 3 H₂ on right (since 3×2=6)
But then P=2 → so need 2 AlPO₄ → then Al=2 → so left needs 2 Al.
Check: 2Al + 2H₃PO₄ → 3H₂ + 2AlPO₄
Left: Al=2, H=6, P=2, O=8
Right: H=6, Al=2, P=2, O=8 ✔
✔ Balanced: 2Al + 2H₃PO₄ → 3H₂ + 2AlPO₄
---
9) NaNO₃ → NaNO₂ + O₂
Left: Na=1, N=1, O=3
Right: Na=1, N=1, O=2+2=4? Wait: NaNO₂ has O=2, O₂ has O=2 → total O=4 → too many.
Actually: NaNO₃ → NaNO₂ + ½O₂ → but we want whole numbers.
Multiply all by 2: 2NaNO₃ → 2NaNO₂ + O₂
Check: Left: Na=2, N=2, O=6
Right: Na=2, N=2, O=4 + 2 = 6 ✔
✔ Balanced: 2NaNO₃ → 2NaNO₂ + O₂
---
10) Mg(ClO₃)₂ → MgCl₂ + O₂
Left: Mg=1, Cl=2, O=6
Right: Mg=1, Cl=2, O=2 → need more O₂
Oxygen: 6 on left → need 3 O₂ on right (since 3×2=6)
✔ Balanced: Mg(ClO₃)₂ → MgCl₂ + 3O₂
---
11) H₂O₂ → H₂O + O₂
Left: H=2, O=2
Right: H=2, O=1+2=3 → not balanced
Try 2H₂O₂ → 2H₂O + O₂
Left: H=4, O=4
Right: H=4, O=2+2=4 ✔
✔ Balanced: 2H₂O₂ → 2H₂O + O₂
---
12) BaO₂ → BaO + O₂
Left: Ba=1, O=2
Right: Ba=1, O=1+2=3 → no
Try 2BaO₂ → 2BaO + O₂
Left: Ba=2, O=4
Right: Ba=2, O=2+2=4 ✔
✔ Balanced: 2BaO₂ → 2BaO + O₂
---
13) Pb(NO₃)₂ + KCl → PbCl₂ + KNO₃
Left: Pb=1, N=2, O=6, K=1, Cl=1
Right: Pb=1, Cl=2, K=1, N=1, O=3
Need 2 KCl → then Cl=2, K=2 → so right needs 2 KNO₃ → then N=2, O=6 → matches left.
✔ Balanced: Pb(NO₃)₂ + 2KCl → PbCl₂ + 2KNO₃
---
14) P + O₂ → P₂O₅
Left: P=1, O=2
Right: P=2, O=5
Find LCM for O: 2 and 5 → 10 → so 5 O₂ (10 O), 2 P₂O₅ (10 O and 4 P) → so left needs 4 P.
✔ Balanced: 4P + 5O₂ → 2P₂O₅
---
15) NH₄NO₂ → N₂ + H₂O
Left: N=2, H=4, O=2
Right: N=2, H=2, O=1
Need 2 H₂O → then H=4, O=2 → matches.
✔ Balanced: NH₄NO₂ → N₂ + 2H₂O
---
16) H₂ + N₂ → NH₃
Left: H=2, N=2
Right: N=1, H=3
LCM for H: 2 and 3 → 6 → so 3 H₂ (6 H), 2 NH₃ (6 H and 2 N) → left N₂ already has 2 N.
✔ Balanced: 3H₂ + N₂ → 2NH₃
---
17) Cl₂ + KBr → KCl + Br₂
Left: Cl=2, K=1, Br=1
Right: K=1, Cl=1, Br=2
Need 2 KBr → then Br=2, K=2 → so right needs 2 KCl → then Cl=2 → matches left Cl₂.
✔ Balanced: Cl₂ + 2KBr → 2KCl + Br₂
---
18) BaCl₂ + (NH₄)₂CO₃ → BaCO₃ + NH₄Cl
Left: Ba=1, Cl=2, N=2, H=8, C=1, O=3
Right: Ba=1, C=1, O=3, N=1, H=4, Cl=1
Need 2 NH₄Cl → then N=2, H=8, Cl=2 → matches left.
✔ Balanced: BaCl₂ + (NH₄)₂CO₃ → BaCO₃ + 2NH₄Cl
---
19) MgCO₃ + HCl → MgCl₂ + CO₂ + H₂O
Left: Mg=1, C=1, O=3, H=1, Cl=1
Right: Mg=1, Cl=2, C=1, O=2+1=3, H=2
Need 2 HCl → then H=2, Cl=2 → matches right.
✔ Balanced: MgCO₃ + 2HCl → MgCl₂ + CO₂ + H₂O
---
20) H₂CO₃ → H₂O + CO₂
Left: H=2, C=1, O=3
Right: H=2, O=1+2=3, C=1 → already balanced!
✔ Balanced: H₂CO₃ → H₂O + CO₂
---
Final Answer:
1) 4Na + O₂ → 2Na₂O
2) 4P + 3O₂ → 2P₂O₃
3) Sb₂S₃ + 6HCl → 2SbCl₃ + 3H₂S
4) 2NH₃ + H₂SO₄ → (NH₄)₂SO₄
5) CuO + 2HCl → CuCl₂ + H₂O
6) 2AgNO₃ + H₂S → Ag₂S + 2HNO₃
7) 2Cu + S → Cu₂S
8) 2Al + 2H₃PO₄ → 3H₂ + 2AlPO₄
9) 2NaNO₃ → 2NaNO₂ + O₂
10) Mg(ClO₃)₂ → MgCl₂ + 3O₂
11) 2H₂O₂ → 2H₂O + O₂
12) 2BaO₂ → 2BaO + O₂
13) Pb(NO₃)₂ + 2KCl → PbCl₂ + 2KNO₃
14) 4P + 5O₂ → 2P₂O₅
15) NH₄NO₂ → N₂ + 2H₂O
16) 3H₂ + N₂ → 2NH₃
17) Cl₂ + 2KBr → 2KCl + Br₂
18) BaCl₂ + (NH₄)₂CO₃ → BaCO₃ + 2NH₄Cl
19) MgCO₃ + 2HCl → MgCl₂ + CO₂ + H₂O
20) H₂CO₃ → H₂O + CO₂
Parent Tip: Review the logic above to help your child master the concept of balancing equations chemistry worksheet.
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