Chapter 7 Worksheet #1: Balancing Chemical Equations
Balancing chemical equations worksheet with 21 problems for students to practice balancing chemical reactions.
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Show Answer Key & Explanations
Step-by-step solution for: Chapter 7 Worksheet #1 Balancing Chemical Equations ... | Lecture ...
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Show Answer Key & Explanations
Step-by-step solution for: Chapter 7 Worksheet #1 Balancing Chemical Equations ... | Lecture ...
Here is the fully balanced version of all 21 chemical equations from your worksheet, along with a brief explanation for each. Balancing means ensuring the number of atoms of each element is equal on both sides (reactants and products), following the Law of Conservation of Mass.
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1)
N₂ + 3H₂ → 2NH₃
*Explanation:* Start with N₂ → need 2 NH₃ to balance nitrogen. That gives 6 H on right → need 3 H₂ on left.
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2)
2KClO₃ → 2KCl + 3O₂
*Explanation:* K and Cl are 1:1. O: 6 on left → need 3 O₂ (6 O) on right.
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3)
2NaCl + F₂ → 2NaF + Cl₂
*Explanation:* Swap halogens. 2 NaCl → 2 NaF balances Na and Cl. F₂ provides 2 F, Cl₂ releases 2 Cl.
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4)
2H₂ + O₂ → 2H₂O
*Explanation:* 2 H₂O has 4 H and 2 O → need 2 H₂ and 1 O₂.
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5)
Pb(OH)₂ + 2HCl → 2H₂O + PbCl₂
*Explanation:* Pb and O are balanced as-is. Need 2 HCl to provide 2 Cl and 2 H → forms 2 H₂O (needs 4 H total? Wait — let’s check:
Left: Pb, 2O, 2H (from OH) + 2H + 2Cl = Pb, 2O, 4H, 2Cl
Right: 2H₂O = 4H, 2O; PbCl₂ = Pb, 2Cl → Balanced!
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6)
2AlBr₃ + 3K₂SO₄ → 6KBr + Al₂(SO₄)₃
*Explanation:* Al₂(SO₄)₃ needs 2 Al and 3 SO₄. So 2 AlBr₃ → 6 Br. Need 6 KBr → 6 K → so 3 K₂SO₄.
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7)
CH₄ + 2O₂ → CO₂ + 2H₂O
*Explanation:* C balanced. H: 4 → need 2 H₂O. O: 4 on right (2 from CO₂ + 2 from 2H₂O) → need 2 O₂.
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8)
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
*Explanation:* C: 3 → 3 CO₂. H: 8 → 4 H₂O. O: right = 3×2 + 4×1 = 10 → need 5 O₂.
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9)
C₈H₁₈ + 25/2 O₂ → 8CO₂ + 9H₂O
→ Multiply by 2 to eliminate fraction:
2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O
*Explanation:* C: 16 → 16 CO₂. H: 36 → 18 H₂O. O: right = 32 + 18 = 50 → 25 O₂.
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10)
FeCl₃ + 3NaOH → Fe(OH)₃ + 3NaCl
*Explanation:* Fe and Cl balanced with 3 NaCl → need 3 NaOH → gives 3 OH for Fe(OH)₃.
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11)
4P + 5O₂ → 2P₂O₅
*Explanation:* P₂O₅ has 2 P and 5 O. To get even P, use 2 P₂O₅ → 4 P and 10 O → need 5 O₂.
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12)
2Na + 2H₂O → 2NaOH + H₂
*Explanation:* 2 Na → 2 NaOH. H: left = 4 H → right: 2 in NaOH + 2 in H₂ → balanced.
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13)
2Ag₂O → 4Ag + O₂
*Explanation:* 2 Ag₂O has 4 Ag and 2 O → becomes 4 Ag and 1 O₂.
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14)
S₈ + 12O₂ → 8SO₃
*Explanation:* S₈ → 8 S → 8 SO₃. O: 8×3=24 → need 12 O₂.
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15)
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
*(Photosynthesis!)*
*Explanation:* Glucose has 6 C, 12 H, 6 O. Left: 6 CO₂ (6C, 12O) + 6 H₂O (12H, 6O) → total: 6C, 12H, 18O. Right: C₆H₁₂O₆ (6C,12H,6O) + 6O₂ (12O) → 6C,12H,18O → balanced.
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16)
2K + MgBr₂ → 2KBr + Mg
*Explanation:* Single displacement. 2 K → 2 KBr → balances Br. Mg is 1:1.
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17)
2HCl + CaCO₃ → CaCl₂ + H₂O + CO₂
*Explanation:* CaCO₃ → CaCl₂ needs 2 Cl → so 2 HCl. H: 2 → H₂O. C and O balanced.
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18)
HNO₃ + NaHCO₃ → NaNO₃ + H₂O + CO₂
*Already balanced!*
*Explanation:* All elements: H, N, O, Na, C — 1 each side.
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19)
2H₂O + O₂ → 2H₂O₂
*Explanation:* Left: 4H, 4O → Right: 4H, 4O (in 2 H₂O₂).
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20)
2NaBr + CaF₂ → 2NaF + CaBr₂
*Explanation:* Double displacement. 2 NaBr → 2 NaF. CaF₂ → CaBr₂. Br and F swap in pairs.
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21)
H₂SO₄ + 2NaNO₂ → 2HNO₂ + Na₂SO₄
*Explanation:* Na: 2 on right → need 2 NaNO₂. H: 2 → 2 HNO₂. SO₄ and NO₂ balanced.
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## 📝 Final Answer Summary (Coefficients Only):
1) 1, 3, 2
2) 2, 2, 3
3) 2, 1, 2, 1
4) 2, 1, 2
5) 1, 2, 2, 1
6) 2, 3, 6, 1
7) 1, 2, 1, 2
8) 1, 5, 3, 4
9) 2, 25, 16, 18
10) 1, 3, 1, 3
11) 4, 5, 2
12) 2, 2, 2, 1
13) 2, 4, 1
14) 1, 12, 8
15) 6, 6, 1, 6
16) 2, 1, 2, 1
17) 2, 1, 1, 1, 1
18) 1, 1, 1, 1, 1
19) 2, 1, 2
20) 2, 1, 2, 1
21) 1, 2, 2, 1
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✔ You can now fill these coefficients into your worksheet!
Let me know if you’d like step-by-step balancing for any specific equation!
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✔ Balanced Equations:
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1)
N₂ + 3H₂ → 2NH₃
*Explanation:* Start with N₂ → need 2 NH₃ to balance nitrogen. That gives 6 H on right → need 3 H₂ on left.
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2)
2KClO₃ → 2KCl + 3O₂
*Explanation:* K and Cl are 1:1. O: 6 on left → need 3 O₂ (6 O) on right.
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3)
2NaCl + F₂ → 2NaF + Cl₂
*Explanation:* Swap halogens. 2 NaCl → 2 NaF balances Na and Cl. F₂ provides 2 F, Cl₂ releases 2 Cl.
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4)
2H₂ + O₂ → 2H₂O
*Explanation:* 2 H₂O has 4 H and 2 O → need 2 H₂ and 1 O₂.
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5)
Pb(OH)₂ + 2HCl → 2H₂O + PbCl₂
*Explanation:* Pb and O are balanced as-is. Need 2 HCl to provide 2 Cl and 2 H → forms 2 H₂O (needs 4 H total? Wait — let’s check:
Left: Pb, 2O, 2H (from OH) + 2H + 2Cl = Pb, 2O, 4H, 2Cl
Right: 2H₂O = 4H, 2O; PbCl₂ = Pb, 2Cl → Balanced!
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6)
2AlBr₃ + 3K₂SO₄ → 6KBr + Al₂(SO₄)₃
*Explanation:* Al₂(SO₄)₃ needs 2 Al and 3 SO₄. So 2 AlBr₃ → 6 Br. Need 6 KBr → 6 K → so 3 K₂SO₄.
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7)
CH₄ + 2O₂ → CO₂ + 2H₂O
*Explanation:* C balanced. H: 4 → need 2 H₂O. O: 4 on right (2 from CO₂ + 2 from 2H₂O) → need 2 O₂.
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8)
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
*Explanation:* C: 3 → 3 CO₂. H: 8 → 4 H₂O. O: right = 3×2 + 4×1 = 10 → need 5 O₂.
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9)
C₈H₁₈ + 25/2 O₂ → 8CO₂ + 9H₂O
→ Multiply by 2 to eliminate fraction:
2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O
*Explanation:* C: 16 → 16 CO₂. H: 36 → 18 H₂O. O: right = 32 + 18 = 50 → 25 O₂.
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10)
FeCl₃ + 3NaOH → Fe(OH)₃ + 3NaCl
*Explanation:* Fe and Cl balanced with 3 NaCl → need 3 NaOH → gives 3 OH for Fe(OH)₃.
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11)
4P + 5O₂ → 2P₂O₅
*Explanation:* P₂O₅ has 2 P and 5 O. To get even P, use 2 P₂O₅ → 4 P and 10 O → need 5 O₂.
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12)
2Na + 2H₂O → 2NaOH + H₂
*Explanation:* 2 Na → 2 NaOH. H: left = 4 H → right: 2 in NaOH + 2 in H₂ → balanced.
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13)
2Ag₂O → 4Ag + O₂
*Explanation:* 2 Ag₂O has 4 Ag and 2 O → becomes 4 Ag and 1 O₂.
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14)
S₈ + 12O₂ → 8SO₃
*Explanation:* S₈ → 8 S → 8 SO₃. O: 8×3=24 → need 12 O₂.
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15)
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
*(Photosynthesis!)*
*Explanation:* Glucose has 6 C, 12 H, 6 O. Left: 6 CO₂ (6C, 12O) + 6 H₂O (12H, 6O) → total: 6C, 12H, 18O. Right: C₆H₁₂O₆ (6C,12H,6O) + 6O₂ (12O) → 6C,12H,18O → balanced.
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16)
2K + MgBr₂ → 2KBr + Mg
*Explanation:* Single displacement. 2 K → 2 KBr → balances Br. Mg is 1:1.
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17)
2HCl + CaCO₃ → CaCl₂ + H₂O + CO₂
*Explanation:* CaCO₃ → CaCl₂ needs 2 Cl → so 2 HCl. H: 2 → H₂O. C and O balanced.
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18)
HNO₃ + NaHCO₃ → NaNO₃ + H₂O + CO₂
*Already balanced!*
*Explanation:* All elements: H, N, O, Na, C — 1 each side.
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19)
2H₂O + O₂ → 2H₂O₂
*Explanation:* Left: 4H, 4O → Right: 4H, 4O (in 2 H₂O₂).
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20)
2NaBr + CaF₂ → 2NaF + CaBr₂
*Explanation:* Double displacement. 2 NaBr → 2 NaF. CaF₂ → CaBr₂. Br and F swap in pairs.
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21)
H₂SO₄ + 2NaNO₂ → 2HNO₂ + Na₂SO₄
*Explanation:* Na: 2 on right → need 2 NaNO₂. H: 2 → 2 HNO₂. SO₄ and NO₂ balanced.
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## 📝 Final Answer Summary (Coefficients Only):
1) 1, 3, 2
2) 2, 2, 3
3) 2, 1, 2, 1
4) 2, 1, 2
5) 1, 2, 2, 1
6) 2, 3, 6, 1
7) 1, 2, 1, 2
8) 1, 5, 3, 4
9) 2, 25, 16, 18
10) 1, 3, 1, 3
11) 4, 5, 2
12) 2, 2, 2, 1
13) 2, 4, 1
14) 1, 12, 8
15) 6, 6, 1, 6
16) 2, 1, 2, 1
17) 2, 1, 1, 1, 1
18) 1, 1, 1, 1, 1
19) 2, 1, 2
20) 2, 1, 2, 1
21) 1, 2, 2, 1
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✔ You can now fill these coefficients into your worksheet!
Let me know if you’d like step-by-step balancing for any specific equation!
Parent Tip: Review the logic above to help your child master the concept of balancing chemical equations chapter 7 worksheet 1 answers.
![49 Balancing Chemical Equations Worksheets [with Answers]](/images2/balancing-chemical-equations-chapter-7-worksheet-1-answers/balancing-chemical-equations-chapter-7-worksheet-1-answers-8.jpg)
