Free Printable Balancing and Classifying Chemical Equations Worksheets - Free Printable
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Step-by-step solution for: Free Printable Balancing and Classifying Chemical Equations Worksheets
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Show Answer Key & Explanations
Step-by-step solution for: Free Printable Balancing and Classifying Chemical Equations Worksheets
Let’s go through each reaction 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:
1) N₂ + H₂ → NH₃
Left: 2 N, 2 H
Right: 1 N, 3 H
To balance nitrogen: put a 2 in front of NH₃ → now right has 2 N and 6 H
Now hydrogen: left has 2 H, need 6 → so put 3 in front of H₂ (3×2=6)
✔ Balanced: 1 N₂ + 3 H₂ → 2 NH₃
---
2) K + Cl₂ → KCl
Left: 1 K, 2 Cl
Right: 1 K, 1 Cl
Need 2 Cl on right → put 2 in front of KCl → now right has 2 K and 2 Cl
So left needs 2 K → put 2 in front of K
✔ Balanced: 2 K + 1 Cl₂ → 2 KCl
---
3) N₂ + F₂ → NF₃
Left: 2 N, 2 F
Right: 1 N, 3 F
Balance N first: put 2 in front of NF₃ → right has 2 N, 6 F
Now F: left has 2 F → need 6 → put 3 in front of F₂ (3×2=6)
✔ Balanced: 1 N₂ + 3 F₂ → 2 NF₃
---
4) C + S₈ → CS₂
Left: 1 C, 8 S
Right: 1 C, 2 S
Sulfur: right has 2 S per molecule → to get 8 S, need 4 CS₂ (4×2=8)
Then carbon: 4 CS₂ needs 4 C → put 4 in front of C
✔ Balanced: 4 C + 1 S₈ → 4 CS₂
---
5) Na + O₂ → Na₂O
Left: 1 Na, 2 O
Right: 2 Na, 1 O
Oxygen: right has 1 O → to match 2 O from left, put 2 in front of Na₂O → now right has 4 Na and 2 O
Na: left has 1 → need 4 → put 4 in front of Na
✔ Balanced: 4 Na + 1 O₂ → 2 Na₂O
---
6) N₂ + O₂ → N₂O₅
Left: 2 N, 2 O
Right: 2 N, 5 O
Oxygen: right has 5 O → least common multiple of 2 and 5 is 10
Put 5 in front of O₂ → 5×2=10 O on left
Put 2 in front of N₂O₅ → 2×5=10 O, and 2×2=4 N on right
Now N: left has 2 N → need 4 → put 2 in front of N₂
✔ Balanced: 2 N₂ + 5 O₂ → 2 N₂O₅
---
7) Al + S₈ → Al₂S₃
Left: 1 Al, 8 S
Right: 2 Al, 3 S
Find LCM for S: 8 and 3 → 24
So S₈ × 3 = 24 S → put 3 in front of S₈
Al₂S₃ × 8 = 24 S → put 8 in front of Al₂S₃ → that gives 16 Al on right
So left needs 16 Al → put 16 in front of Al
✔ Balanced: 16 Al + 3 S₈ → 8 Al₂S₃
---
8) Cs + N₂ → Cs₃N
Left: 1 Cs, 2 N
Right: 3 Cs, 1 N
Nitrogen: right has 1 N → to match 2 N from left, put 2 in front of Cs₃N → now right has 6 Cs and 2 N
Cs: left has 1 → need 6 → put 6 in front of Cs
✔ Balanced: 6 Cs + 1 N₂ → 2 Cs₃N
---
9) Na + Cl₂ → NaCl
Same as #2 but with Na instead of K.
Left: 1 Na, 2 Cl
Right: 1 Na, 1 Cl
Put 2 in front of NaCl → 2 Na, 2 Cl
Put 2 in front of Na
✔ Balanced: 2 Na + 1 Cl₂ → 2 NaCl
---
10) Rb + S₈ → Rb₂S
Left: 1 Rb, 8 S
Right: 2 Rb, 1 S
Sulfur: right has 1 S → to get 8 S, put 8 in front of Rb₂S → 16 Rb and 8 S
Rb: left has 1 → need 16 → put 16 in front of Rb
✔ Balanced: 16 Rb + 1 S₈ → 8 Rb₂S
---
11) Al + O₂ → Al₂O₃
Left: 1 Al, 2 O
Right: 2 Al, 3 O
LCM for O: 2 and 3 → 6
Put 3 in front of O₂ → 6 O
Put 2 in front of Al₂O₃ → 4 Al and 6 O
Al: left has 1 → need 4 → put 4 in front of Al
✔ Balanced: 4 Al + 3 O₂ → 2 Al₂O₃
---
12) Sb + Cl₂ → SbCl₃
Left: 1 Sb, 2 Cl
Right: 1 Sb, 3 Cl
Cl: LCM of 2 and 3 is 6
Put 3 in front of Cl₂ → 6 Cl
Put 2 in front of SbCl₃ → 2 Sb and 6 Cl
Sb: left has 1 → need 2 → put 2 in front of Sb
✔ Balanced: 2 Sb + 3 Cl₂ → 2 SbCl₃
---
13) Mg + O₂ → MgO
Left: 1 Mg, 2 O
Right: 1 Mg, 1 O
Oxygen: put 2 in front of MgO → 2 Mg, 2 O
Mg: left has 1 → need 2 → put 2 in front of Mg
✔ Balanced: 2 Mg + 1 O₂ → 2 MgO
---
14) Sn + Cl₂ → SnCl₄
Left: 1 Sn, 2 Cl
Right: 1 Sn, 4 Cl
Chlorine: right has 4 Cl → left has 2 → put 2 in front of Cl₂ (2×2=4)
✔ Balanced: 1 Sn + 2 Cl₂ → 1 SnCl₄
---
15) P + O₂ → P₄O₆
Left: 1 P, 2 O
Right: 4 P, 6 O
Phosphorus: right has 4 → put 4 in front of P
Oxygen: right has 6 → left has 2 → put 3 in front of O₂ (3×2=6)
✔ Balanced: 4 P + 3 O₂ → 1 P₄O₆
---
Final Answer:
1) 1 N₂ + 3 H₂ → 2 NH₃
2) 2 K + 1 Cl₂ → 2 KCl
3) 1 N₂ + 3 F₂ → 2 NF₃
4) 4 C + 1 S₈ → 4 CS₂
5) 4 Na + 1 O₂ → 2 Na₂O
6) 2 N₂ + 5 O₂ → 2 N₂O₅
7) 16 Al + 3 S₈ → 8 Al₂S₃
8) 6 Cs + 1 N₂ → 2 Cs₃N
9) 2 Na + 1 Cl₂ → 2 NaCl
10) 16 Rb + 1 S₈ → 8 Rb₂S
11) 4 Al + 3 O₂ → 2 Al₂O₃
12) 2 Sb + 3 Cl₂ → 2 SbCl₃
13) 2 Mg + 1 O₂ → 2 MgO
14) 1 Sn + 2 Cl₂ → 1 SnCl₄
15) 4 P + 3 O₂ → 1 P₄O₆
We’ll start with #1:
1) N₂ + H₂ → NH₃
Left: 2 N, 2 H
Right: 1 N, 3 H
To balance nitrogen: put a 2 in front of NH₃ → now right has 2 N and 6 H
Now hydrogen: left has 2 H, need 6 → so put 3 in front of H₂ (3×2=6)
✔ Balanced: 1 N₂ + 3 H₂ → 2 NH₃
---
2) K + Cl₂ → KCl
Left: 1 K, 2 Cl
Right: 1 K, 1 Cl
Need 2 Cl on right → put 2 in front of KCl → now right has 2 K and 2 Cl
So left needs 2 K → put 2 in front of K
✔ Balanced: 2 K + 1 Cl₂ → 2 KCl
---
3) N₂ + F₂ → NF₃
Left: 2 N, 2 F
Right: 1 N, 3 F
Balance N first: put 2 in front of NF₃ → right has 2 N, 6 F
Now F: left has 2 F → need 6 → put 3 in front of F₂ (3×2=6)
✔ Balanced: 1 N₂ + 3 F₂ → 2 NF₃
---
4) C + S₈ → CS₂
Left: 1 C, 8 S
Right: 1 C, 2 S
Sulfur: right has 2 S per molecule → to get 8 S, need 4 CS₂ (4×2=8)
Then carbon: 4 CS₂ needs 4 C → put 4 in front of C
✔ Balanced: 4 C + 1 S₈ → 4 CS₂
---
5) Na + O₂ → Na₂O
Left: 1 Na, 2 O
Right: 2 Na, 1 O
Oxygen: right has 1 O → to match 2 O from left, put 2 in front of Na₂O → now right has 4 Na and 2 O
Na: left has 1 → need 4 → put 4 in front of Na
✔ Balanced: 4 Na + 1 O₂ → 2 Na₂O
---
6) N₂ + O₂ → N₂O₅
Left: 2 N, 2 O
Right: 2 N, 5 O
Oxygen: right has 5 O → least common multiple of 2 and 5 is 10
Put 5 in front of O₂ → 5×2=10 O on left
Put 2 in front of N₂O₅ → 2×5=10 O, and 2×2=4 N on right
Now N: left has 2 N → need 4 → put 2 in front of N₂
✔ Balanced: 2 N₂ + 5 O₂ → 2 N₂O₅
---
7) Al + S₈ → Al₂S₃
Left: 1 Al, 8 S
Right: 2 Al, 3 S
Find LCM for S: 8 and 3 → 24
So S₈ × 3 = 24 S → put 3 in front of S₈
Al₂S₃ × 8 = 24 S → put 8 in front of Al₂S₃ → that gives 16 Al on right
So left needs 16 Al → put 16 in front of Al
✔ Balanced: 16 Al + 3 S₈ → 8 Al₂S₃
---
8) Cs + N₂ → Cs₃N
Left: 1 Cs, 2 N
Right: 3 Cs, 1 N
Nitrogen: right has 1 N → to match 2 N from left, put 2 in front of Cs₃N → now right has 6 Cs and 2 N
Cs: left has 1 → need 6 → put 6 in front of Cs
✔ Balanced: 6 Cs + 1 N₂ → 2 Cs₃N
---
9) Na + Cl₂ → NaCl
Same as #2 but with Na instead of K.
Left: 1 Na, 2 Cl
Right: 1 Na, 1 Cl
Put 2 in front of NaCl → 2 Na, 2 Cl
Put 2 in front of Na
✔ Balanced: 2 Na + 1 Cl₂ → 2 NaCl
---
10) Rb + S₈ → Rb₂S
Left: 1 Rb, 8 S
Right: 2 Rb, 1 S
Sulfur: right has 1 S → to get 8 S, put 8 in front of Rb₂S → 16 Rb and 8 S
Rb: left has 1 → need 16 → put 16 in front of Rb
✔ Balanced: 16 Rb + 1 S₈ → 8 Rb₂S
---
11) Al + O₂ → Al₂O₃
Left: 1 Al, 2 O
Right: 2 Al, 3 O
LCM for O: 2 and 3 → 6
Put 3 in front of O₂ → 6 O
Put 2 in front of Al₂O₃ → 4 Al and 6 O
Al: left has 1 → need 4 → put 4 in front of Al
✔ Balanced: 4 Al + 3 O₂ → 2 Al₂O₃
---
12) Sb + Cl₂ → SbCl₃
Left: 1 Sb, 2 Cl
Right: 1 Sb, 3 Cl
Cl: LCM of 2 and 3 is 6
Put 3 in front of Cl₂ → 6 Cl
Put 2 in front of SbCl₃ → 2 Sb and 6 Cl
Sb: left has 1 → need 2 → put 2 in front of Sb
✔ Balanced: 2 Sb + 3 Cl₂ → 2 SbCl₃
---
13) Mg + O₂ → MgO
Left: 1 Mg, 2 O
Right: 1 Mg, 1 O
Oxygen: put 2 in front of MgO → 2 Mg, 2 O
Mg: left has 1 → need 2 → put 2 in front of Mg
✔ Balanced: 2 Mg + 1 O₂ → 2 MgO
---
14) Sn + Cl₂ → SnCl₄
Left: 1 Sn, 2 Cl
Right: 1 Sn, 4 Cl
Chlorine: right has 4 Cl → left has 2 → put 2 in front of Cl₂ (2×2=4)
✔ Balanced: 1 Sn + 2 Cl₂ → 1 SnCl₄
---
15) P + O₂ → P₄O₆
Left: 1 P, 2 O
Right: 4 P, 6 O
Phosphorus: right has 4 → put 4 in front of P
Oxygen: right has 6 → left has 2 → put 3 in front of O₂ (3×2=6)
✔ Balanced: 4 P + 3 O₂ → 1 P₄O₆
---
Final Answer:
1) 1 N₂ + 3 H₂ → 2 NH₃
2) 2 K + 1 Cl₂ → 2 KCl
3) 1 N₂ + 3 F₂ → 2 NF₃
4) 4 C + 1 S₈ → 4 CS₂
5) 4 Na + 1 O₂ → 2 Na₂O
6) 2 N₂ + 5 O₂ → 2 N₂O₅
7) 16 Al + 3 S₈ → 8 Al₂S₃
8) 6 Cs + 1 N₂ → 2 Cs₃N
9) 2 Na + 1 Cl₂ → 2 NaCl
10) 16 Rb + 1 S₈ → 8 Rb₂S
11) 4 Al + 3 O₂ → 2 Al₂O₃
12) 2 Sb + 3 Cl₂ → 2 SbCl₃
13) 2 Mg + 1 O₂ → 2 MgO
14) 1 Sn + 2 Cl₂ → 1 SnCl₄
15) 4 P + 3 O₂ → 1 P₄O₆
Parent Tip: Review the logic above to help your child master the concept of decomposition and synthesis reactions worksheet.
