Classifying & Balancing Chemical Reactions Name | PDF | Chemical ... - Free Printable
Educational worksheet: Classifying & Balancing Chemical Reactions Name | PDF | Chemical .... Download and print for classroom or home learning activities.
JPG
768×1024
90.3 KB
Free · Personal Use
Quality Assured by Worksheets Library Team
Reviewed for educational accuracy and age-appropriateness
ID: #949286
⭐
Show Answer Key & Explanations
Step-by-step solution for: Classifying & Balancing Chemical Reactions Name | PDF | Chemical ...
▼
Show Answer Key & Explanations
Step-by-step solution for: Classifying & Balancing Chemical Reactions Name | PDF | Chemical ...
Let’s go through each reaction one by one. For each, we’ll:
1. Balance the equation (make sure atoms of each element are equal on both sides).
2. Identify the reaction type using the abbreviations from the box:
- SY = Synthesis
- DC = Decomposition
- C = Combustion
- SD = Single Displacement
- DD = Double Displacement
- AB = Acid-Base
We’ll do this carefully for all 18 reactions.
---
1) HBr + Mg(OH)₂ → MgBr₂ + H₂O
Balance:
Left: H=1+2=3, Br=1, Mg=1, O=2
Right: Mg=1, Br=2, H=2, O=1 → Not balanced.
Try 2 HBr:
2HBr + Mg(OH)₂ → MgBr₂ + 2H₂O
Now: Left: H=2+2=4, Br=2, Mg=1, O=2
Right: Mg=1, Br=2, H=4, O=2 → Balanced!
Type: Acid (HBr) + Base (Mg(OH)₂) → Salt + Water → AB
---
2) PbBr₂ + HCl → HBr + PbCl₂
Balance:
Left: Pb=1, Br=2, H=1, Cl=1
Right: H=1, Br=1, Pb=1, Cl=2 → Not balanced.
Need 2 HCl and 2 HBr:
PbBr₂ + 2HCl → 2HBr + PbCl₂
Now: Pb=1, Br=2, H=2, Cl=2 on both sides → Balanced!
Type: Two compounds swap partners → DD
---
3) CoBr₃ + CaSO₄ → CaBr₂ + Co₂(SO₄)₃
Balance:
Left: Co=1, Br=3, Ca=1, S=1, O=4
Right: Ca=1, Br=2, Co=2, S=3, O=12 → Way off.
Start with Co: need 2 CoBr₃ → gives 6 Br
Then need 3 CaBr₂ to take 6 Br → so 3 Ca on right → need 3 CaSO₄ on left
That gives 3 S and 12 O → matches Co₂(SO₄)₃ on right.
So:
2CoBr₃ + 3CaSO₄ → 3CaBr₂ + Co₂(SO₄)₃
Check: Co=2, Br=6, Ca=3, S=3, O=12 → Balanced!
Type: Swap ions → DD
---
4) C₄H₁₂ + O₂ → CO₂ + H₂O
Wait — C₄H₁₂? That’s unusual. Normally alkanes are CₙH₂ₙ₊₂ → C₄H₁₀. But let’s assume it’s written as is.
Balance:
C: 4 on left → need 4 CO₂
H: 12 on left → need 6 H₂O (since 6×2=12 H)
O: Right side: 4×2 + 6×1 = 8+6=14 O → so need 7 O₂ on left.
Equation:
C₄H₁₂ + 7O₂ → 4CO₂ + 6H₂O
Check: C=4, H=12, O=14 → Balanced!
Type: Hydrocarbon + O₂ → CO₂ + H₂O → C
*(Note: If this was meant to be C₄H₁₀, answer would differ slightly, but we follow what’s written.)*
---
5) B₂O₃ + H₂O → H₃BO₃
Balance:
Left: B=2, O=3+1=4, H=2
Right: H=3, B=1, O=3 → Not balanced.
Need 2 H₃BO₃ on right → B=2, H=6, O=6
Left: B₂O₃ has 3 O, so need 3 H₂O to give 6 H and 3 O → total O=3+3=6, H=6.
So:
B₂O₃ + 3H₂O → 2H₃BO₃
Balanced!
Type: Two things combine to make one → SY
---
6) H₂O → H₂ + O₂
Balance:
Left: H=2, O=1
Right: H=2, O=2 → Not balanced.
Need 2 H₂O → gives 4 H and 2 O
Then need 2 H₂ and 1 O₂ → but O₂ has 2 O → good.
2H₂O → 2H₂ + O₂
Balanced!
Type: One compound breaks into two → DC
---
7) C₂H₄ + O₂ → CO₂ + H₂O
Balance:
C: 2 → need 2 CO₂
H: 4 → need 2 H₂O
O: Right: 2×2 + 2×1 = 6 → need 3 O₂
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
Balanced!
Type: Hydrocarbon + O₂ → C
---
8) Mg + Fe₂O₃ → MgO + Fe
Balance:
Left: Mg=1, Fe=2, O=3
Right: Mg=1, O=1, Fe=1 → Not balanced.
Need 3 MgO to get 3 O → so 3 Mg on left
Fe: 2 on left → need 2 Fe on right
3Mg + Fe₂O₃ → 3MgO + 2Fe
Balanced!
Type: Element replaces another in compound → SD
---
9) H₃PO₄ + Ca(OH)₂ → Ca₃(PO)₂ + H₂O
Balance:
Left: H=3+2=5, P=1, O=4+2=6, Ca=1
Right: Ca=3, P=2, O=8+1=9, H=2 → Messy.
Start with PO₄: need 2 H₃PO₄ → gives 2 P, 6 H, 8 O
Ca₃(PO)₂ needs 3 Ca → so 3 Ca(OH)₂ → gives 3 Ca, 6 O, 6 H
Total left: H=6+6=12, O=8+6=14, Ca=3, P=2
Right: Ca₃(PO₄)₂ has 3 Ca, 2 P, 8 O → plus water.
Water must take remaining H and O: 12 H → 6 H₂O → uses 6 O → total O used: 8+6=14 → perfect.
So:
2H₃PO₄ + 3Ca(OH)₂ → Ca₃(PO₄)₂ + 6H₂O
Balanced!
Type: Acid + Base → AB
---
10) Cl₂ + KBr → KCl + Br₂
Balance:
Left: Cl=2, K=1, Br=1
Right: K=1, Cl=1, Br=2 → Not balanced.
Need 2 KBr and 2 KCl:
Cl₂ + 2KBr → 2KCl + Br₂
Balanced!
Type: Element displaces another → SD
---
11) H₃AsO → As₂O₅ + H₂O
Balance:
Left: H=3, As=1, O=4
Right: As=2, O=5+1=6, H=2 → Not balanced.
Need 2 H₃AsO₄ → gives 6 H, 2 As, 8 O
Right: As₂O₅ has 2 As, 5 O → need 3 H₂O to use 6 H and 3 O → total O=5+3=8 → good.
2H₃AsO₄ → As₂O₅ + 3H₂O
Balanced!
Type: One compound breaks down → DC
---
12) S₈ + O₂ → SO₃
Balance:
S₈ has 8 S → need 8 SO₃ → that’s 8 S and 24 O
O₂ provides 2 O per molecule → need 12 O₂
S₈ + 12O₂ → 8SO₃
Balanced!
Type: Two elements combine → SY
---
13) Bi(NO₃)₃ + Al₂(SO₄)₃ → Bi₂(SO₄)₃ + Al(NO₃)₃
Balance:
Left: Bi=1, N=3, O=9+12=21, Al=2, S=3
Right: Bi=2, S=3, Al=1, N=3, O=12+9=21 → Not balanced.
Need 2 Bi(NO₃)₃ → gives 2 Bi, 6 N, 18 O
Al₂(SO₄)₃ stays → 2 Al, 3 S, 12 O
Total left: Bi=2, Al=2, S=3, N=6, O=30
Right: Bi₂(SO₄)₃ → 2 Bi, 3 S, 12 O
Al(NO₃)₃ → need 2 to get 2 Al, 6 N, 18 O → total O=12+18=30 → good.
So:
2Bi(NO₃)₃ + Al₂(SO₄)₃ → Bi₂(SO₄)₃ + 2Al(NO₃)₃
Balanced!
Type: Swap ions → DD
---
14) HgO → Hg + O₂
Balance:
Left: Hg=1, O=1
Right: Hg=1, O=2 → Not balanced.
Need 2 HgO → gives 2 Hg, 2 O → then O₂ takes 2 O.
2HgO → 2Hg + O₂
Balanced!
Type: Breaks down → DC
---
15) P + O₂ → P₂O₃
Balance:
Left: P=1, O=2
Right: P=2, O=3 → Not balanced.
Need 2 P and 3/2 O₂ → but better to use whole numbers.
Multiply by 2:
4P + 3O₂ → 2P₂O₃
Check: P=4, O=6 → Balanced!
Type: Elements combine → SY
---
16) HBr + Ba(OH)₂ → BaBr₂ + H₂O
Balance:
Left: H=1+2=3, Br=1, Ba=1, O=2
Right: Ba=1, Br=2, H=2, O=1 → Not balanced.
Need 2 HBr → gives 2 H, 2 Br
Ba(OH)₂ gives 2 OH → makes 2 H₂O
2HBr + Ba(OH)₂ → BaBr₂ + 2H₂O
Balanced!
Type: Acid + Base → AB
---
17) Fe + H₂O → Fe₃O₄ + H₂
Balance:
Left: Fe=1, H=2, O=1
Right: Fe=3, O=4, H=2 → Not balanced.
Need 3 Fe on left → 3Fe
Need 4 O → so 4 H₂O → gives 8 H → need 4 H₂ on right
3Fe + 4H₂O → Fe₃O₄ + 4H₂
Check: Fe=3, H=8, O=4 → Balanced!
Type: Element reacts with compound, displacing H → SD
---
18) C₃H₈ + O₂ → H₂O + CO₂
Balance:
C: 3 → 3 CO₂
H: 8 → 4 H₂O
O: Right: 3×2 + 4×1 = 6+4=10 → need 5 O₂
C₃H₈ + 5O₂ → 4H₂O + 3CO₂
Balanced!
Type: Hydrocarbon + O₂ → C
---
Final Answers:
1) 2HBr + Mg(OH)₂ → MgBr₂ + 2H₂O → AB
2) PbBr₂ + 2HCl → 2HBr + PbCl₂ → DD
3) 2CoBr₃ + 3CaSO₄ → 3CaBr₂ + Co₂(SO₄)₃ → DD
4) C₄H₁₂ + 7O₂ → 4CO₂ + 6H₂O → C
5) B₂O₃ + 3H₂O → 2H₃BO₃ → SY
6) 2H₂O → 2H₂ + O₂ → DC
7) C₂H₄ + 3O₂ → 2CO₂ + 2H₂O → C
8) 3Mg + Fe₂O₃ → 3MgO + 2Fe → SD
9) 2H₃PO₄ + 3Ca(OH)₂ → Ca₃(PO₄)₂ + 6H₂O → AB
10) Cl₂ + 2KBr → 2KCl + Br₂ → SD
11) 2H₃AsO₄ → As₂O₅ + 3H₂O → DC
12) S₈ + 12O₂ → 8SO₃ → SY
13) 2Bi(NO₃)₃ + Al₂(SO₄)₃ → Bi₂(SO₄)₃ + 2Al(NO₃)₃ → DD
14) 2HgO → 2Hg + O₂ → DC
15) 4P + 3O₂ → 2P₂O₃ → SY
16) 2HBr + Ba(OH)₂ → BaBr₂ + 2H₂O → AB
17) 3Fe + 4H₂O → Fe₃O₄ + 4H₂ → SD
18) C₃H₈ + 5O₂ → 4H₂O + 3CO₂ → C
──────────────────────────────────────
Final Answer:
1) AB
2) DD
3) DD
4) C
5) SY
6) DC
7) C
8) SD
9) AB
10) SD
11) DC
12) SY
13) DD
14) DC
15) SY
16) AB
17) SD
18) C
1. Balance the equation (make sure atoms of each element are equal on both sides).
2. Identify the reaction type using the abbreviations from the box:
- SY = Synthesis
- DC = Decomposition
- C = Combustion
- SD = Single Displacement
- DD = Double Displacement
- AB = Acid-Base
We’ll do this carefully for all 18 reactions.
---
1) HBr + Mg(OH)₂ → MgBr₂ + H₂O
Balance:
Left: H=1+2=3, Br=1, Mg=1, O=2
Right: Mg=1, Br=2, H=2, O=1 → Not balanced.
Try 2 HBr:
2HBr + Mg(OH)₂ → MgBr₂ + 2H₂O
Now: Left: H=2+2=4, Br=2, Mg=1, O=2
Right: Mg=1, Br=2, H=4, O=2 → Balanced!
Type: Acid (HBr) + Base (Mg(OH)₂) → Salt + Water → AB
---
2) PbBr₂ + HCl → HBr + PbCl₂
Balance:
Left: Pb=1, Br=2, H=1, Cl=1
Right: H=1, Br=1, Pb=1, Cl=2 → Not balanced.
Need 2 HCl and 2 HBr:
PbBr₂ + 2HCl → 2HBr + PbCl₂
Now: Pb=1, Br=2, H=2, Cl=2 on both sides → Balanced!
Type: Two compounds swap partners → DD
---
3) CoBr₃ + CaSO₄ → CaBr₂ + Co₂(SO₄)₃
Balance:
Left: Co=1, Br=3, Ca=1, S=1, O=4
Right: Ca=1, Br=2, Co=2, S=3, O=12 → Way off.
Start with Co: need 2 CoBr₃ → gives 6 Br
Then need 3 CaBr₂ to take 6 Br → so 3 Ca on right → need 3 CaSO₄ on left
That gives 3 S and 12 O → matches Co₂(SO₄)₃ on right.
So:
2CoBr₃ + 3CaSO₄ → 3CaBr₂ + Co₂(SO₄)₃
Check: Co=2, Br=6, Ca=3, S=3, O=12 → Balanced!
Type: Swap ions → DD
---
4) C₄H₁₂ + O₂ → CO₂ + H₂O
Wait — C₄H₁₂? That’s unusual. Normally alkanes are CₙH₂ₙ₊₂ → C₄H₁₀. But let’s assume it’s written as is.
Balance:
C: 4 on left → need 4 CO₂
H: 12 on left → need 6 H₂O (since 6×2=12 H)
O: Right side: 4×2 + 6×1 = 8+6=14 O → so need 7 O₂ on left.
Equation:
C₄H₁₂ + 7O₂ → 4CO₂ + 6H₂O
Check: C=4, H=12, O=14 → Balanced!
Type: Hydrocarbon + O₂ → CO₂ + H₂O → C
*(Note: If this was meant to be C₄H₁₀, answer would differ slightly, but we follow what’s written.)*
---
5) B₂O₃ + H₂O → H₃BO₃
Balance:
Left: B=2, O=3+1=4, H=2
Right: H=3, B=1, O=3 → Not balanced.
Need 2 H₃BO₃ on right → B=2, H=6, O=6
Left: B₂O₃ has 3 O, so need 3 H₂O to give 6 H and 3 O → total O=3+3=6, H=6.
So:
B₂O₃ + 3H₂O → 2H₃BO₃
Balanced!
Type: Two things combine to make one → SY
---
6) H₂O → H₂ + O₂
Balance:
Left: H=2, O=1
Right: H=2, O=2 → Not balanced.
Need 2 H₂O → gives 4 H and 2 O
Then need 2 H₂ and 1 O₂ → but O₂ has 2 O → good.
2H₂O → 2H₂ + O₂
Balanced!
Type: One compound breaks into two → DC
---
7) C₂H₄ + O₂ → CO₂ + H₂O
Balance:
C: 2 → need 2 CO₂
H: 4 → need 2 H₂O
O: Right: 2×2 + 2×1 = 6 → need 3 O₂
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
Balanced!
Type: Hydrocarbon + O₂ → C
---
8) Mg + Fe₂O₃ → MgO + Fe
Balance:
Left: Mg=1, Fe=2, O=3
Right: Mg=1, O=1, Fe=1 → Not balanced.
Need 3 MgO to get 3 O → so 3 Mg on left
Fe: 2 on left → need 2 Fe on right
3Mg + Fe₂O₃ → 3MgO + 2Fe
Balanced!
Type: Element replaces another in compound → SD
---
9) H₃PO₄ + Ca(OH)₂ → Ca₃(PO)₂ + H₂O
Balance:
Left: H=3+2=5, P=1, O=4+2=6, Ca=1
Right: Ca=3, P=2, O=8+1=9, H=2 → Messy.
Start with PO₄: need 2 H₃PO₄ → gives 2 P, 6 H, 8 O
Ca₃(PO)₂ needs 3 Ca → so 3 Ca(OH)₂ → gives 3 Ca, 6 O, 6 H
Total left: H=6+6=12, O=8+6=14, Ca=3, P=2
Right: Ca₃(PO₄)₂ has 3 Ca, 2 P, 8 O → plus water.
Water must take remaining H and O: 12 H → 6 H₂O → uses 6 O → total O used: 8+6=14 → perfect.
So:
2H₃PO₄ + 3Ca(OH)₂ → Ca₃(PO₄)₂ + 6H₂O
Balanced!
Type: Acid + Base → AB
---
10) Cl₂ + KBr → KCl + Br₂
Balance:
Left: Cl=2, K=1, Br=1
Right: K=1, Cl=1, Br=2 → Not balanced.
Need 2 KBr and 2 KCl:
Cl₂ + 2KBr → 2KCl + Br₂
Balanced!
Type: Element displaces another → SD
---
11) H₃AsO → As₂O₅ + H₂O
Balance:
Left: H=3, As=1, O=4
Right: As=2, O=5+1=6, H=2 → Not balanced.
Need 2 H₃AsO₄ → gives 6 H, 2 As, 8 O
Right: As₂O₅ has 2 As, 5 O → need 3 H₂O to use 6 H and 3 O → total O=5+3=8 → good.
2H₃AsO₄ → As₂O₅ + 3H₂O
Balanced!
Type: One compound breaks down → DC
---
12) S₈ + O₂ → SO₃
Balance:
S₈ has 8 S → need 8 SO₃ → that’s 8 S and 24 O
O₂ provides 2 O per molecule → need 12 O₂
S₈ + 12O₂ → 8SO₃
Balanced!
Type: Two elements combine → SY
---
13) Bi(NO₃)₃ + Al₂(SO₄)₃ → Bi₂(SO₄)₃ + Al(NO₃)₃
Balance:
Left: Bi=1, N=3, O=9+12=21, Al=2, S=3
Right: Bi=2, S=3, Al=1, N=3, O=12+9=21 → Not balanced.
Need 2 Bi(NO₃)₃ → gives 2 Bi, 6 N, 18 O
Al₂(SO₄)₃ stays → 2 Al, 3 S, 12 O
Total left: Bi=2, Al=2, S=3, N=6, O=30
Right: Bi₂(SO₄)₃ → 2 Bi, 3 S, 12 O
Al(NO₃)₃ → need 2 to get 2 Al, 6 N, 18 O → total O=12+18=30 → good.
So:
2Bi(NO₃)₃ + Al₂(SO₄)₃ → Bi₂(SO₄)₃ + 2Al(NO₃)₃
Balanced!
Type: Swap ions → DD
---
14) HgO → Hg + O₂
Balance:
Left: Hg=1, O=1
Right: Hg=1, O=2 → Not balanced.
Need 2 HgO → gives 2 Hg, 2 O → then O₂ takes 2 O.
2HgO → 2Hg + O₂
Balanced!
Type: Breaks down → DC
---
15) P + O₂ → P₂O₃
Balance:
Left: P=1, O=2
Right: P=2, O=3 → Not balanced.
Need 2 P and 3/2 O₂ → but better to use whole numbers.
Multiply by 2:
4P + 3O₂ → 2P₂O₃
Check: P=4, O=6 → Balanced!
Type: Elements combine → SY
---
16) HBr + Ba(OH)₂ → BaBr₂ + H₂O
Balance:
Left: H=1+2=3, Br=1, Ba=1, O=2
Right: Ba=1, Br=2, H=2, O=1 → Not balanced.
Need 2 HBr → gives 2 H, 2 Br
Ba(OH)₂ gives 2 OH → makes 2 H₂O
2HBr + Ba(OH)₂ → BaBr₂ + 2H₂O
Balanced!
Type: Acid + Base → AB
---
17) Fe + H₂O → Fe₃O₄ + H₂
Balance:
Left: Fe=1, H=2, O=1
Right: Fe=3, O=4, H=2 → Not balanced.
Need 3 Fe on left → 3Fe
Need 4 O → so 4 H₂O → gives 8 H → need 4 H₂ on right
3Fe + 4H₂O → Fe₃O₄ + 4H₂
Check: Fe=3, H=8, O=4 → Balanced!
Type: Element reacts with compound, displacing H → SD
---
18) C₃H₈ + O₂ → H₂O + CO₂
Balance:
C: 3 → 3 CO₂
H: 8 → 4 H₂O
O: Right: 3×2 + 4×1 = 6+4=10 → need 5 O₂
C₃H₈ + 5O₂ → 4H₂O + 3CO₂
Balanced!
Type: Hydrocarbon + O₂ → C
---
Final Answers:
1) 2HBr + Mg(OH)₂ → MgBr₂ + 2H₂O → AB
2) PbBr₂ + 2HCl → 2HBr + PbCl₂ → DD
3) 2CoBr₃ + 3CaSO₄ → 3CaBr₂ + Co₂(SO₄)₃ → DD
4) C₄H₁₂ + 7O₂ → 4CO₂ + 6H₂O → C
5) B₂O₃ + 3H₂O → 2H₃BO₃ → SY
6) 2H₂O → 2H₂ + O₂ → DC
7) C₂H₄ + 3O₂ → 2CO₂ + 2H₂O → C
8) 3Mg + Fe₂O₃ → 3MgO + 2Fe → SD
9) 2H₃PO₄ + 3Ca(OH)₂ → Ca₃(PO₄)₂ + 6H₂O → AB
10) Cl₂ + 2KBr → 2KCl + Br₂ → SD
11) 2H₃AsO₄ → As₂O₅ + 3H₂O → DC
12) S₈ + 12O₂ → 8SO₃ → SY
13) 2Bi(NO₃)₃ + Al₂(SO₄)₃ → Bi₂(SO₄)₃ + 2Al(NO₃)₃ → DD
14) 2HgO → 2Hg + O₂ → DC
15) 4P + 3O₂ → 2P₂O₃ → SY
16) 2HBr + Ba(OH)₂ → BaBr₂ + 2H₂O → AB
17) 3Fe + 4H₂O → Fe₃O₄ + 4H₂ → SD
18) C₃H₈ + 5O₂ → 4H₂O + 3CO₂ → C
──────────────────────────────────────
Final Answer:
1) AB
2) DD
3) DD
4) C
5) SY
6) DC
7) C
8) SD
9) AB
10) SD
11) DC
12) SY
13) DD
14) DC
15) SY
16) AB
17) SD
18) C
Parent Tip: Review the logic above to help your child master the concept of classifying chemical reactions worksheet.
