Balancing Equations Practice - Free Printable
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Step-by-step solution for: Balancing Equations Practice
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Show Answer Key & Explanations
Step-by-step solution for: Balancing Equations Practice
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:
1) ___ AlBr₃ + ___ K → ___ KBr + ___ Al
Left side: Al, Br, K
Right side: K, Br, Al
Al: 1 on left → need 1 on right → put 1 in front of Al
Br: 3 on left → need 3 on right → put 3 in front of KBr
K: now 3 on right → need 3 on left → put 3 in front of K
✔ Balanced: 1 AlBr₃ + 3 K → 3 KBr + 1 Al
---
2) ___ FeO + ___ PdF₂ → ___ FeF₂ + ___ PdO
Fe: 1 left → 1 right → OK
O: 1 left → 1 right → OK
Pd: 1 left → 1 right → OK
F: 2 left → 2 right → OK
✔ Already balanced! 1 FeO + 1 PdF₂ → 1 FeF₂ + 1 PdO
---
3) ___ P₄ + ___ Br₂ → ___ PBr₃
P: 4 on left → need 4 on right → so 4 PBr₃ → that gives 12 Br on right
Br: 12 on right → need 6 Br₂ on left (since each Br₂ has 2 Br)
✔ 1 P₄ + 6 Br₂ → 4 PBr₃
---
4) ___ LiCl + ___ Br₂ → ___ LiBr + ___ Cl₂
Li: 1 left → 1 right → OK for now
Cl: 1 left → but Cl₂ on right → need even number → try 2 LiCl → then 2 Cl on left → makes 1 Cl₂ on right
Br: Br₂ on left → 2 Br → need 2 LiBr on right → which needs 2 Li → matches 2 LiCl
✔ 2 LiCl + 1 Br₂ → 2 LiBr + 1 Cl₂
---
5) ___ PbBr₂ + ___ HCl → ___ HBr + ___ PbCl₂
Pb: 1 left → 1 right → OK
Br: 2 left → need 2 HBr on right
H: 2 from 2 HBr → need 2 HCl on left
Cl: 2 from 2 HCl → matches PbCl₂
✔ 1 PbBr₂ + 2 HCl → 2 HBr + 1 PbCl₂
---
6) ___ CoBr₃ + ___ CaSO₄ → ___ CaBr₂ + ___ Co₂(SO₄)₃
Co: 1 left → need 2 on right → so 2 CoBr₃
Br: 2×3=6 on left → need 3 CaBr₂ on right (each has 2 Br → 3×2=6)
Ca: 3 on right → need 3 CaSO₄ on left
SO₄: 3 on left → matches Co₂(SO₄)₃ on right
✔ 2 CoBr₃ + 3 CaSO₄ → 3 CaBr₂ + 1 Co₂(SO₄)
---
7) ___ Na₃P + ___ CaF₂ → ___ NaF + ___ Ca₃P₂
Na: 3 left → need 3 NaF? But Ca₃P₂ has 2 P → so need 2 Na₃P → 6 Na → need 6 NaF
P: 2 left → matches Ca₃P₂
Ca: need 3 Ca on right → so 3 CaF₂ → gives 6 F → matches 6 NaF
✔ 2 Na₃P + 3 CaF₂ → 6 NaF + 1 Ca₃P₂
---
8) ___ Mn + ___ HI → ___ H₂ + ___ MnI₃
Mn: 1 left → 1 right → OK
I: 3 on right → need 3 HI on left → but H: 3 on left → H₂ on right → need even H → so double everything
Try 2 Mn → 2 MnI₃ → 6 I → need 6 HI → 6 H → makes 3 H₂
✔ 2 Mn + 6 HI → 3 H₂ + 2 MnI₃
---
9) ___ Li₃PO₄ + ___ NaBr → ___ Na₃PO₄ + ___ LiBr
Li: 3 left → need 3 LiBr on right
Na: 3 on right → need 3 NaBr on left
PO₄: 1 each side → OK
✔ 1 Li₃PO₄ + 3 NaBr → 1 Na₃PO₄ + 3 LiBr
---
10) ___ CaF₂ + ___ Li₂SO₄ → ___ CaSO₄ + ___ LiF
Ca: 1 each → OK
F: 2 left → need 2 LiF on right
Li: 2 on right → need 1 Li₂SO₄ on left → matches SO₄
✔ 1 CaF₂ + 1 Li₂SO₄ → 1 CaSO₄ + 2 LiF
---
11) ___ HBr + ___ Mg(OH)₂ → ___ MgBr₂ + ___ H₂O
Mg: 1 each → OK
Br: 2 on right → need 2 HBr on left
H: 2 from HBr + 2 from Mg(OH)₂ = 4 H → need 2 H₂O on right (each has 2 H → 4 total)
O: 2 from Mg(OH)₂ → 2 from 2 H₂O → OK
✔ 2 HBr + 1 Mg(OH)₂ → 1 MgBr₂ + 2 H₂O
---
12) ___ LiNO₃ + ___ CaBr₂ → ___ Ca(NO₃)₂ + ___ LiBr
Ca: 1 each → OK
NO₃: 2 on right → need 2 LiNO₃ on left
Li: 2 on left → need 2 LiBr on right
Br: 2 on left → matches 2 LiBr
✔ 2 LiNO₃ + 1 CaBr₂ → 1 Ca(NO₃)₂ + 2 LiBr
---
13) ___ AgNO₃ + ___ Li → ___ LiNO₃ + ___ Ag
Ag: 1 each → OK
NO₃: 1 each → OK
Li: 1 each → OK
✔ 1 AgNO₃ + 1 Li → 1 LiNO₃ + 1 Ag
---
14) ___ Si(OH)₄ + ___ NaBr → ___ SiBr₄ + ___ NaOH
Si: 1 each → OK
OH: 4 on left → need 4 NaOH on right → so 4 Na on right → need 4 NaBr on left
Br: 4 on left → matches SiBr₄
✔ 1 Si(OH)₄ + 4 NaBr → 1 SiBr₄ + 4 NaOH
---
15) ___ NaCN + ___ CuCO₃ → ___ Na₂CO₃ + ___ Cu(CN)₂
Na: 2 on right → need 2 NaCN on left
CN: 2 on left → matches Cu(CN)₂
Cu: 1 each → OK
CO₃: 1 each → OK
✔ 2 NaCN + 1 CuCO₃ → 1 Na₂CO₃ + 1 Cu(CN)₂
---
Final Answer:
1) 1 AlBr₃ + 3 K → 3 KBr + 1 Al
2) 1 FeO + 1 PdF₂ → 1 FeF₂ + 1 PdO
3) 1 P₄ + 6 Br₂ → 4 PBr₃
4) 2 LiCl + 1 Br₂ → 2 LiBr + 1 Cl₂
5) 1 PbBr₂ + 2 HCl → 2 HBr + 1 PbCl₂
6) 2 CoBr₃ + 3 CaSO₄ → 3 CaBr₂ + 1 Co₂(SO₄)₃
7) 2 Na₃P + 3 CaF₂ → 6 NaF + 1 Ca₃P₂
8) 2 Mn + 6 HI → 3 H₂ + 2 MnI₃
9) 1 Li₃PO₄ + 3 NaBr → 1 Na₃PO₄ + 3 LiBr
10) 1 CaF₂ + 1 Li₂SO₄ → 1 CaSO₄ + 2 LiF
11) 2 HBr + 1 Mg(OH)₂ → 1 MgBr₂ + 2 H₂O
12) 2 LiNO₃ + 1 CaBr₂ → 1 Ca(NO₃)₂ + 2 LiBr
13) 1 AgNO₃ + 1 Li → 1 LiNO₃ + 1 Ag
14) 1 Si(OH)₄ + 4 NaBr → 1 SiBr₄ + 4 NaOH
15) 2 NaCN + 1 CuCO₃ → 1 Na₂CO₃ + 1 Cu(CN)₂
We’ll start with #1:
1) ___ AlBr₃ + ___ K → ___ KBr + ___ Al
Left side: Al, Br, K
Right side: K, Br, Al
Al: 1 on left → need 1 on right → put 1 in front of Al
Br: 3 on left → need 3 on right → put 3 in front of KBr
K: now 3 on right → need 3 on left → put 3 in front of K
✔ Balanced: 1 AlBr₃ + 3 K → 3 KBr + 1 Al
---
2) ___ FeO + ___ PdF₂ → ___ FeF₂ + ___ PdO
Fe: 1 left → 1 right → OK
O: 1 left → 1 right → OK
Pd: 1 left → 1 right → OK
F: 2 left → 2 right → OK
✔ Already balanced! 1 FeO + 1 PdF₂ → 1 FeF₂ + 1 PdO
---
3) ___ P₄ + ___ Br₂ → ___ PBr₃
P: 4 on left → need 4 on right → so 4 PBr₃ → that gives 12 Br on right
Br: 12 on right → need 6 Br₂ on left (since each Br₂ has 2 Br)
✔ 1 P₄ + 6 Br₂ → 4 PBr₃
---
4) ___ LiCl + ___ Br₂ → ___ LiBr + ___ Cl₂
Li: 1 left → 1 right → OK for now
Cl: 1 left → but Cl₂ on right → need even number → try 2 LiCl → then 2 Cl on left → makes 1 Cl₂ on right
Br: Br₂ on left → 2 Br → need 2 LiBr on right → which needs 2 Li → matches 2 LiCl
✔ 2 LiCl + 1 Br₂ → 2 LiBr + 1 Cl₂
---
5) ___ PbBr₂ + ___ HCl → ___ HBr + ___ PbCl₂
Pb: 1 left → 1 right → OK
Br: 2 left → need 2 HBr on right
H: 2 from 2 HBr → need 2 HCl on left
Cl: 2 from 2 HCl → matches PbCl₂
✔ 1 PbBr₂ + 2 HCl → 2 HBr + 1 PbCl₂
---
6) ___ CoBr₃ + ___ CaSO₄ → ___ CaBr₂ + ___ Co₂(SO₄)₃
Co: 1 left → need 2 on right → so 2 CoBr₃
Br: 2×3=6 on left → need 3 CaBr₂ on right (each has 2 Br → 3×2=6)
Ca: 3 on right → need 3 CaSO₄ on left
SO₄: 3 on left → matches Co₂(SO₄)₃ on right
✔ 2 CoBr₃ + 3 CaSO₄ → 3 CaBr₂ + 1 Co₂(SO₄)
---
7) ___ Na₃P + ___ CaF₂ → ___ NaF + ___ Ca₃P₂
Na: 3 left → need 3 NaF? But Ca₃P₂ has 2 P → so need 2 Na₃P → 6 Na → need 6 NaF
P: 2 left → matches Ca₃P₂
Ca: need 3 Ca on right → so 3 CaF₂ → gives 6 F → matches 6 NaF
✔ 2 Na₃P + 3 CaF₂ → 6 NaF + 1 Ca₃P₂
---
8) ___ Mn + ___ HI → ___ H₂ + ___ MnI₃
Mn: 1 left → 1 right → OK
I: 3 on right → need 3 HI on left → but H: 3 on left → H₂ on right → need even H → so double everything
Try 2 Mn → 2 MnI₃ → 6 I → need 6 HI → 6 H → makes 3 H₂
✔ 2 Mn + 6 HI → 3 H₂ + 2 MnI₃
---
9) ___ Li₃PO₄ + ___ NaBr → ___ Na₃PO₄ + ___ LiBr
Li: 3 left → need 3 LiBr on right
Na: 3 on right → need 3 NaBr on left
PO₄: 1 each side → OK
✔ 1 Li₃PO₄ + 3 NaBr → 1 Na₃PO₄ + 3 LiBr
---
10) ___ CaF₂ + ___ Li₂SO₄ → ___ CaSO₄ + ___ LiF
Ca: 1 each → OK
F: 2 left → need 2 LiF on right
Li: 2 on right → need 1 Li₂SO₄ on left → matches SO₄
✔ 1 CaF₂ + 1 Li₂SO₄ → 1 CaSO₄ + 2 LiF
---
11) ___ HBr + ___ Mg(OH)₂ → ___ MgBr₂ + ___ H₂O
Mg: 1 each → OK
Br: 2 on right → need 2 HBr on left
H: 2 from HBr + 2 from Mg(OH)₂ = 4 H → need 2 H₂O on right (each has 2 H → 4 total)
O: 2 from Mg(OH)₂ → 2 from 2 H₂O → OK
✔ 2 HBr + 1 Mg(OH)₂ → 1 MgBr₂ + 2 H₂O
---
12) ___ LiNO₃ + ___ CaBr₂ → ___ Ca(NO₃)₂ + ___ LiBr
Ca: 1 each → OK
NO₃: 2 on right → need 2 LiNO₃ on left
Li: 2 on left → need 2 LiBr on right
Br: 2 on left → matches 2 LiBr
✔ 2 LiNO₃ + 1 CaBr₂ → 1 Ca(NO₃)₂ + 2 LiBr
---
13) ___ AgNO₃ + ___ Li → ___ LiNO₃ + ___ Ag
Ag: 1 each → OK
NO₃: 1 each → OK
Li: 1 each → OK
✔ 1 AgNO₃ + 1 Li → 1 LiNO₃ + 1 Ag
---
14) ___ Si(OH)₄ + ___ NaBr → ___ SiBr₄ + ___ NaOH
Si: 1 each → OK
OH: 4 on left → need 4 NaOH on right → so 4 Na on right → need 4 NaBr on left
Br: 4 on left → matches SiBr₄
✔ 1 Si(OH)₄ + 4 NaBr → 1 SiBr₄ + 4 NaOH
---
15) ___ NaCN + ___ CuCO₃ → ___ Na₂CO₃ + ___ Cu(CN)₂
Na: 2 on right → need 2 NaCN on left
CN: 2 on left → matches Cu(CN)₂
Cu: 1 each → OK
CO₃: 1 each → OK
✔ 2 NaCN + 1 CuCO₃ → 1 Na₂CO₃ + 1 Cu(CN)₂
---
Final Answer:
1) 1 AlBr₃ + 3 K → 3 KBr + 1 Al
2) 1 FeO + 1 PdF₂ → 1 FeF₂ + 1 PdO
3) 1 P₄ + 6 Br₂ → 4 PBr₃
4) 2 LiCl + 1 Br₂ → 2 LiBr + 1 Cl₂
5) 1 PbBr₂ + 2 HCl → 2 HBr + 1 PbCl₂
6) 2 CoBr₃ + 3 CaSO₄ → 3 CaBr₂ + 1 Co₂(SO₄)₃
7) 2 Na₃P + 3 CaF₂ → 6 NaF + 1 Ca₃P₂
8) 2 Mn + 6 HI → 3 H₂ + 2 MnI₃
9) 1 Li₃PO₄ + 3 NaBr → 1 Na₃PO₄ + 3 LiBr
10) 1 CaF₂ + 1 Li₂SO₄ → 1 CaSO₄ + 2 LiF
11) 2 HBr + 1 Mg(OH)₂ → 1 MgBr₂ + 2 H₂O
12) 2 LiNO₃ + 1 CaBr₂ → 1 Ca(NO₃)₂ + 2 LiBr
13) 1 AgNO₃ + 1 Li → 1 LiNO₃ + 1 Ag
14) 1 Si(OH)₄ + 4 NaBr → 1 SiBr₄ + 4 NaOH
15) 2 NaCN + 1 CuCO₃ → 1 Na₂CO₃ + 1 Cu(CN)₂
Parent Tip: Review the logic above to help your child master the concept of balancing equation practice worksheet.
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