Practice worksheet for balancing chemical equations with blank spaces for answers.

Let’s solve each equation step by step. We’ll balance them one at a time, making sure the number of atoms for each element is the same on both sides.

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1) _AlBr₃ + _K → _KBr + _Al

Left: Al=1, Br=3, K=1
Right: K=1, Br=1, Al=1

Br is unbalanced (3 vs 1). Put 3 in front of KBr:

→ AlBr₃ + _K → 3KBr + _Al

Now right has K=3, so put 3 in front of K on left:

→ AlBr₃ + 3K → 3KBr + Al

Check:
Left: Al=1, Br=3, K=3
Right: K=3, Br=3, Al=1 ✔

Answer: 1, 3, 3, 1

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2) _FeO + _PdF₂ → _FeF₂ + _PdO

Left: Fe=1, O=1, Pd=1, F=2
Right: Fe=1, F=2, Pd=1, O=1

Already balanced! All elements match.

Answer: 1, 1, 1, 1

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3) _P₄ + _Br₂ → _PBr₃

Left: P=4, Br=2
Right: P=1, Br=3

Need to balance P first. Put 4 in front of PBr₃:

→ P₄ + _Br₂ → 4PBr₃

Now right has Br = 4×3 = 12 → need 6 Br₂ on left (since each Br₂ gives 2 Br)

→ P₄ + 6Br₂ → 4PBr₃

Check:
Left: P=4, Br=12
Right: P=4, Br=12 ✔

Answer: 1, 6, 4

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4) _LiCl + _Br₂ → _LiBr + _Cl₂

Left: Li=1, Cl=1, Br=2
Right: Li=1, Br=1, Cl=2

Cl and Br are swapped. Need even numbers.

Try putting 2 in front of LiCl and 2 in front of LiBr:

→ 2LiCl + _Br₂ → 2LiBr + _Cl₂

Now left: Li=2, Cl=2, Br=2
Right: Li=2, Br=2, Cl=2 ✔

Answer: 2, 1, 2, 1

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5) _PbBr₂ + _HCl → _HBr + _PbCl₂

Left: Pb=1, Br=2, H=1, Cl=1
Right: H=1, Br=1, Pb=1, Cl=2

Br and Cl are off. Need 2 HBr and 2 HCl:

→ PbBr₂ + 2HCl → 2HBr + PbCl₂

Check:
Left: Pb=1, Br=2, H=2, Cl=2
Right: H=2, Br=2, Pb=1, Cl=2 ✔

Answer: 1, 2, 2, 1

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6) _CoBr₂ + _CaSO₄ → _CaBr₂ + _CoSO₄

Left: Co=1, Br=2, Ca=1, S=1, O=4
Right: Ca=1, Br=2, Co=1, S=1, O=4

Already balanced!

Answer: 1, 1, 1, 1

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7) _Na₃P + _CaF₂ → _NaF + _Ca₃P₂

Left: Na=3, P=1, Ca=1, F=2
Right: Na=1, F=1, Ca=3, P=2

Start with P: need 2 Na₃P on left → gives 6 Na and 2 P

→ 2Na₃P + _CaF₂ → _NaF + Ca₃P₂

Now Ca: right has 3 Ca → need 3 CaF₂ on left → gives 6 F

→ 2Na₃P + 3CaF₂ → _NaF + Ca₃P₂

Now F: left has 6 F → need 6 NaF on right

→ 2Na₃P + 3CaF₂ → 6NaF + Ca₃P₂

Check Na: left=6, right=6 ✔
Ca: left=3, right=3 ✔
F: left=6, right=6 ✔
P: left=2, right=2 ✔

Answer: 2, 3, 6, 1

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8) _KI + _MgCl₂ → _MgI₂ + _KCl

Left: K=1, I=1, Mg=1, Cl=2
Right: Mg=1, I=2, K=1, Cl=1

I and Cl are off. Need 2 KI and 2 KCl:

→ 2KI + _MgCl₂ → _MgI₂ + 2KCl

Now check:
Left: K=2, I=2, Mg=1, Cl=2
Right: Mg=1, I=2, K=2, Cl=2 ✔

Answer: 2, 1, 1, 2

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9) _C₄H₈O₂ + _O₂ → _CO₂ + _H₂O

This is combustion. Balance C, then H, then O.

Left: C=4, H=8, O=2+? from O₂
Right: C=1, H=2, O=2+1=3 per CO₂ and H₂O

Put 4 in front of CO₂ → C balanced
Put 4 in front of H₂O → H=8 on right

→ C₄H₈O₂ + _O₂ → 4CO₂ + 4H₂O

Now count O on right:
4CO₂ → 8 O
4H₂O → 4 O
Total = 12 O

Left: C₄H₈O₂ has 2 O → need 10 more from O₂ → 5 O₂ molecules

→ C₄H₈O₂ + 5O₂ → 4CO₂ + 4H₂O

Check:
C: 4=4
H: 8=8
O: 2 + 10 = 12; right: 8+4=12 ✔

Answer: 1, 5, 4, 4

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10) _C₇H₁₆ + _O₂ → _CO₂ + _H₂O

Combustion again.

C: 7 → put 7 CO₂
H: 16 → put 8 H₂O (since 8×2=16 H)

→ C₇H₁ + _O₂ → 7CO₂ + 8H₂O

O on right:
7CO₂ → 14 O
8H₂O → 8 O
Total = 22 O

Left: O₂ only → need 11 O₂

→ C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O

Check:
C:7=7, H:16=16, O:22=22 ✔

Answer: 1, 11, 7, 8

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11) _Ca + _LiNO₃ → _Ca(NO₃)₂ + _Li

Left: Ca=1, Li=1, N=1, O=3
Right: Ca=1, N=2, O=6, Li=1

N and O are double on right → need 2 LiNO₃ on left

→ Ca + 2LiNO₃ → _Ca(NO₃)₂ + _Li

Now Li: left=2 → need 2 Li on right

→ Ca + 2LiNO₃ → Ca(NO₃)₂ + 2Li

Check:
Ca:1=1, Li:2=2, N:2=2, O:6=6 ✔

Answer: 1, 2, 1, 2

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12) _SeCl₆ + _O₂ → _SeO₂ + _Cl₂

Left: Se=1, Cl=6, O=2
Right: Se=1, O=2, Cl=2

Cl: 6 vs 2 → need 3 Cl₂ on right

→ SeCl₆ + _O₂ → SeO₂ + 3Cl₂

O: right has 2 → left O₂ already gives 2 → good

Check:
Se:1=1, Cl:6=6, O:2=2 ✔

Answer: 1, 1, 1, 3

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13) _MgI₂ + _Mn(SO₃)₂ → _MgSO₃ + _MnI₄

Left: Mg=1, I=2, Mn=1, S=2, O=6
Right: Mg=1, S=1, O=3, Mn=1, I=4

I: 2 vs 4 → need 2 MgI₂ on left → gives 4 I and 2 Mg

→ 2MgI₂ + _Mn(SO₃)₂ → _MgSO₃ + _MnI₄

Now Mg: left=2 → need 2 MgSO₃ on right

→ 2MgI₂ + _Mn(SO₃)₂ → 2MgSO₃ + MnI₄

S: left=2, right=2 ✔
O: left=6, right=6 ✔
Mn:1=1 ✔
I:4=4 ✔

Answer: 2, 1, 2, 1

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14) _Zr + _O₂ + _Nb → _ZrO₂ + _Nb₂O₅

This is tricky. Let’s look at products.

ZrO₂ → needs 1 Zr and 2 O
Nb₂O₅ → needs 2 Nb and 5 O

So total O needed: 2 + 5 = 7 → but O₂ comes in pairs → need even O? Wait, we can use fractions or find LCM.

Better to balance Nb first.

Set Nb₂O₅ coefficient to 1 → needs 2 Nb and 5 O

ZrO₂ set to 1 → needs 1 Zr and 2 O

Total O: 5 + 2 = 7 → so O₂ must be 7/2 → multiply entire equation by 2 to eliminate fraction.

Try:

2Zr + ?O₂ + 4Nb → 2ZrO₂ + 2Nb₂O₅

Wait — let’s do it properly.

Assume:

a Zr + b O₂ + c Nb → d ZrO₂ + e Nb₂O₅

Balance:

Zr: a = d
Nb: c = 2e
O: 2b = 2d + 5e

Pick e=2 → then c=4, d=a, and 2b = 2a + 10 → b = a + 5

Pick a=2 → then d=2, b=7

So:

2Zr + 7O₂ + 4Nb → 2ZrO₂ + 2Nb₂O₅

Check:

Zr:2=2
Nb:4=4
O:14 = 4 + 10 = 14 ✔

Answer: 2, 7, 4, 2, 2

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15) _NaBr + _CaF₂ → _NaF + _CaBr₂

Left: Na=1, Br=1, Ca=1, F=2
Right: Na=1, F=1, Ca=1, Br=2

Br and F off. Need 2 NaBr and 2 NaF:

→ 2NaBr + _CaF₂ → 2NaF + _CaBr₂

Check:
Na:2=2, Br:2=2, Ca:1=1, F:2=2 ✔

Answer: 2, 1, 2, 1

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Final Answer:

1) 1, 3, 3, 1
2) 1, 1, 1, 1
3) 1, 6, 4
4) 2, 1, 2, 1
5) 1, 2, 2, 1
6) 1, 1, 1, 1
7) 2, 3, 6, 1
8) 2, 1, 1, 2
9) 1, 5, 4, 4
10) 1, 11, 7, 8
11) 1, 2, 1, 2
12) 1, 1, 1, 3
13) 2, 1, 2, 1
14) 2, 7, 4, 2, 2
15) 2, 1, 2, 1