Chapter 7 Worksheet #1 Balancing Chemical Equations ... | Lecture ... - Free Printable

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 and work our way down, checking each as we go.

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1) N₂ + H₂ → NH₃

Left: 2 N, 2 H
Right: 1 N, 3 H

Try putting a 2 in front of NH₃ → now right has 2 N, 6 H
So left needs 3 H₂ (to get 6 H) and 1 N₂ (already has 2 N)

✔ Balanced: 1 N₂ + 3 H₂ → 2 NH₃

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2) KClO₃ → KCl + O₂

Left: 1 K, 1 Cl, 3 O
Right: 1 K, 1 Cl, 2 O

Need to make oxygen match. LCM of 3 and 2 is 6.

Put 2 in front of KClO₃ → 6 O on left
Then put 3 in front of O₂ → 6 O on right
Now K and Cl are 2 on left → need 2 KCl on right

✔ Balanced: 2 KClO₃ → 2 KCl + 3 O₂

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3) NaCl + F₂ → NaF + Cl₂

Left: 1 Na, 1 Cl, 2 F
Right: 1 Na, 1 F, 2 Cl

Fluorine: need even number on right → try 2 NaF → then need 2 NaCl on left
Chlorine: 2 Cl on left → need 1 Cl₂ on right
Fluorine: 2 F on left → matches 2 F in 2 NaF

✔ Balanced: 2 NaCl + 1 F₂ → 2 NaF + 1 Cl₂

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4) H₂ + O₂ → H₂O

Left: 2 H, 2 O
Right: 2 H, 1 O

Need 2 H₂O on right → 4 H, 2 O
Then left needs 2 H₂ (for 4 H) and 1 O₂ (for 2 O)

✔ Balanced: 2 H₂ + 1 O₂ → 2 H₂O

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5) Pb(OH)₂ + HCl → H₂O + PbCl₂

Left: Pb=1, O=2, H=2+1=3? Wait — Pb(OH)₂ has Pb, 2 O, 2 H; plus HCl adds more H and Cl.

Better to count per side:

Pb(OH)₂: Pb=1, O=2, H=2
HCl: H=1, Cl=1 → total left: Pb=1, O=2, H=3, Cl=1

Right: H₂O: H=2, O=1; PbCl₂: Pb=1, Cl=2 → total: Pb=1, Cl=2, H=2, O=1

Not balanced.

Try 2 HCl on left → now H=2 (from OH) + 2 (from HCl) = 4 H; Cl=2
Right: need 2 H₂O → gives 4 H and 2 O → matches left O=2

✔ Balanced: 1 Pb(OH)₂ + 2 HCl → 2 H₂O + 1 PbCl₂

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6) AlBr₃ + K₂SO₄ → KBr + Al₂(SO₄)₃

Left: Al=1, Br=3, K=2, S=1, O=4
Right: K=1, Br=1, Al=2, S=3, O=12

Big jump. Start with Al: right has 2 Al → so 2 AlBr₃ on left → Al=2, Br=6
Sulfate: right has 3 SO₄ → so 3 K₂SO₄ on left → K=6, S=3, O=12
Now right: KBr must be 6 to match K=6 → Br=6 → matches left Br=6
Al₂(SO₄)₃ already set.

✔ Balanced: 2 AlBr₃ + 3 K₂SO₄ → 6 KBr + 1 Al₂(SO₄)₃

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7) CH₄ + O₂ → CO₂ + H₂O

Left: C=1, H=4, O=2
Right: C=1, O=2+1=3? Wait: CO₂ has 2 O, H₂O has 1 O → total 3 O if 1 H₂O

But H: right has 2 H per H₂O → need 2 H₂O for 4 H → then O from products: CO₂ (2) + 2 H₂O (2) = 4 O → so need 2 O₂ on left

✔ Balanced: 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O

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8) C₃H₈ + O₂ → CO₂ + H₂O

Left: C=3, H=8, O=2
Right: C=1, H=2, O=3 (if 1 CO₂ and 1 H₂O)

Set C: 3 CO₂ → C=3
Set H: 4 H₂O → H=8
Now O on right: 3×2 + 4×1 = 6+4=10 → so need 5 O₂ on left

✔ Balanced: 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

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9) C₈H₁₈ + O₂ → CO₂ + H₂O

Left: C=8, H=18, O=2
Right: set C=8 → 8 CO₂
Set H=18 → 9 H₂O (since 9×2=18)
O on right: 8×2 + 9×1 = 16+9=25 → so need 25/2 O₂ → multiply all by 2 to eliminate fraction

Multiply entire equation by 2:

2 C₈H₁₈ + ? O₂ → 16 CO₂ + 18 H₂O
O on right: 16×2 + 18×1 = 32+18=50 → so 25 O₂ on left

✔ Balanced: 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O

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10) FeCl₃ + NaOH → Fe(OH)₃ + NaCl

Left: Fe=1, Cl=3, Na=1, O=1, H=1
Right: Fe=1, O=3, H=3, Na=1, Cl=1

Cl: left 3, right 1 → need 3 NaCl on right → then Na=3 on right → need 3 NaOH on left
Now O and H: 3 NaOH → 3 O, 3 H → matches Fe(OH)₃

✔ Balanced: 1 FeCl₃ + 3 NaOH → 1 Fe(OH)₃ + 3 NaCl

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11) P + O₂ → P₂O₅

Left: P=1, O=2
Right: P=2, O=5

LCM for O: 10 → so 2 P₂O₅ → O=10, P=4
Then left: 4 P and 5 O₂ (since 5×2=10 O)

✔ Balanced: 4 P + 5 O₂ → 2 P₂O₅

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12) Na + H₂O → NaOH + H₂

Left: Na=1, H=2, O=1
Right: Na=1, O=1, H=1+2=3? Wait: NaOH has 1 H, H₂ has 2 H → total 3 H

Not matching. Try 2 Na on left → then 2 NaOH on right → Na=2, O=2, H=2 from NaOH
Plus H₂ → say 1 H₂ → total H=2+2=4? No.

Wait: let’s write:

Try 2 Na + 2 H₂O → 2 NaOH + 1 H₂
Left: Na=2, H=4, O=2
Right: Na=2, O=2, H=2 (from NaOH) + 2 (from H₂) = 4 → yes!

✔ Balanced: 2 Na + 2 H₂O → 2 NaOH + 1 H₂

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13) Ag₂O → Ag + O₂

Left: Ag=2, O=1
Right: Ag=1, O=2

Need 2 Ag on right → so 2 Ag
O: left 1, right 2 → so 2 Ag₂O on left → Ag=4, O=2
Then right: 4 Ag and 1 O₂

✔ Balanced: 2 Ag₂O → 4 Ag + 1 O₂

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14) S₈ + O₂ → SO₃

Left: S=8, O=2
Right: S=1, O=3

Set S: 8 SO₃ → S=8, O=24
Then O₂ needed: 24/2 = 12

✔ Balanced: 1 S₈ + 12 O₂ → 8 SO₃

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15) CO₂ + H₂O → C₆H₁₂O₆ + O₂

This is photosynthesis reverse? Let’s balance.

Left: C=1, O=2+1=3, H=2
Right: C=6, H=12, O=6+2=8? Wait: C₆H₁₂O₆ has 6 O, O₂ has 2 → total 8 O

To get C=6 on left → 6 CO₂
H=12 → 6 H₂O (since 6×2=12 H)
Now left O: 6×2 + 6×1 = 12+6=18
Right: C₆H₁₂O₆ has 6 O, so remaining O must be in O₂ → 18-6=12 O → so 6 O₂

✔ Balanced: 6 CO₂ + 6 H₂O → 1 C₆H₁₂O₆ + 6 O₂

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16) K + MgBr₂ → KBr + Mg

Left: K=1, Mg=1, Br=2
Right: K=1, Br=1, Mg=1

Br: left 2, right 1 → need 2 KBr → then K=2 on right → need 2 K on left

✔ Balanced: 2 K + 1 MgBr₂ → 2 KBr + 1 Mg

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17) HCl + CaCO₃ → CaCl₂ + H₂O + CO₂

Left: H=1, Cl=1, Ca=1, C=1, O=3
Right: Ca=1, Cl=2, H=2, O=1+2=3, C=1

Cl: left 1, right 2 → need 2 HCl
Then H: 2 H on left → need 1 H₂O on right → H=2
Everything else matches.

✔ Balanced: 2 HCl + 1 CaCO₃ → 1 CaCl₂ + 1 H₂O + 1 CO₂

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18) HNO₃ + NaHCO₃ → NaNO₃ + H₂O + CO₂

Left: H=1+1=2, N=1, O=3+3=6, Na=1, C=1
Right: Na=1, N=1, O=3+1+2=6, H=2, C=1

Already balanced! All elements match.

✔ Balanced: 1 HNO₃ + 1 NaHCO₃ → 1 NaNO₃ + 1 H₂O + 1 CO₂

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19) H₂O + O₂ → H₂O₂

Left: H=2, O=1+2=3
Right: H=2, O=2

Not balanced. Try 2 H₂O₂ on right → H=4, O=4
Then left: need 2 H₂O → H=4, O=2; plus O₂ → O=2 → total O=4 → perfect

✔ Balanced: 2 H₂O + 1 O₂ → 2 H₂O₂

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20) NaBr + CaF₂ → NaF + CaBr₂

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

Br: left 1, right 2 → need 2 NaBr
Then Na=2 → need 2 NaF
F: left 2, right 2 → good
Ca: 1 on both sides

✔ Balanced: 2 NaBr + 1 CaF₂ → 2 NaF + 1 CaBr₂

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21) H₂SO₄ + NaNO₂ → HNO₂ + Na₂SO₄

Left: H=2, S=1, O=4+2=6, Na=1, N=1
Right: H=1, N=1, O=2+4=6, Na=2, S=1

Na: left 1, right 2 → need 2 NaNO₂ on left → Na=2, N=2
Then right: need 2 HNO₂ → H=2, N=2
H: left 2 (from H₂SO₄), right 2 → good
S and O also match.

✔ Balanced: 1 H₂SO₄ + 2 NaNO₂ → 2 HNO₂ + 1 Na₂SO₄

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Final Answer:
1) 1 N₂ + 3 H₂ → 2 NH₃
2) 2 KClO₃ → 2 KCl + 3 O₂
3) 2 NaCl + 1 F₂ → 2 NaF + 1 Cl₂
4) 2 H₂ + 1 O₂ → 2 H₂O
5) 1 Pb(OH)₂ + 2 HCl → 2 H₂O + 1 PbCl₂
6) 2 AlBr₃ + 3 K₂SO₄ → 6 KBr + 1 Al₂(SO₄)₃
7) 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
8) 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
9) 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O
10) 1 FeCl₃ + 3 NaOH → 1 Fe(OH)₃ + 3 NaCl
11) 4 P + 5 O₂ → 2 P₂O₅
12) 2 Na + 2 H₂O → 2 NaOH + 1 H₂
13) 2 Ag₂O → 4 Ag + 1 O₂
14) 1 S₈ + 12 O₂ → 8 SO₃
15) 6 CO₂ + 6 H₂O → 1 C₆H₁₂O₆ + 6 O₂
16) 2 K + 1 MgBr₂ → 2 KBr + 1 Mg
17) 2 HCl + 1 CaCO₃ → 1 CaCl₂ + 1 H₂O + 1 CO₂
18) 1 HNO₃ + 1 NaHCO₃ → 1 NaNO₃ + 1 H₂O + 1 CO₂
19) 2 H₂O + 1 O₂ → 2 H₂O₂
20) 2 NaBr + 1 CaF₂ → 2 NaF + 1 CaBr₂
21) 1 H₂SO₄ + 2 NaNO₂ → 2 HNO₂ + 1 Na₂SO₄