Solved Balancing Chemical Equations - Lab 7 Balancing | Chegg.com - Free Printable

Let’s go through each equation one by one and balance them step by step. Balancing means making sure the number of atoms of each element is the same on both sides of the arrow.

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1) Br₂ + LiF → LiBr + F₂

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

We need 2 LiBr to get 2 Br on right → so put 2 in front of LiBr
Now right has 2 Li → so we need 2 LiF on left
Now left has 2 F → right needs 1 F₂ (which gives 2 F)

✔ Balanced: 1 Br₂ + 2 LiF → 2 LiBr + 1 F₂

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2) H₃PO₄ + Fe(OH)₂ → H₂O + Fe₃(PO₄)₂

Right has Fe₃ → so need 3 Fe(OH)₂ on left
Right has PO₄ twice → so need 2 H₃PO₄ on left
Now count H and O:

Left:
H from 2 H₃PO₄ = 6H
H from 3 Fe(OH)₂ = 3×2 = 6H → total H = 12
O from 2 H₃PO₄ = 8O
O from 3 Fe(OH)₂ = 3×2 = 6O → total O = 14

Right:
Fe₃(PO₄)₂ has no H or extra O beyond what’s in water
So all H must go to H₂O → 12 H → makes 6 H₂O
Check O in 6 H₂O = 6O
Plus O in Fe₃(PO₄)₂: 8O → total O = 14 ✔

✔ Balanced: 2 H₃PO₄ + 3 Fe(OH)₂ → 6 H₂O + 1 Fe₃(PO₄)₂

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3) C₃H₇OH + O₂ → CO₂ + H₂O

C₃H₇OH is propanol → C₃H₈O (since OH adds one H)

Left: C=3, H=8, O=1 (from alcohol) + ? from O₂
Right: CO₂ and H₂O

Start with C: 3 CO₂
H: 8 H → 4 H₂O
Now O on right: 3×2 = 6 from CO₂ + 4×1 = 4 from H₂O → total 10 O
Left has 1 O from alcohol → need 9 more from O₂ → so 9/2 O₂ → multiply whole thing by 2

Try:
2 C₃H₈O + ? O₂ → 6 CO₂ + 8 H₂O
O on right: 12 + 8 = 20
Left: 2 O from alcohol → need 18 from O₂ → 9 O₂

✔ Balanced: 2 C₃H₇OH + 9 O₂ → 6 CO₂ + 8 H₂O

*(Note: C₃H₇OH is same as C₃H₈O)*

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4) Ni(OH)₃ → Ni₂O₃ + H₂O

Left: Ni=1, O=3, H=3
Right: Ni=2, O=3+?=3 from oxide + ? from water, H=?

Need 2 Ni on left → 2 Ni(OH)₃
Then H = 6 → makes 3 H₂O
O on left: 2×3 = 6
On right: Ni₂O₃ has 3 O, 3 H₂O has 3 O → total 6 ✔

✔ Balanced: 2 Ni(OH)₃ → 1 Ni₂O₃ + 3 H₂O

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5) K₂SO₃ + Mn(OH)₂ → KOH + MnSO₃

Left: K=2, S=1, O=3+2=5? Wait — K₂SO₃ has 3 O, Mn(OH)₂ has 2 O → total O=5? Actually better to count per compound.

Actually, let’s look at ions:

K₂SO₃ → 2K⁺ + SO₃²
Mn(OH)₂ → Mn²⁺ + 2OH⁻
Products: KOH → K⁺ + OH⁻; MnSO₃ → Mn²⁺ + SO₃²⁻

So it’s a double displacement. Just swap partners.

To balance:
Need 2 KOH to match 2 K from K₂SO₃
Then OH: 2 from Mn(OH)₂ → matches 2 OH in 2 KOH
Mn and SO₃ are 1 each side.

✔ Balanced: 1 K₂SO₃ + 1 Mn(OH)₂ → 2 KOH + 1 MnSO₃

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6) NaOH + H₂SO₄ → H₂O + Na₂SO₄

Right has Na₂ → need 2 NaOH on left
Then H: left = 2 (from NaOH) + 2 (from H₂SO₄) = 4 H → makes 2 H₂O
S and O will balance.

Check:
Left: Na=2, O=2+4=6, H=2+2=4, S=1
Right: 2 H₂O → H=4, O=2; Na₂SO₄ → Na=2, S=1, O=4 → total O=6 ✔

✔ Balanced: 2 NaOH + 1 H₂SO₄ → 2 H₂O + 1 Na₂SO₄

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7) Li + Pb(OH)₂ → Pb + LiOH

Left: Li=1, Pb=1, O=2, H=2
Right: Pb=1, Li=1, O=1, H=1 → not balanced

Need 2 LiOH on right → then Li=2, O=2, H=2
So need 2 Li on left

✔ Balanced: 2 Li + 1 Pb(OH)₂ → 1 Pb + 2 LiOH

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8) C₄H₈ + O₂ → CO₂ + H₂O

C=4 → 4 CO₂
H=8 → 4 H₂O
O on right: 4×2 + 4×1 = 8 + 4 = 12 → need 6 O₂ on left

✔ Balanced: 1 C₄H₈ + 6 O₂ → 4 CO₂ + 4 H₂O

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9) Ga(OH)₃ + KF → KOH + GaF₃

Ga: 1 each side
F: need 3 on right → so 3 KF on left
Then K: 3 on left → need 3 KOH on right
OH: 3 on left → 3 on right ✔

✔ Balanced: 1 Ga(OH)₃ + 3 KF → 3 KOH + 1 GaF₃

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10) V + ZnBr₂ → VBr₃ + Zn

V: 1 each
Zn: 1 each
Br: 2 on left, 3 on right → LCM is 6

So: 3 ZnBr₂ → 6 Br → need 2 VBr₃ (which has 6 Br)
Then V: 2 on right → 2 V on left
Zn: 3 on left → 3 Zn on right

✔ Balanced: 2 V + 3 ZnBr₂ → 2 VBr₃ + 3 Zn

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11) As₂O₅ + H₂O → H₃AsO₄

Left: As=2, O=5+1=6, H=2
Right: H₃AsO₄ → if we have 2 of them: As=2, H=6, O=8

Not matching. Try:

As₂O₅ + H₂O → 2 H₃AsO₄
Right: H=6, O=8, As=2
Left: As=2, O=5 + ? from H₂O, H=?

Need 6 H → so 3 H₂O → gives 6 H and 3 O
Total O left: 5 + 3 = 8 ✔

✔ Balanced: 1 As₂O₅ + 3 H₂O → 2 H₃AsO₄

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12) NH₃ + O₂ → NO + H₂O

N: 1 each
H: 3 on left → need 3/2 H₂O on right → multiply by 2 later
O: on right: 1 (NO) + 1.5 (from 1.5 H₂O) = 2.5 → so 5/4 O₂? Messy.

Standard way:

Set NH₃ coefficient = 4 → then N=4 → 4 NO
H=12 → 6 H₂O
O on right: 4 (from NO) + 6 (from H₂O) = 10 → so 5 O₂

✔ Balanced: 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

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13) C₃H₈ + O₂ → CO₂ + H₂O

C=3 → 3 CO₂
H=8 → 4 H₂O
O on right: 6 + 4 = 10 → need 5 O₂

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

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14) NaClO₃ → NaCl + O₂

Left: Na=1, Cl=1, O=3
Right: Na=1, Cl=1, O=2 → not balanced

Need even O on right → try 2 NaClO₃ → O=6 → makes 3 O₂
Then Na and Cl: 2 each → so 2 NaCl

✔ Balanced: 2 NaClO₃ → 2 NaCl + 3 O₂

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15) Ca + O₂ → CaO

Left: Ca=1, O=2
Right: Ca=1, O=1 → need 2 CaO → then Ca=2 → so 2 Ca on left

✔ Balanced: 2 Ca + 1 O₂ → 2 CaO

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Final Answer:

1) 1 Br₂ + 2 LiF → 2 LiBr + 1 F₂
2) 2 H₃PO₄ + 3 Fe(OH)₂ → 6 H₂O + 1 Fe₃(PO₄)₂
3) 2 C₃H₇OH + 9 O₂ → 6 CO₂ + 8 H₂O
4) 2 Ni(OH)₃ → 1 Ni₂O₃ + 3 H₂O
5) 1 K₂SO₃ + 1 Mn(OH)₂ → 2 KOH + 1 MnSO₃
6) 2 NaOH + 1 H₂SO₄ → 2 H₂O + 1 Na₂SO₄
7) 2 Li + 1 Pb(OH)₂ → 1 Pb + 2 LiOH
8) 1 C₄H₈ + 6 O₂ → 4 CO₂ + 4 H₂O
9) 1 Ga(OH)₃ + 3 KF → 3 KOH + 1 GaF₃
10) 2 V + 3 ZnBr₂ → 2 VBr₃ + 3 Zn
11) 1 As₂O₅ + 3 H₂O → 2 H₃AsO₄
12) 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O
13) 1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
14) 2 NaClO₃ → 2 NaCl + 3 O₂
15) 2 Ca + 1 O₂ → 2 CaO