Solutions to a balancing equations worksheet, showing balanced chemical reactions with coefficients.

Let’s go through each equation one by one and check if the coefficients (the numbers in front) balance the atoms on both sides.

We’ll count atoms for each element on left and right side of the arrow → to make sure they match.

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1) _1_ H₃PO₄ + _3_ KOH → _1_ K₃PO₄ + _3_ H₂O

Left:
H = 3 (from H₃PO₄) + 3×1 (from 3KOH) = 6
P = 1
O = 4 + 3×1 = 7
K = 3

Right:
K = 3 (from K₃PO₄)
P = 1
O = 4 (from K₃PO₄) + 3×1 (from 3H₂O) = 7
H = 3×2 = 6 ✔ Balanced!

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2) _6_ K + _1_ B₂O₃ → _3_ K₂O + _2_ B

Left:
K = 6
B = 2
O = 3

Right:
K = 3×2 = 6
O = 3×1 = 3
B = 2 ✔ Balanced!

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3) _1_ HCl + _1_ NaOH → _1_ NaCl + _1_ H₂O

Left:
H = 1 + 1 = 2
Cl = 1
Na = 1
O = 1

Right:
Na = 1
Cl = 1
H = 2
O = 1 ✔ Balanced!

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4) _10_ Na + _2_ NaNO₃ → _6_ Na₂O + _1_ N₂

Left:
Na = 10 + 2×1 = 12
N = 2×1 = 2
O = 2×3 = 6

Right:
Na = 6×2 = 12
O = 6×1 = 6
N = 2 ✔ Balanced!

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5) _4_ C + _1_ S₈ → _4_ CS₂

Left:
C = 4
S = 8

Right:
C = 4
S = 4×2 = 8 ✔ Balanced!

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6) _4_ Na + _1_ O₂ → _2_ Na₂O

Left:
Na = 4
O = 2

Right:
Na = 2×2 = 4
O = 2×1 = 2 ✔ Balanced!

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7) _2_ N₂ + _5_ O₂ → _2_ N₂O₅

Left:
N = 2×2 = 4
O = 5×2 = 10

Right:
N = 2×2 = 4
O = 2×5 = 10 ✔ Balanced!

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8) _2_ H₃PO₄ + _3_ Mg(OH)₂ → _1_ Mg₃(PO₄)₂ + _6_ H₂O

Left:
H = 2×3 + 3×2 = 6 + 6 = 12
P = 2
O = 2×4 + 3×2 = 8 + 6 = 14
Mg = 3

Right:
Mg = 3
P = 2
O = 8 (from PO₄) + 6 (from H₂O) = 14? Wait — Mg₃(PO₄)₂ has 8 O from phosphate, plus 6 H₂O has 6 O → total 14 O
H = 6×2 = 12 ✔ Balanced!

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9) _2_ NaOH + _1_ H₂CO₃ → _1_ Na₂CO₃ + _2_ H₂O

Left:
Na = 2
O = 2 + 3 = 5
H = 2 + 2 = 4
C = 1

Right:
Na = 2
C = 1
O = 3 + 2 = 5
H = 4 ✔ Balanced!

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10) _1_ KOH + _1_ HBr → _1_ KBr + _1_ H₂O

Left:
K = 1
O = 1
H = 1 + 1 = 2
Br = 1

Right:
K = 1
Br = 1
H = 2
O = 1 ✔ Balanced!

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11) _4_ Na + _1_ O₂ → _2_ Na₂O ← Same as #6 ✔

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12) _2_ Al(OH)₃ + _3_ H₂CO₃ → _1_ Al₂(CO₃)₃ + _6_ H₂O

Left:
Al = 2
O = 2×3 + 3×3 = 6 + 9 = 15
H = 2×3 + 3×2 = 6 + 6 = 12
C = 3

Right:
Al = 2
C = 3
O = 9 (from CO₃) + 6 (from H₂O) = 15
H = 12 ✔ Balanced!

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13) _16_ Al + _3_ S₈ → _8_ Al₂S₃

Left:
Al = 16
S = 3×8 = 24

Right:
Al = 8×2 = 16
S = 8×3 = 24 ✔ Balanced!

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14) _6_ Cs + _1_ N₂ → _2_ Cs₃N

Left:
Cs = 6
N = 2

Right:
Cs = 2×3 = 6
N = 2×1 = 2 ✔ Balanced!

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15) _1_ Mg + _1_ Cl₂ → _1_ MgCl₂

Left:
Mg = 1
Cl = 2

Right:
Mg = 1
Cl = 2 ✔ Balanced!

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16) _10_ Rb + _2_ RbNO₃ → _6_ Rb₂O + _1_ N₂

Left:
Rb = 10 + 2×1 = 12
N = 2
O = 2×3 = 6

Right:
Rb = 6×2 = 12
O = 6
N = 2 ✔ Balanced!

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17) _2_ C₆H + _15_ O₂ → _12_ CO₂ + _6_ H₂O

Left:
C = 2×6 = 12
H = 2×6 = 12
O = 15×2 = 30

Right:
C = 12
H = 6×2 = 12
O = 12×2 + 6×1 = 24 + 6 = 30 ✔ Balanced!

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18) _1_ N₂ + _3_ H₂ → _2_ NH₃

Left:
N = 2
H = 6

Right:
N = 2
H = 6 ✔ Balanced!

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19) _2_ C₁₀H₂₂ + _31_ O₂ → _20_ CO₂ + _22_ H₂O

Left:
C = 2×10 = 20
H = 2×22 = 44
O = 31×2 = 62

Right:
C = 20
H = 22×2 = 44
O = 20×2 + 22×1 = 40 + 22 = 62 ✔ Balanced!

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20) _1_ Al(OH)₃ + _3_ HBr → _1_ AlBr₃ + _3_ H₂O

Left:
Al = 1
O = 3
H = 3 + 3 = 6
Br = 3

Right:
Al = 1
Br = 3
H = 6
O = 3 ✔ Balanced!

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21) _2_ CH₃CH₂CH₂CH₃ + _13_ O₂ → _8_ CO₂ + _10_ H₂O

Note: CH₃CH₂CH₂CH₃ is C₄H₁₀

So: 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

Left:
C = 8
H = 20
O = 26

Right:
C = 8
H = 20
O = 16 + 10 = 26 ✔ Balanced!

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22) _1_ C₃H₈ + _5_ O₂ → _3_ CO₂ + _4_ H₂O

Left:
C = 3
H = 8
O = 10

Right:
C = 3
H = 8
O = 6 + 4 = 10 ✔ Balanced!

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23) _3_ Li + _1_ AlCl₃ → _3_ LiCl + _1_ Al

Left:
Li = 3
Al = 1
Cl = 3

Right:
Li = 3
Cl = 3
Al = 1 ✔ Balanced!

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24) _2_ C₂H₆ + _7_ O₂ → _4_ CO₂ + _6_ H₂O

Left:
C = 4
H = 12
O = 14

Right:
C = 4
H = 12
O = 8 + 6 = 14 ✔ Balanced!

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25) _3_ NH₄OH + _1_ H₃PO₄ → _1_ (NH₄)₃PO₄ + _3_ H₂O

Left:
N = 3
H = 3×5 + 3 = 15 + 3 = 18? Wait — NH₄OH has 5 H? Actually: NH₄⁺ and OH⁻ → so NH₄OH = N, 5H, O

Better:
NH₄OH: N=1, H=5, O=1
H₃PO₄: H=3, P=1, O=4

So left:
N = 3×1 = 3
H = 3×5 + 3 = 15 + 3 = 18
O = 3×1 + 4 = 7
P = 1

Right:
(NH₄)₃PO₄: N=3, H=12, P=1, O=4
H₂O: 3×2=6 H, 3 O → total H=12+6=18, O=4+3=7 ✔ Balanced!

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26) _3_ Rb + _1_ P → _1_ Rb₃P

Left:
Rb = 3
P = 1

Right:
Rb = 3
P = 1 ✔ Balanced!

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27) _1_ CH₄ + _2_ O₂ → _1_ CO₂ + _2_ H₂O

Left:
C = 1
H = 4
O = 4

Right:
C = 1
H = 4
O = 2 + 2 = 4 ✔ Balanced!

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28) _2_ Al(OH)₃ + _3_ H₂SO₄ → _1_ Al₂(SO₄) + _6_ H₂O

Left:
Al = 2
O = 2×3 + 3×4 = 6 + 12 = 18
H = 2×3 + 3×2 = 6 + 6 = 12
S = 3

Right:
Al = 2
S = 3
O = 12 (from SO₄) + 6 (from H₂O) = 18
H = 12 ✔ Balanced!

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29) _2_ Na + _1_ Cl₂ → _2_ NaCl

Left:
Na = 2
Cl = 2

Right:
Na = 2
Cl = 2 ✔ Balanced!

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30) _16_ Rb + _1_ S₈ → _8_ Rb₂S

Left:
Rb = 16
S = 8

Right:
Rb = 8×2 = 16
S = 8 ✔ Balanced!

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31) _2_ H₃PO₄ + _3_ Ca(OH)₂ → _1_ Ca₃(PO₄)₂ + _6_ H₂O

Left:
H = 2×3 + 3×2 = 6 + 6 = 12
P = 2
O = 2×4 + 3×2 = 8 + 6 = 14
Ca = 3

Right:
Ca = 3
P = 2
O = 8 (from PO₄) + 6 (from H₂O) = 14
H = 12 ✔ Balanced!

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32) _1_ NH₃ + _1_ HCl → _1_ NH₄Cl

Left:
N = 1
H = 3 + 1 = 4
Cl = 1

Right:
N = 1
H = 4
Cl = 1 ✔ Balanced!

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33) _2_ Li + _2_ H₂O → _2_ LiOH + _1_ H₂

Left:
Li = 2
H = 4
O = 2

Right:
Li = 2
O = 2
H = 2 (from LiOH) + 2 (from H₂) = 4 ✔ Balanced!

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34) _1_ Ca₃(PO₄)₂ + _3_ SiO₂ + _5_ C → _3_ CaSiO₃ + _5_ CO + _2_ P

Left:
Ca = 3
P = 2
O = 8 (from PO₄) + 3×2 = 6 → total O = 14? Wait:

Ca₃(PO₄)₂: Ca=3, P=2, O=8
SiO₂: 3×Si, 6×O
C: 5

Total left:
Ca=3, P=2, Si=3, O=8+6=14, C=5

Right:
CaSiO₃: 3×Ca, 3×Si, 9×O
CO: 5×C, 5×O
P: 2

Total right:
Ca=3, Si=3, O=9+5=14, C=5, P=2 ✔ Balanced!

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35) _4_ NH₃ + _3_ O₂ → _2_ N₂ + _6_ H₂O

Left:
N = 4
H = 12
O = 6

Right:
N = 4
H = 12
O = 6 ✔ Balanced!

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36) _4_ FeS₂ + _11_ O₂ → _2_ Fe₂O₃ + _8_ SO₂

Left:
Fe = 4
S = 8
O = 22

Right:
Fe = 4
S = 8
O = 6 (from Fe₂O₃) + 16 (from SO₂) = 22 ✔ Balanced!

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37) _5_ C + _2_ SO₂ → _1_ CS₂ + _4_ CO

Left:
C = 5
S = 2
O = 4

Right:
C = 1 + 4 = 5
S = 2
O = 4 ✔ Balanced!

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All equations are correctly balanced as shown in the worksheet.

Final Answer: All 37 chemical equations are correctly balanced with the given coefficients.