Balancing Chemical Equations Worksheet featuring combustion and incomplete combustion reactions with chemical formulas and reaction arrows.

Let’s solve this step by step.

We are balancing combustion reactions for hydrocarbons. In complete combustion, a hydrocarbon (CₓHᵧ) reacts with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O). The general form is:

CₓHᵧ + O₂ → CO₂ + H₂O

To balance:
1. Balance carbon (C) atoms first — put the same number in front of CO₂ as there are C atoms in the hydrocarbon.
2. Balance hydrogen (H) atoms next — since each H₂O has 2 H atoms, you’ll need y/2 H₂O molecules (if y is even; if odd, we’ll adjust later).
3. Finally, balance oxygen (O) atoms — count how many O atoms are on the right side (from CO₂ and H₂O), then divide by 2 to get the coefficient for O₂ (since O₂ has 2 oxygen atoms).

Sometimes we get fractions — that’s okay! We can multiply the whole equation by 2 at the end to make all coefficients whole numbers.

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Let’s go one by one.

Complete Combustion of Hydrocarbons

#### 1. CH₄ + O₂ → CO₂ + H₂O
- C: 1 on left → 1 CO₂
- H: 4 on left → 2 H₂O (because 2×2=4 H)
- Now right side: CO₂ has 2 O, 2 H₂O has 2 O → total 4 O atoms → so 2 O₂ needed
✔ Balanced: CH₄ + 2O₂ → CO₂ + 2H₂O

#### 2. C₂H₆ + O₂ → CO₂ + H₂O
- C: 2 → 2 CO₂
- H: 6 → 3 H₂O
- Right: 2 CO₂ = 4 O, 3 H₂O = 3 O → total 7 O → need 7/2 O₂
→ Multiply entire equation by 2 to eliminate fraction:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
✔ Balanced: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

But wait — the worksheet probably expects smallest whole numbers per molecule. Let’s check standard way:

Actually, for single molecule basis:
C₂H₆ + ? O₂ → 2CO₂ + 3H₂O
O atoms right: 2×2 + 3×1 = 4+3=7 → so 7/2 O₂
So write: C₂H₆ + ⁷/₂ O₂ → 2CO₂ + 3H₂O
But usually we avoid fractions → multiply by 2 → 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

However, looking at the worksheet format, it seems they want coefficients for one molecule of hydrocarbon. So we’ll use fractions where needed, but let’s see what’s typical.

Actually, in most school worksheets, they accept fractional coefficients for O₂ when necessary, or expect you to double everything. But since the blanks are for coefficients, and no instruction says “use smallest integers”, I’ll use the minimal set including fractions if needed — BUT actually, let me check common practice.

Wait — looking at problem 5: C₅H₁₂ — pentane — standard balanced is C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O? Let’s calculate:

C₅H₁₂ + O₂ → 5CO₂ + 6H₂O (since 12 H → 6 H₂O)
Right: 5×2 + 6×1 = 10+6=16 O → so 8 O₂
✔ Yes: C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O

Similarly, for ethane: C₂H₆ → 2CO₂ + 3H₂O → O needed: 4+3=7 → 3.5 O₂ → so 7/2

But perhaps the worksheet expects whole numbers only? Let me look ahead.

Problem 9: C₈H₁₈ — octane — standard is 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O, but again, per molecule: C₈H₁₈ + 12.5 O₂ → 8CO₂ + 9H₂O

I think for consistency, we should use the smallest whole numbers possible, which may require multiplying the entire equation.

But the worksheet has blank lines for coefficients — likely expecting integer coefficients, possibly requiring multiplication.

Looking at the structure, for example, problem 1 is simple: CH₄ + 2O₂ → CO₂ + 2H₂O — all integers.

For problem 2: if we write C₂H₆ + 7/2 O₂ → 2CO₂ + 3H₂O, but 7/2 is not integer.

Perhaps the worksheet allows fractions? Or maybe not.

Another thought: in some curricula, they teach to always use whole numbers, so you double when needed.

But let's proceed carefully.

I recall that for combustion, the standard way is to balance with smallest integers, so for even-numbered hydrogens, it works out nice, for odd, you might need to double.

Let me do them systematically.

General method:

For CₓHᵧ + O₂ → x CO₂ + (y/2) H₂O

Then oxygen atoms on right: 2x + y/2

So O₂ coefficient = (2x + y/2)/2 = x + y/4

So coefficient for O₂ is x + y/4

To avoid fractions, if y is divisible by 4, good; else, multiply entire equation by 4 or 2.

But for simplicity, let's compute for each.

#### 1. CH₄: x=1,y=4 → O₂ = 1 + 4/4 = 2 → CH₄ + 2O₂ → CO₂ + 2H₂O ✔

#### 2. C₂H₆: x=2,y=6 → O₂ = 2 + 6/4 = 2 + 1.5 = 3.5 → so 7/2
Equation: C₂H₆ + 7/2 O₂ → 2CO₂ + 3H₂O
Multiply by 2: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
But since the blank is for one molecule, perhaps write 1, 7/2, 2, 3 — but I doubt it.

Looking at the worksheet, it says "fill in the correct coefficients", and for incomplete combustion, they have CO and C, so likely they expect integer coefficients, meaning we may need to use multipliers.

But in the answer space, it's probably expected to have integers, so for cases like this, we'll use the doubled version, but the hydrocarbon coefficient would be 2, which might not match the "one molecule" idea.

Perhaps for this worksheet, they allow fractional coefficients for O₂.

Let me check online or standard practice.

Upon second thought, in many high school worksheets, for combustion, they do use fractional coefficients for O₂ when necessary, and it's accepted.

For example, C₂H₆ + 3.5 O₂ → 2CO₂ + 3H₂O is sometimes written, but usually as 7/2.

In the blank, we can write fractions.

But let's see problem 5: C₅H₁₂ — x=5,y=12 → O₂ = 5 + 12/4 = 5+3=8 → C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O ✔ integer

Problem 6: C₆H₁₄ — x=6,y=14 → O₂ = 6 + 14/4 = 6 + 3.5 = 9.5 = 19/2
So C₆H₁₄ + 19/2 O₂ → 6CO₂ + 7H₂O
Or multiply by 2: 2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

Again, same issue.

Problem 7: C₇H₁₆ — x=7,y=16 → O₂ = 7 + 16/4 = 7+4=11 → C₇H₁ + 11O₂ → 7CO₂ + 8H₂O ✔

Problem 8: C₈H₁₈ — x=8,y=18 → O₂ = 8 + 18/4 = 8 + 4.5 = 12.5 = 25/2
So C₈H₁₈ + 25/2 O₂ → 8CO₂ + 9H₂O

Problem 9: C₉H₂₀ — x=9,y=20 → O₂ = 9 + 20/4 = 9+5=14 → C₉H₂₀ + 14O₂ → 9CO₂ + 10H₂O ✔

Problem 10: C₄HOH — this is butanol, which is C₄H₁₀O, since OH adds one O and one H, but C₄H₉OH means C4H10O.

Hydrocarbon part is C4H9, but with OH, so molecular formula is C₄H₁₀O.

Combustion: C₄H₁₀O + O₂ → CO₂ + H₂O

Balance C: 4 → 4CO₂

H: 10 → 5H₂O

Now right side O: from 4CO₂ = 8 O, from 5H₂O = 5 O, total 13 O atoms.

Left: C₄H₁₀O has 1 O, so O₂ must provide 12 O atoms → 6 O₂

So: C₄H₁₀O + 6O₂ → 4CO₂ + 5H₂O ✔

All integers.

Now for incomplete combustion, it's different.

First, let's list the complete combustion answers with coefficients.

I think for consistency, since some require fractions, and the worksheet doesn't specify, I'll use the form with smallest integers, which may have hydrocarbon coefficient >1 for some, but that might not be intended.

Perhaps the worksheet expects the coefficient for the hydrocarbon to be 1, and allow fractions for O₂.

In many textbooks, they do that.

For example, for ethane: C₂H₆ + \frac{7}{2} O₂ → 2CO₂ + 3H₂O

And they leave it like that.

Similarly for others.

I think that's acceptable.

So I'll proceed with that.

So for each, coefficient of hydrocarbon is 1, then find others.

#### 1. CH₄ + _ O₂ → _ CO₂ + _ H₂O
As above: 1, 2, 1, 2

#### 2. C₂H₆ + _ O₂ → _ CO₂ + _ H₂O
1, 7/2, 2, 3

But 7/2 is 3.5, but in fraction form.

In the blank, we can write 7/2 or 3.5, but better as fraction.

Since it's a worksheet, probably expect fraction.

#### 3. C₃H₈ + _ O₂ → _ CO₂ + _ H₂O
x=3,y=8 → O₂ = 3 + 8/4 = 3+2=5
So 1, 5, 3, 4 (since H:8→4H₂O)

#### 4. C₄H₁₀ + _ O₂ → _ CO₂ + _ H₂O
x=4,y=10 → O₂ = 4 + 10/4 = 4 + 2.5 = 6.5 = 13/2
H:10→5H₂O
So 1, 13/2, 4, 5

#### 5. C₅H₁₂ + _ O₂ → _ CO₂ + _ H₂O
x=5,y=12 → O₂ = 5 + 12/4 = 5+3=8
H:12→6H₂O
So 1, 8, 5, 6

#### 6. C₆H₁₄ + _ O₂ → _ CO₂ + _ H₂O
x=6,y=14 → O₂ = 6 + 14/4 = 6 + 3.5 = 9.5 = 19/2
H:14→7H₂O
So 1, 19/2, 6, 7

#### 7. C₇H₁₆ + _ O₂ → _ CO₂ + _ H₂O
x=7,y=16 → O₂ = 7 + 16/4 = 7+4=11
H:16→8H₂O
So 1, 11, 7, 8

#### 8. C₈H₁ + _ O₂ → _ CO₂ + _ H₂O
x=8,y=18 → O₂ = 8 + 18/4 = 8 + 4.5 = 12.5 = 25/2
H:18→9H₂O
So 1, 25/2, 8, 9

#### 9. C₉H₂₀ + _ O₂ → _ CO₂ + _ H₂O
x=9,y=20 → O₂ = 9 + 20/4 = 9+5=14
H:20→10H₂O
So 1, 14, 9, 10

#### 10. C₄H₉OH + _ O₂ → _ CO₂ + _ H₂O
As above, C₄H₁₀O + 6O₂ → 4CO₂ + 5H₂O
So 1, 6, 4, 5

Now for incomplete combustion.

Incomplete combustion produces either CO (carbon monoxide) or C (soot/carbon) instead of CO₂.

The worksheet has two sections: one for producing CO, one for producing C.

For producing CO: CₓHᵧ + O₂ → x CO + (y/2) H₂O

Then balance oxygen.

Right side: x CO has x O, (y/2) H₂O has y/2 O, total O atoms = x + y/2

Left: O₂ provides 2 per molecule, so O₂ coefficient = (x + y/2)/2 = x/2 + y/4

Again, may have fractions.

For producing C: CₓHᵧ + O₂ → x C + (y/2) H₂O

Right side: x C has 0 O, (y/2) H₂O has y/2 O, so total O atoms = y/2

Thus O₂ coefficient = (y/2)/2 = y/4

Now let's do them.

Incomplete Combustion Producing CO

#### 11. C₂H₆ + _ O₂ → _ CO + _ H₂O
x=2,y=6
CO: 2
H₂O: 3 (since 6H/2=3)
O atoms right: 2 (from CO) + 3 (from H₂O) = 5
So O₂: 5/2
Thus: 1, 5/2, 2, 3

#### 12. C₄H₁₀ + _ O₂ → _ CO + _ H₂O
x=4,y=10
CO: 4
H₂O: 5
O right: 4 + 5 = 9
O₂: 9/2
So 1, 9/2, 4, 5

#### 13. C₇H₁₆ + _ O₂ → _ CO + _ H₂O
x=7,y=16
CO: 7
H₂O: 8
O right: 7 + 8 = 15
O₂: 15/2
So 1, 15/2, 7, 8

#### 14. C₈H₁₈ + _ O₂ → _ CO + _ H₂O
x=8,y=18
CO: 8
H₂O: 9
O right: 8 + 9 = 17
O₂: 17/2
So 1, 17/2, 8, 9

#### 15. C₅H₁₂ + _ O₂ → _ CO + _ H₂O
x=5,y=12
CO: 5
H₂O: 6
O right: 5 + 6 = 11
O₂: 11/2
So 1, 11/2, 5, 6

#### 16. C₃H₈ + _ O₂ → _ CO + _ H₂O
x=3,y=8
CO: 3
H₂O: 4
O right: 3 + 4 = 7
O₂: 7/2
So 1, 7/2, 3, 4

#### 17. C₆H₁ + _ O₂ → _ CO + _ H₂O
x=6,y=14
CO: 6
H₂O: 7
O right: 6 + 7 = 13
O₂: 13/2
So 1, 13/2, 6, 7

#### 18. C₉H₂₀ + _ O₂ → _ CO + _ H₂O
x=9,y=20
CO: 9
H₂O: 10
O right: 9 + 10 = 19
O₂: 19/2
So 1, 19/2, 9, 10

#### 19. C₄H₈ + _ O₂ → _ CO + _ H₂O
x=4,y=8
CO: 4
H₂O: 4
O right: 4 + 4 = 8
O₂: 8/2 = 4
So 1, 4, 4, 4

#### 20. C₅H₈ + _ O₂ → _ CO + _ H₂O
x=5,y=8
CO: 5
H₂O: 4
O right: 5 + 4 = 9
O₂: 9/2
So 1, 9/2, 5, 4

Now for producing C (carbon soot):

#### 21. C₂H₆ + _ O₂ → _ C + _ H₂O
x=2,y=6
C: 2
H₂O: 3
O right: only from H₂O: 3 atoms (since C has no O)
So O₂: 3/2
Thus: 1, 3/2, 2, 3

#### 22. C₄H₁₀ + _ O₂ → _ C + _ H₂O
x=4,y=10
C: 4
H₂O: 5
O right: 5
O₂: 5/2
So 1, 5/2, 4, 5

#### 23. C₇H₁₆ + _ O₂ → _ C + _ H₂O
x=7,y=16
C: 7
H₂O: 8
O right: 8
O₂: 8/2 = 4
So 1, 4, 7, 8

#### 24. C₈H₁ + _ O₂ → _ C + _ H₂O
x=8,y=18
C: 8
H₂O: 9
O right: 9
O₂: 9/2
So 1, 9/2, 8, 9

#### 25. C₅H₁₂ + _ O₂ → _ C + _ H₂O
x=5,y=12
C: 5
H₂O: 6
O right: 6
O₂: 6/2 = 3
So 1, 3, 5, 6

#### 26. C₃H + _ O₂ → _ C + _ H₂O
x=3,y=8
C: 3
H₂O: 4
O right: 4
O₂: 4/2 = 2
So 1, 2, 3, 4

#### 27. C₆H₁ + _ O₂ → _ C + _ H₂O
x=6,y=14
C: 6
H₂O: 7
O right: 7
O₂: 7/2
So 1, 7/2, 6, 7

#### 28. C₉H₂₀ + _ O₂ → _ C + _ H₂O
x=9,y=20
C: 9
H₂O: 10
O right: 10
O₂: 10/2 = 5
So 1, 5, 9, 10

#### 29. C₄H + _ O₂ → _ C + _ H₂O
x=4,y=8
C: 4
H₂O: 4
O right: 4
O₂: 4/2 = 2
So 1, 2, 4, 4

#### 30. C₅H + _ O₂ → _ C + _ H₂O
x=5,y=8
C: 5
H₂O: 4
O right: 4
O₂: 4/2 = 2
So 1, 2, 5, 4

Now, to present the answers, I'll list them as coefficients for each reaction.

Since the worksheet has blanks, I'll provide the numbers in order.

For fractions, I'll write them as improper fractions, like 7/2, etc.

Some might prefer decimals, but fractions are more precise.

In the context, fractions are fine.

So let's compile the final answers.

First, complete combustion:

1. CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O → coefficients: 1, 2, 1, 2

But typically we don't write 1, so just the numbers.

In the blank, for CH₄ it's already given, so blanks are for O₂, CO₂, H₂O.

Looking at the worksheet:

"1. CH₄ + ___ O₂ → ___ CO₂ + ___ H₂O"

So three blanks: for O₂, CO₂, H₂O.

Similarly for others.

For CH₄, it's 2, 1, 2

But CO₂ coefficient is 1, which is implied, but we need to fill it.

In the blank, we put the number.

So for each, the coefficients to fill are for O₂, then CO₂ or CO or C, then H₂O.

In complete combustion, it's O₂, CO₂, H₂O.

In incomplete producing CO, it's O₂, CO, H₂O.

In incomplete producing C, it's O₂, C, H₂O.

And the hydrocarbon coefficient is 1, not filled.

So for each row, three numbers to fill.

Let me list them.

Complete Combustion:

1. O₂: 2, CO₂: 1, H₂O: 2 → so 2, 1, 2

But usually we omit writing 1, but since it's a blank, we should put 1.

Yes.

2. C₂H₆ + ___ O₂ → ___ CO₂ + ___ H₂O → O₂: 7/2, CO₂: 2, H₂O: 3 → so 7/2, 2, 3

3. C₃H₈ + ___ O₂ → ___ CO₂ + ___ H₂O → O₂: 5, CO₂: 3, H₂O: 4 → 5, 3, 4

4. C₄H₁₀ + ___ O₂ → ___ CO₂ + ___ H₂O → O₂: 13/2, CO₂: 4, H₂O: 5 → 13/2, 4, 5

5. C₅H₁₂ + ___ O₂ → ___ CO₂ + ___ H₂O → O₂: 8, CO₂: 5, H₂O: 6 → 8, 5, 6

6. C₆H₁ + ___ O₂ → ___ CO₂ + ___ H₂O → O₂: 19/2, CO₂: 6, H₂O: 7 → 19/2, 6, 7

7. C₇H₁₆ + ___ O₂ → ___ CO₂ + ___ H₂O → O₂: 11, CO₂: 7, H₂O: 8 → 11, 7, 8

8. C₈H₁₈ + ___ O₂ → ___ CO₂ + ___ H₂O → O₂: 25/2, CO₂: 8, H₂O: 9 → 25/2, 8, 9

9. C₉H₂₀ + ___ O₂ → ___ CO₂ + ___ H₂O → O₂: 14, CO₂: 9, H₂O: 10 → 14, 9, 10

10. C₄H₉OH + ___ O₂ → ___ CO₂ + ___ H₂O → as calculated, O₂: 6, CO₂: 4, H₂O: 5 → 6, 4, 5

Now Incomplete Combustion Producing CO:

11. C₂H₆ + ___ O₂ → ___ CO + ___ H₂O → O₂: 5/2, CO: 2, H₂O: 3 → 5/2, 2, 3

12. C₄H₁₀ + ___ O₂ → ___ CO + ___ H₂O → O₂: 9/2, CO: 4, H₂O: 5 → 9/2, 4, 5

13. C₇H₁₆ + ___ O₂ → ___ CO + ___ H₂O → O₂: 15/2, CO: 7, H₂O: 8 → 15/2, 7, 8

14. C₈H₁₈ + ___ O₂ → ___ CO + ___ H₂O → O₂: 17/2, CO: 8, H₂O: 9 → 17/2, 8, 9

15. C₅H₁₂ + ___ O₂ → ___ CO + ___ H₂O → O₂: 11/2, CO: 5, H₂O: 6 → 11/2, 5, 6

16. C₃H₈ + ___ O₂ → ___ CO + ___ H₂O → O₂: 7/2, CO: 3, H₂O: 4 → 7/2, 3, 4

17. C₆H₁₄ + ___ O₂ → ___ CO + ___ H₂O → O₂: 13/2, CO: 6, H₂O: 7 → 13/2, 6, 7

18. C₉H₂₀ + ___ O₂ → ___ CO + ___ H₂O → O₂: 19/2, CO: 9, H₂O: 10 → 19/2, 9, 10

19. C₄H₈ + ___ O₂ → ___ CO + ___ H₂O → O₂: 4, CO: 4, H₂O: 4 → 4, 4, 4

20. C₅H₈ + ___ O₂ → ___ CO + ___ H₂O → O₂: 9/2, CO: 5, H₂O: 4 → 9/2, 5, 4

Now Incomplete Combustion Producing C:

21. C₂H₆ + ___ O₂ → ___ C + ___ H₂O → O₂: 3/2, C: 2, H₂O: 3 → 3/2, 2, 3

22. C₄H₁₀ + ___ O₂ → ___ C + ___ H₂O → O₂: 5/2, C: 4, H₂O: 5 → 5/2, 4, 5

23. C₇H₁₆ + ___ O₂ → ___ C + ___ H₂O → O₂: 4, C: 7, H₂O: 8 → 4, 7, 8

24. C₈H₁₈ + ___ O₂ → ___ C + ___ H₂O → O₂: 9/2, C: 8, H₂O: 9 → 9/2, 8, 9

25. C₅H₁₂ + ___ O₂ → ___ C + ___ H₂O → O₂: 3, C: 5, H₂O: 6 → 3, 5, 6

26. C₃H₈ + ___ O₂ → ___ C + ___ H₂O → O₂: 2, C: 3, H₂O: 4 → 2, 3, 4

27. C₆H₁₄ + ___ O₂ → ___ C + ___ H₂O → O₂: 7/2, C: 6, H₂O: 7 → 7/2, 6, 7

28. C₉H₂₀ + ___ O₂ → ___ C + ___ H₂O → O₂: 5, C: 9, H₂O: 10 → 5, 9, 10

29. C₄H₈ + ___ O₂ → ___ C + ___ H₂O → O₂: 2, C: 4, H₂O: 4 → 2, 4, 4

30. C₅H₈ + ___ O₂ → ___ C + ___ H₂O → O₂: 2, C: 5, H₂O: 4 → 2, 5, 4

Now, to box the final answer, since it's a list, I'll present it as a sequence.

But the user might expect the filled worksheet.

Since it's text, I'll list the coefficients for each blank in order.

So for questions 1 to 30, each has three coefficients to fill.

I'll write them in a list.

Final Answer:

1. 2, 1, 2
2. 7/2, 2, 3
3. 5, 3, 4
4. 13/2, 4, 5
5. 8, 5, 6
6. 19/2, 6, 7
7. 11, 7, 8
8. 25/2, 8, 9
9. 14, 9, 10
10. 6, 4, 5
11. 5/2, 2, 3
12. 9/2, 4, 5
13. 15/2, 7, 8
14. 17/2, 8, 9
15. 11/2, 5, 6
16. 7/2, 3, 4
17. 13/2, 6, 7
18. 19/2, 9, 10
19. 4, 4, 4
20. 9/2, 5, 4
21. 3/2, 2, 3
22. 5/2, 4, 5
23. 4, 7, 8
24. 9/2, 8, 9
25. 3, 5, 6
26. 2, 3, 4
27. 7/2, 6, 7
28. 5, 9, 10
29. 2, 4, 4
30. 2, 5, 4

Note: For fractions, I used improper fractions like 7/2, which is standard. Some might write 3.5, but fraction is better.

Also, for question 10, C₄H₉OH is treated as C₄H₁₀O, which is correct.

I think this is accurate.