Balancing Act 2: A worksheet for practicing balancing chemical equations with numbered answer boxes.

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 use coefficients (numbers in front) to balance — never change subscripts!

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1. ___ B₈ + ___ O₂ → ___ BO₂

Left: B=8, O=?
Right: B=1, O=2 per BO₂

To get 8 B on right → put 8 in front of BO₂ → now right has 8 B and 16 O

So left needs 16 O → O₂ gives 2 per molecule → 16/2 = 8 → so 8 O₂

Answer: 1 B₈ + 8 O₂ → 8 BO₂

Check: Left: B=8, O=16; Right: B=8, O=16 ✔

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2. ___ Na + ___ H₂O → ___ NaOH + ___ H₂

Left: Na=?, H=2 per H₂O, O=1 per H₂O
Right: Na=1, O=1, H=1+2=3? Wait — NaOH has H=1, H₂ has H=2 → total H=3 per “set”

Try putting 2 NaOH → then need 2 Na on left → 2 Na

Now H: right = 2 (from 2 NaOH) + 2 (from H₂) = 4 H → so left needs 4 H → that’s 2 H₂O (since each has 2 H)

O: 2 H₂O → 2 O → matches 2 NaOH → good

H₂: we have 2 H from NaOH and 2 H from H₂? Wait no:

Wait — if we have 2 NaOH → that’s 2 H from NaOH, plus H₂ which is 2 H → total 4 H → yes, from 2 H₂O (which gives 4 H and 2 O)

But H₂ is a separate product — how many H₂ molecules?

Total H on right: from 2 NaOH → 2 H, and from x H₂ → 2x H → total H = 2 + 2x

From left: 2 H₂O → 4 H

So 2 + 2x = 4 → 2x=2 → x=1 → so 1 H₂

Final: 2 Na + 2 H₂O → 2 NaOH + 1 H₂

Check:
Na: 2=2
H: left=4, right=2 (in NaOH) + 2 (in H₂) = 4
O: 2=2 ✔

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3. ___ C₄H₆ + ___ O₂ → ___ C + ___ HCl

Wait — this looks odd. C₄H₆ reacting with O₂ to give C and HCl? That doesn’t make sense chemically — where does Cl come from? Probably a typo. Maybe it should be H₂O or something else? But as written, there’s no chlorine on left, but HCl on right → impossible to balance unless we assume error.

Looking back at image — maybe it’s meant to be combustion? Or perhaps it’s C₄H₆ + O₂ → CO₂ + H₂O? But it says C and HCl.

Alternatively — maybe it’s a different reaction? Let me check original problem again.

Actually, looking at the image text: "C₄H₆ + O₂ → C + HCl" — this must be a mistake because no Cl on left. Perhaps it's supposed to be H₂O instead of HCl? Or maybe it's C₄H₆Cl something? But as written, it can't be balanced.

Wait — perhaps it’s a trick question? Or misprint? Since this is for students, likely a typo. Common reaction for hydrocarbons is combustion to CO₂ and H₂O.

But let’s see what the worksheet expects. Maybe it’s C₄H₆ + O₂ → CO₂ + H₂O? But it says C and HCl.

Another possibility: maybe it’s C₄H₆ + Cl₂ → C + HCl? But it says O₂.

I think there might be an error in the problem. However, since we must solve as given, and no Cl on left, we cannot produce HCl. So perhaps coefficient for HCl is 0? Not possible.

Maybe it’s a different compound? Let me look at other problems — #4 has S and F, etc.

Perhaps it’s C₄H₆ + O₂ → CO + H₂O? Still not matching.

Wait — maybe it’s C₄H₆ + O₂ → C (soot) + H₂O? But it says HCl.

I think this is likely a typo, and it should be H₂O instead of HCl. Let me assume that for now, as otherwise unsolvable.

Assume: C₄H₆ + O₂ → C + H₂O

Then balance:

Left: C=4, H=6, O=?

Right: C=1 per C, H=2 per H₂O, O=1 per H₂O

To get 4 C on right → 4 C

To get 6 H on right → 3 H₂O (since 3×2=6 H) → then O=3 on right

So left needs 3 O → O₂ gives 2 per molecule → 3/2 = 1.5 → so multiply all by 2:

2 C₄H₆ + 3 O₂ → 8 C + 6 H₂O

Check: C: 8=8, H: 12=12, O: 6=6 ✔

But the problem says HCl, not H₂O. Hmm.

Perhaps it’s C₄H₆ + Cl₂ → C + HCl? Then:

Left: C=4, H=6, Cl=?

Right: C=1, H=1, Cl=1 per HCl

So for 4 C → 4 C

For 6 H → 6 HCl → so 6 Cl on right → so left needs 6 Cl → 3 Cl₂

So: C₄H₆ + 3 Cl₂ → 4 C + 6 HCl

That works! And makes sense — chlorination producing carbon and HCl.

Probably the O₂ is a typo and should be Cl₂. Because otherwise impossible.

Given that, I'll go with that assumption.

So answer: 1 C₄H₆ + 3 Cl₂ → 4 C + 6 HCl

But the problem says O₂ — wait, in the image it's written as O₂? Let me double-check user input.

User wrote: "3. ___ C₄H₆ + ___ O₂ → ___ C + ___ HCl"

Yes, O₂. But that can't be. Unless... is there oxygen in products? No.

Perhaps it's a different reaction. Another thought: maybe it's incomplete combustion to carbon and water, but still no Cl.

I think for educational purposes, we should note the inconsistency, but since we must provide answer, and assuming it's a common type, perhaps it's meant to be Cl₂.

Maybe the student copied wrong, but based on standard problems, I'll proceed with Cl₂.

So: 1 C₄H₆ + 3 Cl₂ → 4 C + 6 HCl

Coefficients: 1, 3, 4, 6

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4. ___ Sb₂S₃ + ___ O₂ → ___ SbO₂ + ___ H₂O

Wait — left has Sb, S, O; right has Sb, O, H, O — H₂O has H, but no H on left! Again, problem.

Sb₂S₃ + O₂ → SbO₂ + H₂O — no hydrogen on left, but H₂O on right → impossible.

Must be typo. Probably should be SO₂ instead of H₂O? Common reaction: antimony sulfide burns to antimony oxide and sulfur dioxide.

So likely: Sb₂S₃ + O₂ → SbO₂ + SO₂

Let me balance that.

Left: Sb=2, S=3, O=?

Right: Sb=1 per SbO₂, S=1 per SO₂, O=2+2=4 per "pair" but need to match.

Set Sb: 2 SbO₂ → so 2 Sb

S: 3 SO₂ → so 3 S

Now O: right = 2*2 (from SbO₂) + 3*2 (from SO₂) = 4 + 6 = 10 O

Left: O₂ → so 5 O₂ (since 5*2=10)

So: 1 Sb₂S₃ + 5 O₂ → 2 SbO₂ + 3 SO₂

Check: Sb:2=2, S:3=3, O:10=10 ✔

But problem says H₂O — again, probably typo. I'll assume SO₂.

If it must be H₂O, then impossible. So for sake of solving, I'll use SO₂.

Coefficients: 1, 5, 2, 3

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This is taking too long, and there are inconsistencies. Perhaps I should look for a pattern or do only solvable ones.

Let me list all equations as given and try to balance as is, noting issues.

But for the purpose of this exercise, I'll assume common corrections.

Perhaps in #3, it's C₄H₆ + O₂ → CO₂ + H₂O, and "C" is typo for CO₂, "HCl" for H₂O.

Similarly for #4.

Let me try that approach for consistency.

For #3: C₄H₆ + O₂ → CO₂ + H₂O

Balance:

C: 4 on left → 4 CO₂

H: 6 on left → 3 H₂O (6H)

O: right = 4*2 + 3*1 = 8+3=11 O → so left needs 11/2 O₂ → multiply by 2:

2 C₄H₆ + 11 O₂ → 8 CO₂ + 6 H₂O

Check: C:8=8, H:12=12, O:22=16+6=22 ✔

But problem says "C" and "HCl", not CO₂ and H₂O.

Perhaps "C" means carbon monoxide? But usually CO.

I think to move forward, I'll use the corrected versions as they are standard.

For #4: Sb₂S₃ + O₂ → Sb₂O₃ + SO₂ or something. Earlier I had SbO₂, but antimony oxide is usually Sb₂O₃ or Sb₂O₅.

In my earlier calculation, I used SbO₂, which might not be stable, but for balancing, ok.

But with H₂O, impossible.

Another idea: perhaps it's Sb₂S₃ + O₂ + H₂O → but no, only two reactants.

I think there are typos in the worksheet. For educational value, I'll balance the intended reactions.

Let me do #5 to #10 first, as they seem clearer.

5. ___ FeS₂ + ___ O₂ → ___ Fe₂O₃ + ___ SO₂

This is pyrite burning.

Left: Fe=1, S=2, O=?

Right: Fe=2, S=1, O=3+2=5 per Fe₂O₃ and SO₂, but need to match.

Set Fe: 2 FeS₂ → so 2 Fe, 4 S

Then S: 4 SO₂ → so 4 S

Fe: 1 Fe₂O₃ → 2 Fe

O: right = 3 (from Fe₂O₃) + 8 (from 4 SO₂) = 11 O

Left: O₂ → so 11/2 O₂ → multiply by 2:

4 FeS₂ + 11 O₂ → 2 Fe₂O₃ + 8 SO₂

Check: Fe:4=4, S:8=8, O:22=6+16=22 ✔

Coefficients: 4, 11, 2, 8

6. ___ Fe₂O₃ + ___ H₂ → ___ Fe + ___ H₂O

Left: Fe=2, O=3, H=2

Right: Fe=1, H=2, O=1 per H₂O

Set Fe: 2 Fe on right → so 2 Fe

O: 3 on left → 3 H₂O on right → so 3 H₂O

H: right = 6 H (from 3 H₂O) → left needs 6 H → 3 H₂ (since each H₂ has 2H)

So: 1 Fe₂O₃ + 3 H₂ → 2 Fe + 3 H₂O

Check: Fe:2=2, O:3=3, H:6=6 ✔

Coefficients: 1, 3, 2, 3

7. ___ H₂CO₃ → ___ H₂O + ___ CO₂

Decomposition of carbonic acid.

Left: H=2, C=1, O=3

Right: H₂O has H=2,O=1; CO₂ has C=1,O=2; total O=3, H=2, C=1 → already balanced!

So: 1 H₂CO₃ → 1 H₂O + 1 CO₂

Coefficients: 1, 1, 1

8. ___ SiO₂ + ___ HF → ___ SiF₄ + ___ H₂O

Left: Si=1, O=2, H=1, F=1

Right: Si=1, F=4, H=2, O=1

Set Si: 1 SiO₂ → 1 SiF₄

F: 4 on right → so 4 HF on left

H: 4 on left → so 2 H₂O on right (since each has 2H)

O: left=2 (from SiO₂), right=2 (from 2 H₂O) → good

So: 1 SiO₂ + 4 HF → 1 SiF₄ + 2 H₂O

Check: Si:1=1, O:2=2, H:4=4, F:4=4 ✔

Coefficients: 1, 4, 1, 2

9. ___ K + ___ Br₂ → ___ KBr

Left: K=1, Br=2

Right: K=1, Br=1

So need 2 KBr on right → then 2 K on left, and 1 Br₂ (since Br₂ has 2Br)

So: 2 K + 1 Br₂ → 2 KBr

Coefficients: 2, 1, 2

10. ___ Bi + ___ F₂ → ___ BiF₃

Left: Bi=1, F=2

Right: Bi=1, F=3

Need common multiple for F: LCM of 2 and 3 is 6.

So right: 2 BiF₃ → 2 Bi, 6 F

Left: 3 F₂ → 6 F, and 2 Bi

So: 2 Bi + 3 F₂ → 2 BiF₃

Coefficients: 2, 3, 2

Now back to problematic ones.

For #3: C₄H₆ + O₂ → C + HCl — as discussed, likely should be C₄H₆ + Cl₂ → C + HCl or C₄H₆ + O₂ → CO₂ + H₂O.

Since the product is "C" and "HCl", and no Cl on left, perhaps it's a different interpretation. Maybe "C" is carbon, and HCl is hydrogen chloride, but source of Cl missing.

Perhaps it's C₄H₆ + 3Cl₂ -> 4C + 6HCl, as I had.

And for #4: Sb₂S₃ + O₂ -> SbO₂ + H₂O — no H on left, so likely Sb₂S₃ + O₂ -> Sb₂O₃ + SO₂ or something.

Let me assume for #4: Sb₂S₃ + O₂ -> Sb₂O₃ + SO₂

Balance:

Sb:2=2

S:3 on left, so 3 SO₂ on right

O: right = 3 (from Sb₂O₃) + 6 (from 3 SO₂) = 9 O

Left: O₂ -> 9/2 O₂ -> multiply by 2:

2 Sb₂S₃ + 9 O₂ -> 2 Sb₂O₃ + 6 SO₂

Check: Sb:4=4, S:6=6, O:18=6+12=18 ✔

But problem says SbO₂ and H₂O, not Sb₂O₃ and SO₂.

Perhaps SbO₂ is intended, so Sb₂S₃ + O₂ -> 2SbO₂ + 3SO₂, as I had earlier, with 5 O₂.

And ignore H₂O.

For the sake of completing, I'll use the following assumptions:

- #3: C₄H₆ + 3Cl₂ -> 4C + 6HCl (assuming O₂ is typo for Cl₂)

- #4: Sb₂S₃ + 5O₂ -> 2SbO₂ + 3SO₂ (assuming H₂O is typo for SO₂)

But in the answer grid, we need numbers, and for #3 and #4, if we use these, coefficients are:

#3: 1, 3, 4, 6

#4: 1, 5, 2, 3

Now let's list all answers:

1. B₈ + O₂ → BO₂: 1, 8, 8

2. Na + H₂O → NaOH + H₂: 2, 2, 2, 1

3. C₄H₆ + O₂ → C + HCl: assume 1, 3, 4, 6 (with Cl₂)

4. Sb₂S₃ + O₂ → SbO₂ + H₂O: assume 1, 5, 2, 3 (with SO₂)

5. FeS₂ + O₂ → Fe₂O₃ + SO₂: 4, 11, 2, 8

6. Fe₂O₃ + H₂ → Fe + H₂O: 1, 3, 2, 3

7. H₂CO₃ → H₂O + CO₂: 1, 1, 1

8. SiO₂ + HF → SiF₄ + H₂O: 1, 4, 1, 2

9. K + Br₂ → KBr: 2, 1, 2

10. Bi + F₂ → BiF₃: 2, 3, 2

Now, the grid at bottom has letters A to N, and numbers below, probably for matching or something, but the instruction is to balance, and perhaps fill in coefficients.

The user didn't specify what to do with the grid, but since the task is to balance, I'll provide the coefficients for each.

For the final answer, perhaps list the coefficients for each equation.

Since the grid has 14 slots, and we have 10 equations, maybe not related.

Perhaps the grid is for the coefficients, but there are more spaces.

Another thought: perhaps the grid is to be filled with the coefficients in order, but there are 10 equations, each with up to 4 coefficients, so 40 values, but grid has 14 cells.

Looking at the grid: it's a row of boxes with letters A to N, and below each letter a number, like A:4, B:11, C:2, D:8, E:1, F:3, G:2, H:3, I:1, J:1, K:1, L:1, M:4, N:1,2

N has "1,2" which is unusual.

Perhaps it's a key for something else.

Maybe for some equations, the coefficients are to be matched to the letters.

But the user's request is to solve the problem, which is to balance the equations.

So I'll provide the balanced equations with coefficients.

For the final answer, since it's a homework, perhaps list the coefficients for each blank.

So for each equation, the coefficients in order.

Let me write them:

1. 1, 8, 8

2. 2, 2, 2, 1

3. 1, 3, 4, 6 (assuming Cl₂)

4. 1, 5, 2, 3 (assuming SO₂)

5. 4, 11, 2, 8

6. 1, 3, 2, 3

7. 1, 1, 1

8. 1, 4, 1, 2

9. 2, 1, 2

10. 2, 3, 2

Now, to confirm, let's see if any match the grid.

Grid has: A:4, B:11, C:2, D:8, E:1, F:3, G:2, H:3, I:1, J:1, K:1, L:1, M:4, N:1,2

Notice that for equation 5: 4,11,2,8 — which matches A,B,C,D: 4,11,2,8

Equation 6: 1,3,2,3 — matches E,F,G,H: 1,3,2,3

Equation 7: 1,1,1 — but grid has I:1, J:1, K:1 — three 1's

Equation 8: 1,4,1,2 — L:1, M:4, then N:1,2 — so perhaps L,M,N for last three, but N is "1,2" which might mean two numbers.

Perhaps the grid is for the coefficients of equations 5 to 10 or something.

Equation 5: coeffs 4,11,2,8 — A,B,C,D

Equation 6: 1,3,2,3 — E,F,G,H

Equation 7: 1,1,1 — but only three, while I,J,K are three 1's

Equation 8: 1,4,1,2 — L:1, M:4, and N:1,2 — so perhaps N represents the last two coefficients 1 and 2.

Similarly, equation 9: 2,1,2 — not in grid yet.

Equation 10: 2,3,2 — not in grid.

Grid has up to N, 14 items.

Perhaps for all equations, but some have fewer coefficients.

Another idea: perhaps the grid is to be used to fill in the blanks, but the blanks are in the equations, and the grid provides possible numbers.

But the user didn't ask for that.

Perhaps after balancing, the coefficients correspond to the letters, and we need to find which letter corresponds to which coefficient, but that seems complicated.

Given the complexity, and since the main task is to balance the equations, I'll provide the balanced forms.

For the final answer, I'll list the coefficients for each equation in order.

So:

1. 1, 8, 8

2. 2, 2, 2, 1

3. 1, 3, 4, 6

4. 1, 5, 2, 3

5. 4, 11, 2, 8

6. 1, 3, 2, 3

7. 1, 1, 1

8. 1, 4, 1, 2

9. 2, 1, 2

10. 2, 3, 2

And for the grid, it might be that A=4, B=11, etc., but since not specified, I'll omit.

Perhaps the "Final Answer" is to box the coefficients.

But to follow the format, I'll provide the balanced equations.

Since the user said "solve the problem", and the problem is to balance, I'll state the coefficients.

For the response, I'll write the balanced equations with coefficients filled in.

So for each:

1. 1 B₈ + 8 O₂ → 8 BO₂

2. 2 Na + 2 H₂O → 2 NaOH + 1 H₂

3. 1 C₄H₆ + 3 Cl₂ → 4 C + 6 HCl (noting assumption)

4. 1 Sb₂S₃ + 5 O₂ → 2 SbO₂ + 3 SO₂ (noting assumption)

5. 4 FeS₂ + 11 O₂ → 2 Fe₂O₃ + 8 SO₂

6. 1 Fe₂O₃ + 3 H₂ → 2 Fe + 3 H₂O

7. 1 H₂CO₃ → 1 H₂O + 1 CO₂

8. 1 SiO₂ + 4 HF → 1 SiF₄ + 2 H₂O

9. 2 K + 1 Br₂ → 2 KBr

10. 2 Bi + 3 F₂ → 2 BiF₃

For #3 and #4, I added notes, but for the answer, perhaps just the numbers.

Since the grid has specific numbers, and for #5 and #6 it matches A-D and E-H, and #7 matches I-J-K, #8 matches L-M-N with N being 1,2 for the last two, then for #9 and #10, not covered, but perhaps the grid is only for some.

Maybe the grid is for the coefficients of the last few equations.

Another thought: perhaps the "some spaces will not require a number" means that for some blanks, coefficient is 1, and we don't write it, but in balancing, we usually write 1 or leave blank, but here we need to fill.

In the instructions: "Some spaces will not require a number." — probably means that if coefficient is 1, you can leave it blank or write 1, but in this case, we need to provide the number.

For the final answer, I'll provide the list of coefficients as per the equations.

To match the grid, let's see:

If we take equation 5: coefficients 4,11,2,8 — matches A,B,C,D

Equation 6: 1,3,2,3 — matches E,F,G,H

Equation 7: 1,1,1 — matches I,J,K (three 1's)

Equation 8: 1,4,1,2 — L:1, M:4, and N:1,2 — so N represents the last two coefficients 1 and 2.

Then for equations 1-4 and 9-10, not in grid, but perhaps the grid is only for 5-8 or something.

The grid has 14 items, and equations 5-8 have 4+4+3+4=15 coefficients, close.

Equation 5: 4 coeffs, 6:4, 7:3, 8:4 — total 15, but grid has 14, and N is "1,2" which might count as one entry for two numbers.

So perhaps for equation 8, the last two coefficients are combined in N.

So the grid corresponds to the coefficients of equations 5,6,7,8 in order.

Specifically:

- Eq5: coeff1=4=A, coeff2=11=B, coeff3=2=C, coeff4=8=D

- Eq6: coeff1=1=E, coeff2=3=F, coeff3=2=G, coeff4=3=H

- Eq7: coeff1=1=I, coeff2=1=J, coeff3=1=K

- Eq8: coeff1=1=L, coeff2=4=M, coeff3=1 and coeff4=2=N (so N is "1,2")

Perfect match.

So for the other equations, we don't need the grid, but we still need to balance them.

For the final answer, since the grid is provided, and it matches eq5-8, perhaps the task is to balance all, and the grid is for verification or something.

But for the response, I'll provide the balanced equations.

Since the user may expect the coefficients for all, I'll list them.

For the Final Answer section, I'll box the coefficients for each equation.

To save space, perhaps list the sequences.

Given that, I'll write:

The balanced equations have the following coefficients (in order for each equation):

1. 1, 8, 8

2. 2, 2, 2, 1

3. 1, 3, 4, 6 (assuming Cl₂ instead of O₂)

4. 1, 5, 2, 3 (assuming SO₂ instead of H₂O)

5. 4, 11, 2, 8

6. 1, 3, 2, 3

7. 1, 1, 1

8. 1, 4, 1, 2

9. 2, 1, 2

10. 2, 3, 2

And for the grid, it corresponds to equations 5-8 as above.

For the Final Answer, since the problem is to balance, and no specific output format, I'll provide the list.

Perhaps the "Final Answer" is the set of coefficients for the blanks.

So for each blank in the equations, the number to fill in.

So for equation 1: three blanks: 1, 8, 8

etc.

I think that's it.

So in the response, I'll state the coefficients.

For the boxed answer, perhaps list them as a sequence or something.

Since the grid is there, and it matches, but for completeness, I'll include all.

Another idea: perhaps the grid is to be used to fill the blanks, and we need to select from the grid, but the grid has specific numbers, and for example in eq1, we need 1,8,8, and 8 is not in grid, grid has 4,11,2,8,1,3,2,3,1,1,1,1,4,1,2 — so 8 is there (D), 1 is there, but 8 is used, and for eq1, we need another 8, but only one 8 in grid.

So probably not.

I think the grid is for equations 5-8 only, as it matches perfectly.

For the other equations, we balance normally.

So for the final answer, I'll provide the balanced forms for all, and for the grid, it's consistent with eq5-8.

But since the user didn't ask for the grid, I'll focus on balancing.

So in the response, I'll explain the balancing for each, but to keep it concise, I'll list the coefficients.

For the Final Answer section, I'll box the list of coefficient tuples.

So:

Final Answer:
1. (1, 8, 8)
2. (2, 2, 2, 1)
3. (1, 3, 4, 6) [note: assumed Cl₂]
4. (1, 5, 2, 3) [note: assumed SO₂]
5. (4, 11, 2, 8)
6. (1, 3, 2, 3)
7. (1, 1, 1)
8. (1, 4, 1, 2)
9. (2, 1, 2)
10. (2, 3, 2)

To make it neat, perhaps write the equations with coefficients.

Since the assistant should not describe the image, and just solve, I'll provide the answers.

For the sake of time, I'll go with that.

So in the response:

After balancing each chemical equation, the coefficients are as follows:

1. 1, 8, 8
2. 2, 2, 2, 1
3. 1, 3, 4, 6 (assuming the reactant is Cl₂ instead of O₂ due to presence of Cl in products)
4. 1, 5, 2, 3 (assuming the product is SO₂ instead of H₂O due to absence of H in reactants)
5. 4, 11, 2, 8
6. 1, 3, 2, 3
7. 1, 1, 1
8. 1, 4, 1, 2
9. 2, 1, 2
10. 2, 3, 2

Note: For equations 3 and 4, there appear to be typos in the original problem, as the elements do not balance with the given formulas. The assumptions made are based on common chemical reactions.

For the grid at the bottom, it matches the coefficients of equations 5 through 8:
- Equation 5: 4, 11, 2, 8 → A, B, C, D
- Equation 6: 1, 3, 2, 3 → E, F, G, H
- Equation 7: 1, 1, 1 → I, J, K
- Equation 8: 1, 4, 1, 2 → L, M, and N (where N represents the last two coefficients 1 and 2)

But since the primary task is to balance the equations, the above coefficients are the solution.

Final Answer:
The coefficients for the balanced equations are:
1. 1, 8, 8
2. 2, 2, 2, 1
3. 1, 3, 4, 6
4. 1, 5, 2, 3
5. 4, 11, 2, 8
6. 1, 3, 2, 3
7. 1, 1, 1
8. 1, 4, 1, 2
9. 2, 1, 2
10. 2, 3, 2

With the understanding that for #3 and #4, assumptions were made to correct likely typos.

To box it, perhaps list them.

Since the format requires "Final Answer" after reasoning, and reasoning is done, I'll put the list.

So:

Final Answer:
1. 1, 8, 8
2. 2, 2, 2, 1
3. 1, 3, 4, 6
4. 1, 5, 2, 3
5. 4, 11, 2, 8
6. 1, 3, 2, 3
7. 1, 1, 1
8. 1, 4, 1, 2
9. 2, 1, 2
10. 2, 3, 2