Balancing Equations Practice by Teach Simple - Free Printable

It looks like you've uploaded a chemistry worksheet containing balanced chemical equations, and the coefficients (the numbers in front of each compound) have already been filled in with red text. Your request is to "solve the problem and explain the solution," so I’ll go through this step by step.

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🔍 What is being asked?

The task appears to be balancing chemical equations — ensuring that the number of atoms of each element is the same on both sides of the reaction (following the Law of Conservation of Mass).

Even though the answers are already filled in, I will:
1. Verify that each equation is correctly balanced.
2. Explain how to balance them (using a few examples).
3. Confirm if all the given coefficients are correct.

Let’s go through each one briefly:

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✔ Verification & Explanation

#### 1. \( \mathbf{2H_2 + O_2 \rightarrow 2H_2O} \)
- Left: H = 4, O = 2
- Right: H = 4, O = 2 → ✔️ Balanced

> Explanation: Hydrogen and oxygen combine to form water. We need 2 H₂ molecules to provide 4 H atoms, and 1 O₂ molecule gives 2 O atoms → makes 2 H₂O.

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#### 2. \( \mathbf{Se + O_2 \rightarrow SeO_2} \)
- Left: Se = 1, O = 2
- Right: Se = 1, O = 2 → ✔️ Balanced

> Simple combination: one selenium atom reacts with one oxygen molecule.

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#### 3. \( \mathbf{2HgO \rightarrow 2Hg + O_2} \)
- Left: Hg = 2, O = 2
- Right: Hg = 2, O = 2 → ✔️ Balanced

> Decomposition of mercury(II) oxide into mercury and oxygen gas.

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#### 4. \( \mathbf{Zn + 2HCl \rightarrow ZnCl_2 + H_2} \)
- Left: Zn = 1, H = 2, Cl = 2
- Right: Zn = 1, Cl = 2, H = 2 → ✔️ Balanced

> Single displacement: zinc replaces hydrogen in HCl.

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#### 5. \( \mathbf{2Na + 2H_2O \rightarrow 2NaOH + H_2} \)
- Left: Na = 2, H = 4, O = 2
- Right: Na = 2, O = 2, H = 2 (in NaOH) + 2 (in H₂) = 4 → ✔️ Balanced

> Sodium reacts with water to produce sodium hydroxide and hydrogen gas.

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#### 6. \( \mathbf{C_{10}H_{16} + 8Cl_2 \rightarrow 10C + 16HCl} \)
- Left: C = 10, H = 16, Cl = 16
- Right: C = 10, H = 16, Cl = 16 → ✔️ Balanced

> This is a halogenation or decomposition reaction where carbon-hydrogen bonds break and chlorine replaces hydrogen.

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#### 7. \( \mathbf{2SiH_4 + 5O_2 \rightarrow 4SiO_2 + 2H_2O} \)
- Left: Si = 2, H = 8, O = 10
- Right: Si = 4 → ✘ Wait! Problem here?

Wait — let's check:

Left:
- Si = 2 (from 2 SiH₄)
- H = 8 (from 2×4)
- O = 10 (from 5 O₂)

Right:
- Si = 4 (from 4 SiO₂) → too many!
- H = 4 (from 2 H₂O), O = 8 (from 4 SiO₂) + 2 (from 2 H₂O) = 10 → O is OK, but Si and H are off.

✘ This is NOT balanced!

But wait — original says:
\( \mathbf{2SiH_4 + 5O_2 \rightarrow 4SiO_2 + 2H_2O} \)

That can't be right.

Let’s try balancing it properly.

Correct balancing:

We want:
SiH₄ + O₂ → SiO₂ + H₂O

Balance Si: 1 Si on each side → OK
Balance H: 4 H on left → need 2 H₂O on right → H₂O = 2
Now O: 2 from H₂O + 2 from SiO₂ = 4 O → so need 2 O₂

So:
\( \mathbf{SiH_4 + 2O_2 \rightarrow SiO_2 + 2H_2O} \)

Multiply by 2:
\( \mathbf{2SiH_4 + 4O_2 \rightarrow 2SiO_2 + 4H_2O} \)

But the answer shows 5 O₂, 4 SiO₂, and 2 H₂O → clearly wrong.

✔ So the given answer is incorrect.

Corrected version:
\( \mathbf{2SiH_4 + 4O_2 \rightarrow 2SiO_2 + 4H_2O} \)

Or simplified:
\( \mathbf{SiH_4 + 2O_2 \rightarrow SiO_2 + 2H_2O} \)

👉 So #7 is incorrectly balanced in your image.

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#### 8. \( \mathbf{2C_7H_6O_2 + 17O_2 \rightarrow 14CO_2 + 6H_2O} \)
Check:
- Left: C = 14, H = 12, O = 2×2 + 17×2 = 4 + 34 = 38
- Right: C = 14, H = 12, O = 28 (from CO₂) + 6 (from H₂O) = 34 → ✘ Not balanced!

Wait — H₂O has 6 molecules → 6 O and 12 H → OK for H.

But O on right:
- 14 CO₂ → 14×2 = 28 O
- 6 H₂O → 6 O
→ Total O = 34

Left:
- 2 × C₇H₆O₂ → 2×2 = 4 O
- 17 O₂ → 34 O
→ Total O = 4 + 34 = 38

Not equal → not balanced

But let’s recall: benzoic acid (C₇H₆O₂) combustion:

General formula:
C₇H₆O₂ + O₂ → CO₂ + H₂O

Balance:
- C: 7 → 7 CO₂
- H: 6 → 3 H₂O
- O: Left: 2 (from benzoic) + 2×O₂ → Right: 14 (from CO₂) + 3 (from H₂O) = 17 O
- So O needed: 17 - 2 = 15 → 15/2 = 7.5 O₂ → multiply by 2:

→ 2C₇H₆O₂ + 15O₂ → 14CO₂ + 6H₂O

Now check:
- Left: C=14, H=12, O=4 + 30 = 34
- Right: C=14, H=12, O=28+6=34 → ✔️

So correct:
\( \mathbf{2C_7H_6O_2 + 15O_2 \rightarrow 14CO_2 + 6H_2O} \)

But your image says 17 O₂ → ✘ Incorrect.

So #8 is also wrong.

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#### 9. \( \mathbf{Fe_2O_3 + H_2 \rightarrow 2Fe + H_2O} \)
Given: 1 Fe₂O₃ + 1 H₂ → 2 Fe + 2 H₂O

Check:
- Left: Fe=2, O=3, H=2
- Right: Fe=2, H=4, O=2 → ✘ H and O not balanced

Correct way:
Fe₂O₃ + 3H₂ → 2Fe + 3H₂O

Now:
- Left: Fe=2, O=3, H=6
- Right: Fe=2, H=6, O=3 → ✔️

So given answer is incorrect.

Should be:
\( \mathbf{1 Fe_2O_3 + 3 H_2 \rightarrow 2 Fe + 3 H_2O} \)

So #9 is wrong.

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#### 10. \( \mathbf{2K + Br_2 \rightarrow 2KBr} \)
- K: 2 = 2
- Br: 2 = 2 → ✔️ Correct

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#### 11. \( \mathbf{SiO_2 + 4HF \rightarrow SiF_4 + 2H_2O} \)
- Left: Si=1, O=2, H=4, F=4
- Right: Si=1, F=4, H=4, O=2 → ✔️ Correct

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#### 12. \( \mathbf{KClO_2 \rightarrow KCl + O_2} \)
- Left: K=1, Cl=1, O=2
- Right: K=1, Cl=1, O=2 → ✔️ But O₂ is diatomic → must be 1 O₂

So:
\( \mathbf{KClO_2 \rightarrow KCl + O_2} \) → O atoms: 2 on each side → ✔️

But is it stable? Yes, decomposition.

✔ Correct

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#### 13. \( \mathbf{2KClO_2 \rightarrow KClO_4 + KCl} \)
- Left: K=2, Cl=2, O=4
- Right: K=1+1=2, Cl=1+1=2, O=4+0=4 → ✔️ Balanced

This is a disproportionation reaction.

✔ Correct

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#### 14. \( \mathbf{P_4O_{10} + 6H_2O \rightarrow 4H_3PO_4} \)
- Left: P=4, O=10+6=16, H=12
- Right: P=4, H=12, O=16 → ✔️ Correct

✔ Correct

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#### 15. \( \mathbf{4Sb + 3O_2 \rightarrow Sb_4O_6} \)
- Left: Sb=4, O=6
- Right: Sb=4, O=6 → ✔️ Correct

✔ Correct

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#### 16. \( \mathbf{PCl_5 + 4H_2O \rightarrow 5HCl + H_3PO_4} \)
- Left: P=1, Cl=5, H=8, O=4
- Right: H=5+3=8, Cl=5, P=1, O=4 → ✔️ Correct

✔ Correct

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#### 17. \( \mathbf{2Fe + 4H_2O \rightarrow Fe_2O_4 + 4H_2} \)
Note: Fe₂O₄ is magnetite, which is Fe₃O₄? Wait — Fe₂O₄ doesn’t exist. It should be Fe₃O₄.

Wait — Fe₂O₄ implies 2 Fe and 4 O → but Fe₃O₄ is more common.

But actually, Fe₂O₄ is not standard. Let's assume it's a typo and meant Fe₃O₄.

But even then:
If it were:
\( \mathbf{3Fe + 4H_2O \rightarrow Fe_3O_4 + 4H_2} \)

Then:
- Left: Fe=3, H=8, O=4
- Right: Fe=3, O=4, H=8 → ✔️

But here it's written as Fe₂O₄ → which would mean:
- Fe=2, O=4 → but only 2 Fe on left → OK?
But Fe₂O₄ is not chemically valid. The correct formula is Fe₃O₄.

So likely typo.

Assuming it's Fe₃O₄, then:
\( \mathbf{3Fe + 4H_2O \rightarrow Fe_3O_4 + 4H_2} \)

But given: 2Fe + 4H₂O → Fe₂O₄ + 4H₂

Then:
- Left: Fe=2, H=8, O=4
- Right: Fe=2, O=4, H=8 → ✔️ atom-wise

But Fe₂O₄ is not a real compound — it's an invalid formula. Should be Fe₃O₄.

So chemically incorrect, even if atoms balance.

Thus, #17 is flawed due to incorrect formula.

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✔ Summary of Errors in Given Answers:

| Equation | Status | Issue |
|--------|--------|-------|
| 7 | ✘ Wrong | Should be 2SiH₄ + 4O₂ → 2SiO₂ + 4H₂O |
| 8 | ✘ Wrong | Should be 2C₇H₆O₂ + 15O₂ → 14CO₂ + 6H₂O |
| 9 | ✘ Wrong | Should be Fe₂O₃ + 3H₂ → 2Fe + 3H₂O |
| 17 | ⚠️ Invalid | Fe₂O₄ is not a valid compound; should be Fe₃O₄ |

All others seem correctly balanced.

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🧠 How to Balance Equations (General Steps):

1. Write unbalanced equation.
2. Count atoms of each element on both sides.
3. Start with elements that appear in only one compound on each side.
4. Use coefficients (whole numbers) to balance.
5. Balance polyatomic ions as groups if they stay intact.
6. Check all atoms.
7. Reduce coefficients to lowest whole numbers.

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✔ Final Answer:

The majority of the equations are correctly balanced, but there are three major errors:

- Equation 7: Incorrect coefficient for O₂ and SiO₂.
- Equation 8: Incorrect coefficient for O₂ (should be 15, not 17).
- Equation 9: Incorrect stoichiometry (should be 3 H₂, not 1).
- Equation 17: Uses invalid formula Fe₂O₄; should be Fe₃O₄.

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🔧 Corrected Equations:

7. \( \mathbf{2SiH_4 + 4O_2 \rightarrow 2SiO_2 + 4H_2O} \)

8. \( \mathbf{2C_7H_6O_2 + 15O_2 \rightarrow 14CO_2 + 6H_2O} \)

9. \( \mathbf{Fe_2O_3 + 3H_2 \rightarrow 2Fe + 3H_2O} \)

17. \( \mathbf{3Fe + 4H_2O \rightarrow Fe_3O_4 + 4H_2} \)

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Let me know if you'd like a printable version or step-by-step balancing guide!