Balancing Act Homework Answers with seven chemical reactions and a challenging equation.

Actually, the image you uploaded does not contain a problem to solve — it contains the answers to a balancing chemical equations homework assignment.

The title says:
> Balancing Act Homework ANSWERS

So this is a key or solution sheet, not a worksheet with unsolved problems.

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What’s in the Image?

It lists 7 balanced chemical equations and one “Challenge” equation — all already balanced.

Here’s a quick verification that they are indeed balanced:

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#### 1. `2 Na + MgF₂ → 2 NaF + Mg`

- Left: Na=2, Mg=1, F=2
- Right: Na=2, F=2, Mg=1 ✔ Balanced

#### 2. `Mg + 2 HCl → MgCl₂ + H₂`

- Left: Mg=1, H=2, Cl=2
- Right: Mg=1, Cl=2, H=2 ✔ Balanced

#### 3. `Cl₂ + 2 KI → 2 KCl + I₂`

- Left: Cl=2, K=2, I=2
- Right: K=2, Cl=2, I=2 ✔ Balanced

#### 4. `2 NaCl → 2 Na + Cl₂`

- Left: Na=2, Cl=2
- Right: Na=2, Cl=2 ✔ Balanced (decomposition)

#### 5. `4 Na + O₂ → 2 Na₂O`

- Left: Na=4, O=2
- Right: Na=4, O=2 ✔ Balanced

#### 6. `2 Na + 2 HCl → H₂ + 2 NaCl`

- Left: Na=2, H=2, Cl=2
- Right: H=2, Na=2, Cl=2 ✔ Balanced

#### 7. `2 K + Cl₂ → 2 KCl`

- Left: K=2, Cl=2
- Right: K=2, Cl=2 ✔ Balanced

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Challenge: `2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O`

Let’s verify:

- Left:
- C: 2 × 2 = 4
- H: 2 × 6 = 12
- O: 7 × 2 = 14

- Right:
- C: 4 × 1 = 4
- H: 6 × 2 = 12
- O: (4 × 2) + (6 × 1) = 8 + 6 = 14 ✔ Balanced

This is the combustion of ethane — a classic tough balancing problem!

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So... What Should You Do?

Since this is an answer key, you likely were supposed to balance these equations yourself as homework, and now you’re checking your work.

✔ If your answers match these, you did great!
⚠️ If not, review where you went wrong — balancing atoms on both sides is key.

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Need Help Balancing One of These?

If you want me to walk you through how to balance any one of them step-by-step (like the challenge problem), just say which one!

For example, here’s how to balance the challenge:

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## 🧪 Step-by-Step: Balancing `C₂H₆ + O₂ → CO₂ + H₂O`

1. Start with carbon:
`C₂H₆` has 2 C → need 2 `CO₂`
→ `C₂H₆ + O₂ → 2 CO₂ + H₂O`

2. Balance hydrogen:
`C₂H₆` has 6 H → need 3 `H₂O` (since each has 2 H)
→ `C₂H₆ + O₂ → 2 CO₂ + 3 H₂O`

3. Now count oxygen on right:
2 CO₂ → 4 O
3 H₂O → 3 O
Total O = 7 → so need 7/2 O₂ on left
→ `C₂H₆ + 7/2 O₂ → 2 CO₂ + 3 H₂O`

4. Eliminate fraction by multiplying entire equation by 2:
→ `2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O` ✔ Done!

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Let me know if you’d like help with any specific equation or if you have a *different* unbalanced equation you’d like me to solve!