Detailed handwritten work solving algebra problems from a math worksheet.

Let’s go step by step through each problem on the worksheet. I’ll solve them one at a time, check my work, and then give you the final answers.

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Problem 1: Balance the equation

Given:
N₂ + H₂ → NH₃

We need to balance nitrogen (N) and hydrogen (H).

Left side: N = 2, H = 2
Right side: N = 1, H = 3

To balance N: put 2 in front of NH₃ → now right has 2N and 6H
So left needs 6H → that means 3 H₂ molecules (since 3 × 2 = 6)

Final balanced equation:
N₂ + 3H₂ → 2NH₃

✔ Check: Left: N=2, H=6; Right: N=2, H=6 → Balanced!

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Problem 2: How many moles of O₂ are needed to react with 4.0 mol C₂H?

Equation given:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

From the equation:
2 mol C₂H require 7 mol O₂

So for 4.0 mol C₂H₆:

Set up ratio:
(7 mol O₂ / 2 mol C₂H₆) × 4.0 mol C₂H₆ = ?

→ (7/2) × 4 = 14 mol O₂

✔ Answer: 14 mol O₂

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Problem 3: If 8.0 mol CO₂ are produced, how many mol O₂ were used?

Same equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

From equation:
4 mol CO₂ come from 7 mol O₂

So for 8.0 mol CO₂:

Ratio: (7 mol O₂ / 4 mol CO₂) × 8.0 mol CO₂ = ?

→ (7/4) × 8 = 14 mol O₂

✔ Answer: 14 mol O₂

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Problem 4: If 5.0 mol H₂O are produced, how many mol C₂H₆ reacted?

Equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

From equation:
6 mol H₂O come from 2 mol C₂H₆

So for 5.0 mol H₂O:

Ratio: (2 mol C₂H₆ / 6 mol H₂O) × 5.0 mol H₂O = ?

→ (2/6) × 5 = 10/6 ≈ 1.666... → round to 2 sig figs? Wait — 5.0 has 2 sig figs, so answer should have 2.

But let’s keep it as fraction first: 10/6 = 5/3 ≈ 1.67 mol

Actually, looking at the handwritten answer, they wrote “1.7 mol” — which is rounded to 2 sig figs.

✔ So we’ll go with 1.7 mol C₂H₆

Wait — let me double-check:

If 6 mol H₂O ← 2 mol C₂H₆
Then 1 mol H₂O ← 2/6 = 1/3 mol C₂H₆
So 5.0 mol H₂O ← 5.0 × (1/3) = 5/3 = 1.666... → rounds to 1.7 mol

Yes.

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Problem 5: If 2.0 mol C₂H₆ react, how many mol CO₂ are formed?

Equation:
2C₂H₆ → 4CO₂

So 2 mol C₂H₆ produce 4 mol CO₂

Therefore, 2.0 mol C₂H₆ → 4.0 mol CO₂

✔ Straightforward ratio.

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Problem 6: If 3.0 mol O₂ react, how many mol H₂O are formed?

Equation:
7O₂ → 6H₂O

So 7 mol O₂ produce 6 mol H₂O

For 3.0 mol O₂:

(6 mol H₂O / 7 mol O₂) × 3.0 = ?

→ (6/7) × 3 = 18/7 ≈ 2.571...

Rounded to 2 sig figs? 3.0 has 2 sig figs → so 2.6 mol?

Wait — 18 7 = 2.5714… → if we use 2 sig figs, it’s 2.6

But let’s see what the student wrote — they had “2.6 mol H₂O”

✔ So we’ll go with 2.6 mol H₂O

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Problem 7: If 4.0 mol C₂H₆ react, how many mol O₂ are required?

Equation:
2C₂H₆ → 7O₂

So 2 mol C₂H₆ need 7 mol O₂

For 4.0 mol C₂H₆:

(7 mol O₂ / 2 mol C₂H₆) × 4.0 = 14 mol O₂

✔ Same as Problem 2 → 14 mol O₂

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Problem 8: If 6.0 mol H₂O are produced, how many mol CO₂ are also produced?

Equation:
Produces 6H₂O and 4CO₂ together

So ratio: 6 mol H₂O : 4 mol CO₂

Thus, if 6.0 mol H₂O → then 4.0 mol CO₂

Because they’re produced in fixed ratio from same reaction.

✔ Answer: 4.0 mol CO₂

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Now, let’s compile all final answers clearly.

──────────────────────────────────────

Final Answer:
1. N₂ + 3H₂ → 2NH₃
2. 14 mol O₂
3. 14 mol O₂
4. 1.7 mol C₂H₆
5. 4.0 mol CO₂
6. 2.6 mol H₂O
7. 14 mol O₂
8. 4.0 mol CO₂