Chemistry worksheet page focusing on stoichiometric calculations, including dimensional analysis setups for sodium and acetylene reactions.

Let’s solve each part step by step.

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Problem 1: Sodium burning in air

Balanced equation:
4 Na (s) + O₂ (g) → 2 Na₂O (g)

We’ll use molar masses:
- Molar mass of Na = 23.0 g/mol
- Molar mass of Na₂O = 2×23.0 + 16.0 = 62.0 g/mol
- Molar mass of O₂ = 32.0 g/mol (but we may not need it directly here)

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(a) How many moles of oxygen are needed to completely react with 9.5 g of sodium?

Step 1: Convert grams of Na to moles of Na
→ Use: 1 mol Na / 23.0 g Na

Step 2: Use mole ratio from balanced equation to find moles of O₂
→ From equation: 4 mol Na : 1 mol O₂
→ So, multiply by (1 mol O₂ / 4 mol Na)

Setup:

9.5 g Na × (1 mol Na / 23.0 g Na) × (1 mol O₂ / 4 mol Na) = ? mol O₂

Calculate:

First: 9.5 ÷ 23.0 = 0.4130 mol Na
Then: 0.4130 ÷ 4 = 0.10325 mol O₂

Rounded to 2 significant figures (since 9.5 has 2 sig figs):
→ 0.10 mol O₂

But let’s keep more digits for now and round at the end if needed.

Actually, 9.5 has 2 sig figs, so answer should be 0.10 mol O₂

Wait — let me recalculate precisely:

9.5 / 23.0 = 0.413043...
Divide by 4 → 0.10326... ≈ 0.10 mol (if rounding to 2 sig figs)

But sometimes in stoichiometry, we keep intermediate steps precise and round only final answer. Let’s check what the problem expects. The setup boxes suggest filling numbers, so we’ll compute exactly.

Final calculation:

9.5 × (1/23.0) × (1/4) = 9.5 / (23.0 × 4) = 9.5 / 92.0 = 0.10326... mol

Typically, we report as 0.10 mol (2 sig figs), but since 9.5 is 2 sig figs and 23.0 and 4 are exact or have more, we can say 0.10 mol.

Actually, 4 is an exact number from the equation, so sig figs determined by 9.5 and 23.0. 23.0 has 3 sig figs, 9.5 has 2 → so answer has 2 sig figs.

✔ Answer for (a): 0.10 mol O₂

But wait — let’s write the setup as requested:

Boxes:

[9.5] g Na × [1] mol Na / [23.0] g Na × [1] mol O₂ / [4] mol Na = [0.10] mol O₂

Yes.

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(b) How many grams of sodium are needed to produce 12.5 g of sodium oxide?

Given: 12.5 g Na₂O

Molar mass Na₂O = 62.0 g/mol (given in problem)

From balanced equation: 2 mol Na₂O ← 4 mol Na
So, mole ratio: 4 mol Na / 2 mol Na₂O = 2 mol Na per 1 mol Na₂O

Setup:

12.5 g Na₂O × (1 mol Na₂O / 62.0 g Na₂O) × (4 mol Na / 2 mol Na₂O) × (23.0 g Na / 1 mol Na)

Simplify step by step:

First: 12.5 / 62.0 = 0.20161 mol Na₂O

Then: × (4/2) = × 2 → 0.40323 mol Na

Then: × 23.0 = 9.274 g Na

Sig figs: 12.5 has 3 sig figs, 62.0 has 3, 23.0 has 3 → answer should have 3 sig figs.

→ 9.27 g Na

Check calculation again:

12.5 × (1/62.0) × (4/2) × 23.0
= 12.5 × (1/62.0) × 2 × 23.0
= 12.5 × 2 × 23.0 / 62.0
= (12.5 × 46.0) / 62.0
= 575 / 62.0 = 9.27419...

Yes → 9.27 g

Setup as shown in problem:

12.5 g Na₂O × (1 mol Na₂O / 62.0 g Na₂O) × (4 mol Na / 2 mol Na₂O) × (23.0 g Na / 1 mol Na) = 9.27 g Na

The handwritten part says “mol Na₂O” and “g Na”, which matches.

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Problem 2: Acetylene combustion

Unbalanced: C₂H₂(g) + O₂(g) → CO₂(g) + H₂O(g)

Balance it:

Left: C=2, H=2, O=?

Right: C=1, H=2, O=3 (from CO₂ and H₂O)

Try balancing C first: put 2 before CO₂

C₂H₂ + O₂ → 2CO₂ + H₂O

Now H: left 2, right 2 → OK

O: right = 2×2 + 1 = 5 atoms → so need 5/2 O₂ on left

C₂H₂ + ⁵/₂ O₂ → 2CO₂ + H₂O

Multiply all by 2 to eliminate fraction:

2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O

Check:

C: 4 = 4
H: 4 = 4
O: 10 = 8 + 2 → yes!

Balanced equation:
2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O

Molar masses:
- C₂H₂: 2×12.0 + 2×1.0 = 26.0 g/mol
- H₂O: 18.0 g/mol
- CO₂: 44.0 g/mol (not needed yet)

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(a) How many grams of water can form if 113 g of acetylene is burned?

Use balanced equation: 2 mol C₂H₂ → 2 mol H₂O
So, mole ratio: 2 mol H₂O / 2 mol C₂H₂ = 1:1

That means moles of H₂O produced = moles of C₂H₂ burned

Step 1: Moles of C₂H₂ = 113 g / 26.0 g/mol = 4.34615 mol

Step 2: Moles of H₂O = same = 4.34615 mol (because 1:1 ratio)

Step 3: Mass of H₂O = 4.34615 mol × 18.0 g/mol = 78.2307 g

Sig figs: 113 has 3 sig figs, 26.0 and 18.0 have 3 → answer has 3 sig figs.

→ 78.2 g H₂O

Wait — let’s confirm ratio:

From equation: 2 mol C₂H₂ → 2 mol H₂O → yes, 1:1

So:

113 g C₂H₂ × (1 mol C₂H₂ / 26.0 g) × (2 mol H₂O / 2 mol C₂H₂) × (18.0 g H₂O / 1 mol H₂O)

Simplifies to:

113 × (1/26.0) × 1 × 18.0 = (113 × 18.0) / 26.0

= 2034 / 26.0 = 78.2307... → 78.2 g

✔ Correct.

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(b) How many grams of acetylene react if 1.10 mol of CO₂ are produced?

From balanced equation: 2 mol C₂H₂ → 4 mol CO₂

So, mole ratio: 2 mol C₂H₂ / 4 mol CO₂ = 1/2

Meaning: moles of C₂H₂ = (1/2) × moles of CO₂

Given: 1.10 mol CO₂

So, moles C₂H₂ = 1.10 × (2/4) = 1.10 × 0.5 = 0.550 mol

Mass of C₂H₂ = 0.550 mol × 26.0 g/mol = 14.3 g

Sig figs: 1.10 has 3 sig figs, 26.0 has 3 → answer 3 sig figs.

→ 14.3 g

Calculation:

1.10 mol CO₂ × (2 mol C₂H₂ / 4 mol CO₂) × (26.0 g C₂H₂ / 1 mol C₂H₂)

= 1.10 × (2/4) × 26.0 = 1.10 × 0.5 × 26.0 = 0.55 × 26.0 = 14.3

Perfect.

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Now, compiling final answers.

Note: For Problem 1(a), I calculated 0.103 mol, which rounds to 0.10 mol with 2 sig figs. But let's double-check the input: 9.5 g has 2 sig figs, molar mass of Na is typically taken as 23.0 (3 sig figs), and 4 is exact. So limiting is 9.5 → 2 sig figs → 0.10 mol.

However, sometimes in textbooks, they might expect 0.103 or 0.10. Since 9.5 / 92.0 = 0.10326, and 92.0 has 3 sig figs (23.0×4), but 4 is exact, so 23.0 has 3, 9.5 has 2 → quotient has 2 sig figs → 0.10 mol.

Similarly, for others, we used proper sig figs.

Final Answers:

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Final Answer:
1. (a) 0.10 mol O₂
(b) 9.27 g Na
2. Balanced equation: 2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O
(a) 78.2 g H₂O
(b) 14.3 g C₂H₂