Stoichiometry practice problems worksheet with questions on butane combustion, sodium reaction with oxygen, and carbon dioxide production.

Let’s solve each problem step by step.

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Problem 1: Combustion of butane (C₄H₁₀)
Balanced equation:
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

Given: 2.46 grams of water produced.

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(a) How many moles of water formed?

Molar mass of H₂O = 2(1.008) + 16.00 ≈ 18.016 g/mol

Moles of H₂O = mass / molar mass = 2.46 g / 18.016 g/mol ≈ 0.1365 mol

✔ Let’s keep more decimals for accuracy:
2.46 ÷ 18.016 = 0.13654 mol → we’ll round to 0.137 mol at the end if needed, but let’s carry 0.1365 for now.

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(b) How many moles of butane burned?

From the balanced equation:
10 mol H₂O ← from ← 2 mol C₄H₁₀

So mole ratio:
mol C₄H₁₀ = (2/10) × mol H₂O = 0.2 × 0.13654 = 0.027308 mol

→ Approximately 0.0273 mol

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(c) How many grams of butane burned?

Molar mass of C₄H₁₀ = 4(12.01) + 10(1.008) = 48.04 + 10.08 = 58.12 g/mol

Mass = moles × molar mass = 0.027308 mol × 58.12 g/mol ≈ ?

Calculate:
0.027308 × 58.12 = let’s compute:

First, 0.0273 × 58.12 ≈
0.02 × 58.12 = 1.1624
0.0073 × 58.12 ≈ 0.424
Total ≈ 1.586 → better to do exact:

0.027308 × 58.12 =
= (0.027308 × 50) + (0.027308 × 8.12)
= 1.3654 + 0.2217 ≈ 1.5871

Actually, use calculator-style:
0.027308 × 58.12 = 1.587 g (rounded to 3 decimal places)

But let’s check with precise values:

Earlier: moles H₂O = 2.46 / 18.016 = 0.136542...

Then moles C₄H₁₀ = (2/10)*0.136542 = 0.0273084

Mass = 0.0273084 × 58.12 = let’s compute:

0.0273084 × 58.12 =
Break it down:
0.0273084 × 50 = 1.36542
0.0273084 × 8 = 0.2184672
0.0273084 × 0.12 = 0.003277
Sum: 1.36542 + 0.2184672 = 1.5838872 + 0.003277 ≈ 1.58716 g

→ So approximately 1.59 g (if rounding to 3 significant figures — since 2.46 has 3 sig figs)

Wait — 2.46 has 3 sig figs, so answers should be reported to 3 sig figs.

So:

(a) Moles H₂O = 2.46 / 18.016 = 0.1365 → 0.137 mol (3 sig figs)

(b) Moles C₄H₁₀ = (2/10)(0.1365) = 0.0273 → 0.0273 mol (still 3 sig figs? Actually 2.46 has 3, and ratios are exact, so yes)

(c) Mass C₄H₁₀ = 0.0273 mol × 58.12 g/mol = let’s compute exactly:

0.0273 × 58.12 =
0.0273 × 58 = 1.5834
0.0273 × 0.12 = 0.003276
Total = 1.586676 → 1.59 g (rounded to 3 sig figs)

✔ Yes.

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(d) How much oxygen was used up in moles?

From equation:
10 mol H₂O ← from ← 13 mol O₂

So moles O₂ = (13/10) × moles H₂O = 1.3 × 0.13654 = 0.1775 mol

With 3 sig figs: 0.178 mol

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(e) How much oxygen was used up in grams?

Molar mass O₂ = 32.00 g/mol

Mass O₂ = 0.1775 mol × 32.00 g/mol = 5.68 g

Check: 0.1775 × 32 =
0.17 × 32 = 5.44
0.0075 × 32 = 0.24
Total = 5.68 → correct.

To 3 sig figs: 5.68 g

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Problem 2: Sodium burning in air

Equation:
4 Na(s) + O₂(g) → 2 Na₂O(g)

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

Molar mass Na = 22.99 g/mol

Moles Na = 9.5 g / 22.99 g/mol ≈ 0.4132 mol

From equation:
4 mol Na require 1 mol O₂

So moles O₂ = (1/4) × moles Na = 0.4132 / 4 = 0.1033 mol

To 2 sig figs? 9.5 has 2 sig figs → so answer should have 2 sig figs.

0.1033 → 0.10 mol? Wait, 9.5 has two sig figs, but 0.1033 is closer to 0.10 or 0.103?

Actually, 9.5 has two sig figs, so moles Na = 9.5 / 22.99 = 0.4132 → which we report as 0.41 mol (two sig figs)? But 9.5 could be considered as having uncertainty in tenths place.

Better: 9.5 / 22.99 = 0.4132 → but since 9.5 has two sig figs, we take 0.41 mol Na.

Then moles O₂ = 0.41 / 4 = 0.1025 → 0.10 mol (two sig figs)

But let’s see what’s standard. Often in such problems, they expect calculation without premature rounding.

I’ll calculate precisely and then round final answer.

Moles Na = 9.5 / 22.99 = 0.413223...

Moles O₂ = 0.413223 / 4 = 0.10330575 → to two sig figs: 0.10 mol

But wait — 0.10 has two sig figs, but it might be written as 0.103 if we consider 9.5 as having implied precision. However, strictly, 9.5 has two sig figs, so answer should be 0.10 mol.

Alternatively, some teachers accept 0.103 if intermediate steps aren’t rounded. I think for consistency, we'll go with 0.10 mol (since 9.5 limits us).

But let me double-check: 9.5 g — if it's 9.5 ± 0.1, then relative error ~1%, so answer should reflect that.

0.1033 has about 0.3% error if we say 0.10, but actually 0.10 implies ±0.01, which is 10% — too rough.

Perhaps better to report as 0.103 mol, noting that 9.5 has two sig figs but we keep an extra digit for intermediate.

In many textbooks, they would write 0.10 mol.

I’ll go with 0.10 mol for strict sig fig rules.

Wait — let’s look at part (b): 12.5 g has three sig figs, so probably here also we can use three.

Actually, 9.5 has two, 12.5 has three — different.

For (a), I’ll compute and report as 0.10 mol (two sig figs).

But let’s calculate numerically:

9.5 / 22.99 = 0.4132 → /4 = 0.1033 → rounded to two sig figs is 0.10 mol.

Yes.

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

First, molar mass Na₂O = 2(22.99) + 16.00 = 45.98 + 16.00 = 61.98 g/mol

Moles Na₂O = 12.5 g / 61.98 g/mol ≈ 0.20168 mol

From equation:
2 mol Na₂O ← from ← 4 mol Na

So mole ratio: mol Na = (4/2) × mol Na₂O = 2 × 0.20168 = 0.40336 mol

Mass Na = moles × molar mass = 0.40336 mol × 22.99 g/mol ≈ ?

Compute:
0.40336 × 22.99 =
0.4 × 22.99 = 9.196
0.00336 × 22.99 ≈ 0.0772
Total ≈ 9.2732

More accurately:
0.40336 × 20 = 8.0672
0.40336 × 2.99 = let’s compute 0.40336 × 3 = 1.21008 minus 0.40336 × 0.01 = 0.0040336 → so 1.21008 - 0.0040336 = 1.2060464
Total = 8.0672 + 1.2060464 = 9.2732464 g

→ 9.27 g (to three sig figs, since 12.5 has three)

Yes.

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Problem 3: Combustion of butane (C₄H₆?) — wait, typo?

The problem says: “combustion of 54.5 g of butane (C₄H₆)”

But butane is C₄H₁₀, not C₄H₆. C₄H₆ is butyne or something else.

Probably a typo. In context, since previous problems used C₄H₁₀, and combustion typically refers to alkanes, likely it’s meant to be C₄H₁₀.

Moreover, the instruction says “write and balance the equation”, implying it’s not given.

Assuming it’s butane, C₄H₁₀.

Combustion reaction:
C₄H₁₀ + O₂ → CO₂ + H₂O

Balance it:

C₄H₁₀ + O₂ → 4CO₂ + 5H₂O (carbon and hydrogen balanced)

Now oxygen: right side: 4×2 + 5×1 = 8 + 5 = 13 oxygen atoms → so 13/2 O₂ molecules.

Multiply entire equation by 2:

2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

Same as problem 1.

Now, given: 54.5 g of butane (C₄H₁₀)

Find grams of CO₂ produced.

Molar mass C₄H₁₀ = 58.12 g/mol (as before)

Moles C₄H₁₀ = 54.5 / 58.12 ≈ ?

54.5 ÷ 58.12 = 0.9377 mol (let’s compute)

58.12 × 0.9377 ≈ 58.12 × 0.9 = 52.308, 58.12 × 0.0377 ≈ 2.19, total 54.498 → close.

Exactly: 54.5 / 58.12 = 0.937715... mol

From equation:
2 mol C₄H₁₀ produce 8 mol CO₂

So moles CO₂ = (8/2) × moles C₄H₁₀ = 4 × 0.937715 = 3.75086 mol

Molar mass CO₂ = 12.01 + 32.00 = 44.01 g/mol

Mass CO₂ = 3.75086 × 44.01 ≈ ?

Compute:
3.75 × 44.01 = 3.75 × 44 = 165, 3.75 × 0.01 = 0.0375 → 165.0375
But more accurately:
3.75086 × 44.01 =
First, 3.75086 × 40 = 150.0344
3.75086 × 4 = 15.00344
3.75086 × 0.01 = 0.0375086
Sum: 150.0344 + 15.00344 = 165.03784 + 0.0375086 ≈ 165.07535

So approximately 165.1 g

Significant figures: 54.5 has three sig figs, so answer should have three.

165.1 → 165 g? But 165 has three sig figs.

165.1 rounded to three sig figs is 165 g.

But let’s confirm calculation:

Moles C₄H₁₀ = 54.5 / 58.12 = 0.937715

Times 4 = 3.75086 mol CO₂

Times 44.01 = 3.75086 × 44.01

Do 3.75086 × 44 = 165.03784

3.75086 × 0.01 = 0.0375086

Total 165.0753486 → 165 g (to three sig figs)

Yes.

Note: If it were really C₄H₆, we’d need to adjust, but I believe it’s a typo and should be C₄H₁₀.

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Now, compiling all answers:

Problem 1:

(a) 0.137 mol H₂O
(b) 0.0273 mol C₄H₁₀
(c) 1.59 g C₄H₁₀
(d) 0.178 mol O₂
(e) 5.68 g O₂

Problem 2:

(a) 0.10 mol O₂ (or 0.103? I’ll stick with 0.10 for two sig figs)
Wait, let’s reconsider: 9.5 has two sig figs, but in chemistry, sometimes they allow one extra. To be safe, I’ll use 0.10 mol.

But looking back, in problem 1, 2.46 had three, and we used three. Here 9.5 has two, so 0.10 mol is fine.

However, let me check online or standard practice — often for 9.5, they might expect 0.103. But I think for correctness, we'll use two sig figs.

Actually, 9.5 divided by 22.99 is 0.413, which to two sig figs is 0.41, then /4 = 0.1025 → 0.10 mol.

Yes.

(b) 9.27 g Na

Problem 3:

165 g CO₂

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Final Answers:

Problem 1(a): 0.137 mol
Problem 1(b): 0.0273 mol
Problem 1(c): 1.59 g
Problem 1(d): 0.178 mol
Problem 1(e): 5.68 g
Problem 2(a): 0.10 mol
Problem 2(b): 9.27 g
Problem 3: 165 g

But for Problem 2(a), let's write it as 0.10 mol to match sig figs.

Some might argue 0.103, but I think 0.10 is correct.

To be precise, I'll box them as per calculation.

Now, for the response, since ExplainAnswer is false, etc., I just provide final answers after reasoning.

But the user said "ExplainAnswer: false", so I don't need to explain, but I did work through it.

In the response, I should only give the final answers in the specified format.

Looking back at response format:

It says: "If an option is false, completely omit that section."

And "Complete ALL reasoning and explanations BEFORE writing the Final Answer."

But since ExplainAnswer is false, I shouldn't include explanation section.

However, the instruction says: "CRITICAL: First work through the problem step by step, check your calculations carefully, and ONLY THEN provide the final answer."

And "Response format rules: → Always follow the structure below."

The structure includes sections only if true.

Since all options are false, I should only have "Final Answer:" section.

But that seems odd because I need to show work? No, the rule says "omit that section" if false.

Re-reading:

"If ExplainAnswer is true: Explanation: ..."

Similarly for others.

Since all are false, I should not include any of those sections.

But then how to present the answer? The "Final Answer" section is separate.

The response format says:

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

Final Answer:
[Provide only the final correct answer — this should come AFTER all explanations and reasoning.]

But if no explanations are included because options are false, then directly Final Answer.

However, the critical rule says to work through step by step first, but not necessarily include it in output if options are false.

To clarify: the "work through" is for my internal process; the output should only contain the sections corresponding to true options, plus Final Answer.

Since all options are false, only Final Answer section.

But Final Answer should contain all answers.

So I'll list them clearly.

Also, for Problem 3, I assumed C₄H₁₀, but the problem says C₄H₆. Let me double-check.

The problem says: "butane (C₄H₆)" — but butane is C₄H₁₀. C₄H₆ is not butane; it's butadiene or butyne.

This is likely a typo. In educational contexts, it's probably meant to be C₄H₁₀.

If it's C₄H₆, combustion would be different.

Let me verify.

Suppose it's C₄H₆ (which is not butane, but perhaps mislabeled).

Combustion: C₄H₆ + O₂ → CO₂ + H₂O

Balance: C₄H₆ + ? O₂ → 4CO₂ + 3H₂O

Oxygen: right side 8 + 3 = 11 atoms → 11/2 O₂

So 2C₄H₆ + 11O₂ → 8CO₂ + 6H₂O

Molar mass C₄H₆ = 4*12.01 + 6*1.008 = 48.04 + 6.048 = 54.088 g/mol

Moles = 54.5 / 54.088 ≈ 1.0076 mol

From equation: 2 mol C₄H₆ produce 8 mol CO₂, so ratio 4:1

Moles CO₂ = 4 * 1.0076 = 4.0304 mol

Mass CO₂ = 4.0304 * 44.01 ≈ 177.4 g

But the problem says "butane", which is definitely C₄H₁₀. Probably a typo in the formula.

In most sources, butane is C₄H₁₀.

I think it's safe to assume C₄H₁₀.

Moreover, in problem 1, it's given as C₄H₁₀.

So I'll proceed with that.

Final answers:

For Problem 1:
(a) 0.137 mol
(b) 0.0273 mol
(c) 1.59 g
(d) 0.178 mol
(e) 5.68 g

For Problem 2:
(a) 0.10 mol
(b) 9.27 g

For Problem 3:
165 g

Now, to box them.

In Final Answer, I'll list them clearly.