Stoichiometry practice problems worksheet with questions on combustion reactions and chemical calculations.

Let’s work through 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 (H₂O) 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 digits for accuracy:
2.46 ÷ 18.016 = 0.13654 mol

We’ll round to 3 significant figures at the end, but keep extra during calc.

✔ Answer (a): 0.137 mol (rounded to 3 sig figs)

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(b) How many moles of butane burned?

From 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

✔ Answer (b): 0.0273 mol (3 sig figs)

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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:

0.027308 × 58 = 1.583864
0.027308 × 0.12 = 0.003277
Total ≈ 1.58714 g

More accurately:
0.027308 × 58.12 = 1.587 g

✔ Answer (c): 1.59 g (3 sig figs)

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

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

So:
mol O₂ = (13/10) × mol H₂O = 1.3 × 0.13654 = 0.177502 mol

✔ Answer (d): 0.178 mol (3 sig figs)

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

Molar mass O₂ = 32.00 g/mol

Mass = 0.177502 mol × 32.00 g/mol = 5.680064 g

✔ Answer (e): 5.68 g (3 sig figs)

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Problem 2: Sodium burning in air

Balanced 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:
mol O₂ = (1/4) × mol Na = 0.4132 / 4 = 0.1033 mol

✔ Answer (a): 0.103 mol (3 sig figs — since 9.5 has 2 sig figs? Wait — 9.5 has 2 sig figs, so answer should be 0.10 mol? Let’s check.)

Actually, 9.5 has two significant figures → so we should report to 2 sig figs.

0.4132 mol Na → 0.4132 / 4 = 0.1033 → rounds to 0.10 mol O₂

But wait — 9.5 could be interpreted as having 2 sig figs, yes.

However, sometimes in chemistry problems, if it's written as “9.5” without context, we might assume 2 sig figs. But let’s see part (b) uses 12.5 which is 3 sig figs. To be safe, I’ll note both.

Actually, standard rule: 9.5 has 2 sig figs → answer should have 2.

So: 0.1033 → 0.10 mol

But let me recalculate precisely:

9.5 / 22.99 = 0.413223...
÷ 4 = 0.103305...

Rounded to 2 sig figs: 0.10 mol

✔ Answer (a): 0.10 mol

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

Mass Na = 0.40336 mol × 22.99 g/mol ≈ ?

0.40336 × 22.99 = let’s calculate:

0.4 × 22.99 = 9.196
0.00336 × 22.99 ≈ 0.0772
Total ≈ 9.2732 g

More accurately:
0.40336 × 22.99 = 9.273 g

Now, 12.5 has 3 sig figs → so answer should have 3.

✔ Answer (b): 9.27 g

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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! C₄H₆ is not butane — that’s butyne or something else.

Looking back at Problem 1, it correctly says C₄H₁₀.

This must be a typo. Probably meant C₄H₁₀.

Also, instruction: “Write and balance the equation before solving.”

Assuming it’s butane → C₄H₁₀.

Balanced combustion equation for C₄H₁₀:

C₄H₁₀ + O₂ → CO₂ + H₂O

Balance:

C: 4 on left → 4 CO₂
H: 10 on left → 5 H₂O
O: right side: 4×2 + 5×1 = 8+5=13 → need 13/2 O₂

So:
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O ← same as Problem 1.

Now, given: 54.5 g of C₄H₁₀

Find grams of CO₂ produced.

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

Moles C₄H₁₀ = 54.5 / 58.12 ≈ 0.9377 mol

From equation: 2 mol C₄H₁₀ → 8 mol CO₂

So:
mol CO₂ = (8/2) × mol C₄H₁₀ = 4 × 0.9377 = 3.7508 mol

Molar mass CO₂ = 12.01 + 2(16.00) = 44.01 g/mol

Mass CO₂ = 3.7508 × 44.01 ≈ ?

3.7508 × 44 = 165.0352
3.7508 × 0.01 = 0.0375
Total ≈ 165.0727 g

More accurately:
3.7508 × 44.01 = let’s compute:

3.7508 × 40 = 150.032
3.7508 × 4 = 15.0032
3.7508 × 0.01 = 0.037508
Wait — better: 44.01 = 44 + 0.01

Actually:
3.7508 × 44.01 = 3.7508 × (44 + 0.01) = 3.7508×44 + 3.7508×0.01

3.7508 × 44:
3 × 44 = 132
0.7508 × 44 = 33.0352
Total = 165.0352

Plus 0.037508 → total ≈ 165.0727 g

With 54.5 having 3 sig figs → answer should have 3.

✔ Answer (3): 165 g

But wait — let me double-check calculation:

54.5 / 58.12 = 0.937715...
× 4 = 3.75086
× 44.01 = 3.75086 × 44.01

Compute directly:
3.75086 × 44.01 = ?

Use calculator-style:
3.75086 × 44.01 = (3.75086 × 44) + (3.75086 × 0.01)
= 165.03784 + 0.0375086 = 165.0753486 ≈ 165.1 g

Hmm — now I get 165.1, which would round to 165 g with 3 sig figs? 165 has 3 sig figs.

But 165.1 rounded to 3 sig figs is still 165.

Actually, 165.1 — the first three digits are 1,6,5 — and next is 1 <5, so 165.

But let’s use exact values:

Molar masses:
C = 12.01, H = 1.008, O = 16.00

C₄H₁₀ = 4*12.01 + 10*1.008 = 48.04 + 10.08 = 58.12 g/mol — correct

CO₂ = 12.01 + 32.00 = 44.01 g/mol — correct

54.5 / 58.12 = 0.937715...
× 4 = 3.75086...
× 44.01 = 165.075... → 165 g (to 3 sig figs)

Yes.

But note: the problem said "butane (C₄H₆)" — that’s incorrect. If it were really C₄H₆, we’d need to adjust.

C₄H₆ combustion:
C₄H₆ + O₂ → CO₂ + H₂O

Balance:
C: 4 → 4CO₂
H: 6 → 3H₂O
O: right: 8 + 3 = 11 → need 11/2 O₂

So: 2 C₄H₆ + 11 O₂ → 8 CO₂ + 6 H₂O

Then 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 eq: 2 mol C₄H₆ → 8 mol CO₂ → so 4:1 ratio

mol CO₂ = 4 × 1.0076 = 4.0304 mol

Mass CO₂ = 4.0304 × 44.01 ≈ 177.4 g → 177 g

But the problem says "butane", which is universally C₄H₁₀. The formula C₄H₆ is likely a typo.

In educational contexts, especially since Problem 1 uses C₄H₁₀, and this is labeled "butane", I think it’s safe to assume C₄H₁₀.

Moreover, 54.5 g is close to molar mass of C₄H₆ (54.09), so maybe intentional? But the name is wrong.

To resolve: look at the problem statement: “butane (C₄H₆)” — this is chemically inaccurate. Butane is C₄H₁₀.

I think it’s a typo, and should be C₄H₁₀.

Perhaps the student copied wrong? But based on standard knowledge, I’ll proceed with C₄H₁₀.

Final decision: use C₄H₁₀ as per naming.

So answer is 165 g.

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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
(b) 9.27 g Na

Problem 3:
165 g CO₂ (assuming C₄H₁₀)

But for Problem 3, to be thorough, I should note the discrepancy.

Since the instruction is to solve accurately, and "butane" is C₄H₁₀, I'll go with that.

Final Answers:

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