Stoichiometry practice problems worksheet with questions on mole ratios, mass calculations, and chemical reactions.

Let’s solve each problem step by step.

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Problem 1:

Reaction:
2 Al (s) + 3 H₂SO₄ (aq) → Al₂(SO₄)₃ + 3 H₂ (g)

a. How many moles of H₂SO₄ will react with 18 mol of Al?

From the balanced equation:
2 mol Al reacts with 3 mol H₂SO₄

So, set up a ratio:

(3 mol H₂SO₄ / 2 mol Al) × 18 mol Al = ?

→ (3/2) × 18 = 27 mol H₂SO₄

✔ Answer for 1a: 27 moles

b. How many moles of each product will be produced?

Products are: Al₂(SO₄)₃ and H₂

From the equation:
2 mol Al produces → 1 mol Al₂(SO₄)₃ and 3 mol H₂

So for 18 mol Al:

Al₂(SO₄)₃: (1 mol product / 2 mol Al) × 18 mol Al = 9 mol

H₂: (3 mol product / 2 mol Al) × 18 mol Al = 27 mol

✔ Answer for 1b:
9 moles of Al₂(SO₄)₃ and 27 moles of H₂

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Problem 2:

Reaction:
Na₂CO₃ + 2 HNO₃ → 2 NaNO₃ + CO₂ + H₂O

a. How many moles of Na₂CO₃ are required to produce 100 g of NaNO₃?

First, find molar mass of NaNO₃:

Na = 23, N = 14, O = 16×3 = 48 → 23 + 14 + 48 = 85 g/mol

Moles of NaNO₃ in 100 g = mass / molar mass = 100 g / 85 g/mol ≈ 1.1765 mol

From the equation:
2 mol NaNO₃ come from 1 mol Na₂CO₃

So, moles of Na₂CO₃ needed = (1 mol Na₂CO₃ / 2 mol NaNO₃) × 1.1765 mol NaNO₃

= 0.5882 mol ≈ 0.588 mol (we’ll keep 3 sig figs since 100 has 3)

✔ Answer for 2a: 0.588 moles

b. If 7.5 g of Na₂CO₃ reacts, how many moles of CO₂ are produced?

Molar mass of Na₂CO₃:
Na₂ = 23×2 = 46, C = 12, O₃ = 48 → 46 + 12 + 48 = 106 g/mol

Moles of Na₂CO₃ = 7.5 g / 106 g/mol ≈ 0.07075 mol

From the equation:
1 mol Na₂CO₃ produces 1 mol CO₂

So, moles of CO₂ = same as moles of Na₂CO₃ = 0.0708 mol (rounded to 3 sig figs)

✔ Answer for 2b: 0.0708 moles

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Problem 3:

Reaction:
CaC₂ (s) + 2 H₂O (l) → C₂H₂ (g) + Ca(OH)₂ (s)

What mass of acetylene (C₂H₂) can be produced from 90 g of calcium carbide (CaC₂)?

Molar mass of CaC₂:
Ca = 40, C₂ = 24 → 40 + 24 = 64 g/mol

Moles of CaC₂ = 90 g / 64 g/mol = 1.40625 mol

From the equation:
1 mol CaC₂ produces 1 mol C₂H₂

So, moles of C₂H₂ = 1.40625 mol

Molar mass of C₂H₂:
C₂ = 24, H₂ = 2 → 26 g/mol

Mass of C₂H₂ = moles × molar mass = 1.40625 mol × 26 g/mol = 36.5625 g

Round to 2 sig figs? Wait — 90 g has 2 sig figs? Actually, 90 could be 1 or 2. But in chemistry problems like this, we usually assume it’s exact or use given precision. Let’s go with 2 sig figs if 90 is considered 2, but often in such contexts, we treat it as having 2. However, 90 without decimal might be ambiguous. Looking at other numbers in worksheet (like 7.5, 30.0), they use decimals for precision. So perhaps 90 is meant to be 2 sig figs.

But let’s check: 90 ÷ 64 = 1.40625 → times 26 = 36.5625 → round to 2 sig figs? That would be 37 g.

Wait — actually, 90 has two significant figures, so answer should have two.

But let me double-check calculation:

90 / 64 = 1.40625
1.40625 × 26 = let’s compute:
1.4 × 26 = 36.4
0.00625 × 26 = 0.1625
Total = 36.5625 → yes.

Rounded to 2 sig figs: 37 g

But wait — maybe the problem expects more precision? In many textbooks, if no decimal, they still accept 3 sig figs if intermediate steps allow. But to be safe, let’s see: 90 has two sig figs, so answer should be 37 g.

Actually, let me recalculate with exact fractions:

90 / 64 = 45/32
45/32 × 26 = (45 × 26) / 32 = 1170 / 32 = 36.5625

Yes.

In stoichiometry, sometimes we keep one extra digit. But I think for consistency, since 90 has two sig figs, we report 37 g.

However, looking at problem 4, it uses 30.0 g (three sig figs). Here it’s 90 g — probably intended as two sig figs.

But let’s check online or standard practice — actually, in many cases, “90” in such problems is treated as having two sig figs.

I’ll go with 37 g.

Wait — another way: perhaps calculate exactly and see.

Alternatively, maybe the problem allows 36.6 g? But 90 has two sig figs.

Actually, let’s look back: in problem 2b, 7.5 g (two sig figs) gave answer 0.0708 (which is three, but that’s because 7.5/106=0.07075… and we rounded to three). Hmm.

To avoid confusion, let’s do this: 90 g has two sig figs, molar masses are exact for sig fig purposes? No, molar masses are known precisely, so limiting sig figs come from measured quantity — 90 g.

So 90 has two sig figs → answer should have two.

36.5625 → rounds to 37 g.

✔ Answer for 3: 37 grams

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Problem 4:

Reaction:
2 Al₂O₃ → 4 Al + 3 O₂

How many moles of aluminum will be produced from 30.0 g of Al₂O₃?

Molar mass of Al₂O₃:
Al₂ = 27×2 = 54, O₃ = 48 → 54 + 48 = 102 g/mol

Moles of Al₂O₃ = 30.0 g / 102 g/mol = 0.2941176 mol

From the equation:
2 mol Al₂O₃ produce 4 mol Al

So, mole ratio: (4 mol Al / 2 mol Al₂O₃) = 2

Thus, moles of Al = 0.2941176 mol × 2 = 0.588235 mol

Now, 30.0 g has three sig figs, molar mass 102 has three (if we take Al=27.0, O=16.0, then 2*27.0=54.0, 3*16.0=48.0, total 102.0 — so four, but typically we use 102 as three).

Actually, atomic masses: Al is 26.98, O is 16.00, so Al₂O₃ = 2*26.98 + 3*16.00 = 53.96 + 48.00 = 101.96 g/mol ≈ 102.0 g/mol (four sig figs)

But 30.0 has three sig figs, so answer should have three.

0.2941176 mol Al₂O₃ × (4 mol Al / 2 mol Al₂O₃) = 0.2941176 × 2 = 0.588235 mol

Rounded to three sig figs: 0.588 mol

✔ Answer for 4: 0.588 moles

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Final Answers:

1a. 27 moles
1b. 9 moles of Al₂(SO₄)₃ and 27 moles of H₂
2a. 0.588 moles
2b. 0.0708 moles
3. 37 grams
4. 0.588 moles

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Final Answer:
1a. 27 mol
1b. 9 mol Al₂(SO₄)₃ and 27 mol H₂
2a. 0.588 mol
2b. 0.0708 mol
3. 37 g
4. 0.588 mol