Free Printable Mole to Mole Stoichiometry Worksheets - Free Printable

Let’s solve each problem step by step. We’ll use the balanced chemical equations to find mole ratios — that’s the key in stoichiometry!

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Problem 1:

Reaction:
2 Na (s) + 2 H₂O (l) → 2 NaOH (aq) + H₂ (g)

We’re told 4.0 mol of H₂ is produced. From the equation, 1 mol H₂ comes from 2 mol Na.

So, for 4.0 mol H₂:

→ Moles of Na = 4.0 mol H₂ × (2 mol Na / 1 mol H₂) = 8.0 mol Na

✔ Check: Ratio is 2:1 (Na:H₂), so yes, double the H₂ moles.

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Problem 2:

Reaction:
3 H₂ (g) + N₂ (g) → 2 NH₃ (g)

Given: 4.0 mol H₂ react.

From equation: 3 mol H₂ produce 2 mol NH₃

So, moles of NH₃ = 4.0 mol H₂ × (2 mol NH₃ / 3 mol H₂) = 8.0 / 3 ≈ 2.67 mol NH₃

✔ Check: 3 H₂ → 2 NH₃, so 4 H₂ should give a bit less than 3 NH₃? Wait — no: 3 H₂ gives 2 NH₃, so 4 H₂ gives (2/3)*4 = 8/3 = 2.666... → rounds to 2.67 if using 3 sig figs. But since 4.0 has 2 sig figs, maybe we say 2.7? Let’s check instructions — it says “how many moles”, and input was 4.0 (two sig figs). So answer should be 2.7 mol? Actually, let’s keep it as fraction or decimal based on context. In most school settings, they accept 2.67 or 8/3. But let’s go with exact value unless specified.

Wait — 4.0 has two significant figures, so answer should have two: 2.7 mol.

But let’s see what the worksheet expects — probably exact calculation. I’ll write both and decide later. For now, I’ll compute exactly.

Actually, let’s stick to precise math first: 4.0 × (2/3) = 8.0/3 = 2.666... → 2.67 mol (if rounding to three digits, which is common in such problems).

But to be safe, let’s note: 4.0 has two sig figs → answer should be 2.7 mol. However, many textbooks don’t strictly enforce sig figs in early stoich. Since the question didn’t specify, I’ll go with 2.67 mol as intermediate, but final answer can be written as 2.7 if needed. Let me recheck — actually, in Problem 1, 4.0 gave 8.0 (which kept one decimal). So perhaps here too, 2.7 mol.

Wait — 4.0 × 2 / 3 = 8.0 / 3 = 2.666... → to two sig figs is 2.7.

I think for consistency, we'll use 2.7 mol.

But let’s hold off and do all problems first.

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Problem 3a:

Reaction:
4 P (s) + 5 O₂ (g) → P₄O₁₀ (s)

Need to produce 3.25 mol P₄O₁₀.

From equation: 1 mol P₄O₁₀ requires 4 mol P

So, moles of P = 3.25 mol P₄O₁₀ × (4 mol P / 1 mol P₄O₁₀) = 13.0 mol P

✔ Check: 4:1 ratio → 3.25 × 4 = 13.0 → correct.

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Problem 3b:

Same reaction: 4 P → 1 P₄O₁₀

Given: 2.74 mol P

Moles of P₄O₁₀ = 2.74 mol P × (1 mol P₄O₁₀ / 4 mol P) = 2.74 / 4 = 0.685 mol P₄O₁₀

✔ Check: divide by 4 → 2.74 ÷ 4 = 0.685 → correct.

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Problem 4:

Reaction:
2 LiBr (aq) + Cl₂ (g) → 2 LiCl (aq) + Br₂ (g)

Given: 0.046 mol LiBr

From equation: 2 mol LiBr produce 2 mol LiCl → so 1:1 ratio!

Therefore, moles of LiCl = 0.046 mol LiBr × (2 mol LiCl / 2 mol LiBr) = 0.046 mol LiCl

✔ Check: same coefficient → same number of moles.

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Problem 5:

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

Given: 18 mol Al

From equation: 2 mol Al react with 3 mol H₂SO₄

So, moles of H₂SO₄ = 18 mol Al × (3 mol H₂SO₄ / 2 mol Al) = (18 × 3) / 2 = 54 / 2 = 27 mol H₂SO₄

✔ Check: 2 Al : 3 H₂SO₄ → so for 18 Al, which is 9 times 2, then H₂SO₄ = 9 × 3 = 27 → correct.

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Now, going back to Problem 2 — let’s confirm sig figs.

Input: 4.0 mol H₂ → two sig figs.

Calculation: 4.0 × (2/3) = 2.666...

Rounded to two sig figs: 2.7 mol

Yes, that’s appropriate.

In Problem 1, 4.0 → 8.0 (kept one decimal, two sig figs).

Similarly, here 2.7 has two sig figs.

So final answers:

1. 8.0 mol Na
2. 2.7 mol NH₃
3a. 13.0 mol P
3b. 0.685 mol P₄O₁₀
4. 0.046 mol LiCl
5. 27 mol H₂SO₄

All checked.

Final Answer:
1. 8.0 mol
2. 2.7 mol
3a. 13.0 mol
3b. 0.685 mol
4. 0.046 mol
5. 27 mol