Chemistry worksheet with stoichiometry problems involving chemical equations and calculations.

Let’s work through each problem one by one.

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Problem 1:

Equation:
Pb(SO₄)₂ + 4 LiNO₃ → Pb(NO₃)₄ + 2 Li₂SO₄

We are asked: How many grams of lithium nitrate (LiNO₃) are needed to make 250 grams of lithium sulfate (Li₂SO₄)?

Step 1: Find molar masses.

- Molar mass of Li₂SO₄ = (2 × 6.94) + 32.07 + (4 × 16.00) = 13.88 + 32.07 + 64.00 = 109.95 g/mol
- Molar mass of LiNO₃ = 6.94 + 14.01 + (3 × 16.00) = 6.94 + 14.01 + 48.00 = 68.95 g/mol

Step 2: Convert 250 g Li₂SO₄ to moles.

Moles of Li₂SO₄ = 250 / 109.95 ≈ 2.274 mol

Step 3: Use mole ratio from equation.

From the balanced equation:
2 moles Li₂SO₄ are produced from 4 moles LiNO₃
→ So, 1 mole Li₂SO₄ requires 2 moles LiNO₃

Therefore, 2.274 mol Li₂SO₄ requires:
2.274 × 2 = 4.548 mol LiNO₃

Step 4: Convert moles LiNO₃ to grams.

Mass = 4.548 mol × 68.95 g/mol ≈ 313.6 grams

✔ Final Answer for Problem 1: 313.6 grams

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Problem 2:

Equation:
Fe₂O₃ + 3 H₂ → 2 Fe + 3 H₂O

Given: 16.5 grams of Fe₂O₃ → find grams of Fe produced.

Step 1: Molar masses.

- Fe₂O₃ = (2 × 55.85) + (3 × 16.00) = 111.7 + 48.00 = 159.7 g/mol
- Fe = 55.85 g/mol

Step 2: Moles of Fe₂O₃ = 16.5 / 159.7 ≈ 0.1033 mol

Step 3: Mole ratio — 1 mol Fe₂O₃ produces 2 mol Fe
→ 0.1033 mol Fe₂O₃ produces 0.1033 × 2 = 0.2066 mol Fe

Step 4: Mass of Fe = 0.2066 × 55.85 ≈ 11.54 grams

✔ Final Answer for Problem 2: 11.54 grams

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Problem 3:

Equation:
5 KNO₂ + 2 KMnO₄ + 3 H₂SO₄ → 5 KNO₃ + 2 MnSO₄ + K₂SO₄ + 3 H₂O

Wait — the question says: “How many moles and grams of KMnO₄ are needed... on 11.4 grams of KNO₄?”

But in the equation, we have KNO₃, not KNO₄. That must be a typo. Probably meant KNO₃.

Assuming it’s KNO₃ (potassium nitrate), since that’s what appears in the products.

So: Given 11.4 g KNO₃ → find moles and grams of KMnO₄ needed.

Step 1: Molar masses.

- KNO₃ = 39.10 + 14.01 + 48.00 = 101.11 g/mol
- KMnO₄ = 39.10 + 54.94 + 64.00 = 158.04 g/mol

Step 2: Moles of KNO₃ = 11.4 / 101.11 ≈ 0.1127 mol

Step 3: Mole ratio — from equation:
5 mol KNO₃ are produced from 2 mol KMnO₄
→ So, moles KMnO₄ = (2/5) × moles KNO₃ = 0.4 × 0.1127 ≈ 0.04508 mol

Step 4: Grams of KMnO₄ = 0.04508 × 158.04 ≈ 7.125 grams

✔ Final Answer for Problem 3:
Moles of KMnO₄: 0.0451 mol
Grams of KMnO₄: 7.13 grams

*(Note: If the problem truly meant KNO₄, which doesn’t exist commonly, then there’s an error. But based on context, KNO₃ is intended.)*

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Problem 4:

Equation:
2 I₂ + KIO₃ + 6 HCl → 5 ICl + KCl + 3 H₂O

Find: How many grams of iodine (I₂) are needed to prepare 28.6 grams of ICl?

Step 1: Molar masses.

- ICl = 126.90 + 35.45 = 162.35 g/mol
- I₂ = 2 × 126.90 = 253.80 g/mol

Step 2: Moles of ICl = 28.6 / 162.35 ≈ 0.1762 mol

Step 3: Mole ratio — from equation:
5 mol ICl are produced from 2 mol I₂
→ So, moles I₂ = (2/5) × 0.1762 = 0.07048 mol

Step 4: Mass of I₂ = 0.07048 × 253.80 ≈ 17.89 grams

✔ Final Answer for Problem 4: 17.89 grams

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Final Answers:

1. 313.6 grams
2. 11.54 grams
3. 0.0451 mol and 7.13 grams
4. 17.89 grams