Stoichiometry practice worksheet featuring four chemistry problems with balanced equations and questions on mole and gram calculations.

Problem 1:

Using the following equation:
$$
2 \text{I}_2 + \text{KIO}_3 + 6 \text{HCl} \rightarrow 5 \text{ICl} + \text{KCl} + 3 \text{H}_2\text{O}
$$
Calculate how many grams of iodine are needed to prepare 28.6 grams of ICl by this reaction.

#### Solution:
1. Determine the molar mass of ICl:
- Molar mass of ICl = Atomic mass of I + Atomic mass of Cl
- Atomic mass of I = 126.9 g/mol
- Atomic mass of Cl = 35.45 g/mol
- Molar mass of ICl = 126.9 + 35.45 = 162.35 g/mol

2. Calculate the moles of ICl:
- Moles of ICl = Mass of ICl / Molar mass of ICl
- Moles of ICl = 28.6 g / 162.35 g/mol ≈ 0.1761 mol

3. Use the stoichiometry of the balanced equation:
- From the balanced equation, 5 moles of ICl are produced from 2 moles of I₂.
- Therefore, the mole ratio of I₂ to ICl is 2 : 5.
- Moles of I₂ required = (Moles of ICl) × (2 moles of I₂ / 5 moles of ICl)
- Moles of I₂ required = 0.1761 mol × (2/5) ≈ 0.07044 mol

4. Calculate the mass of I₂:
- Molar mass of I₂ = 2 × Atomic mass of I = 2 × 126.9 g/mol = 253.8 g/mol
- Mass of I₂ = Moles of I₂ × Molar mass of I₂
- Mass of I₂ = 0.07044 mol × 253.8 g/mol ≈ 17.87 g

Answer:
$$
\boxed{17.87 \text{ grams}}
$$

---

Problem 2:

Using the following equation:
$$
5 \text{KNO}_2 + 2 \text{KMnO}_4 + 3 \text{H}_2\text{SO}_4 \rightarrow 5 \text{KNO}_3 + 2 \text{MnSO}_4 + \text{K}_2\text{SO}_4 + 3 \text{H}_2\text{O}
$$
How many moles and grams of KMnO₄ are needed for this reaction on 11.4 grams of KNO₂?

#### Solution:
1. Determine the molar mass of KNO₂:
- Molar mass of KNO₂ = Atomic mass of K + Atomic mass of N + 2 × Atomic mass of O
- Atomic mass of K = 39.1 g/mol
- Atomic mass of N = 14.01 g/mol
- Atomic mass of O = 16.0 g/mol
- Molar mass of KNO₂ = 39.1 + 14.01 + 2 × 16.0 = 75.11 g/mol

2. Calculate the moles of KNO₂:
- Moles of KNO₂ = Mass of KNO₂ / Molar mass of KNO₂
- Moles of KNO₂ = 11.4 g / 75.11 g/mol ≈ 0.1518 mol

3. Use the stoichiometry of the balanced equation:
- From the balanced equation, 5 moles of KNO₂ react with 2 moles of KMnO₄.
- Therefore, the mole ratio of KNO₂ to KMnO₄ is 5 : 2.
- Moles of KMnO₄ required = (Moles of KNO₂) × (2 moles of KMnO₄ / 5 moles of KNO₂)
- Moles of KMnO₄ required = 0.1518 mol × (2/5) ≈ 0.06072 mol

4. Calculate the mass of KMnO₄:
- Molar mass of KMnO₄ = Atomic mass of K + Atomic mass of Mn + 4 × Atomic mass of O
- Atomic mass of K = 39.1 g/mol
- Atomic mass of Mn = 54.94 g/mol
- Atomic mass of O = 16.0 g/mol
- Molar mass of KMnO₄ = 39.1 + 54.94 + 4 × 16.0 = 158.04 g/mol
- Mass of KMnO₄ = Moles of KMnO₄ × Molar mass of KMnO₄
- Mass of KMnO₄ = 0.06072 mol × 158.04 g/mol ≈ 9.56 g

Answer:
$$
\boxed{0.0607 \text{ moles}, 9.56 \text{ grams}}
$$

---

Problem 3:

Using the following equation:
$$
4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 4 \text{NO} + 6 \text{H}_2\text{O}
$$
How many moles and grams of oxygen (O₂) are needed to react with 56.8 grams of ammonia by this reaction?

#### Solution:
1. Determine the molar mass of NH₃:
- Molar mass of NH₃ = Atomic mass of N + 3 × Atomic mass of H
- Atomic mass of N = 14.01 g/mol
- Atomic mass of H = 1.01 g/mol
- Molar mass of NH₃ = 14.01 + 3 × 1.01 = 17.04 g/mol

2. Calculate the moles of NH₃:
- Moles of NH₃ = Mass of NH₃ / Molar mass of NH₃
- Moles of NH₃ = 56.8 g / 17.04 g/mol ≈ 3.333 mol

3. Use the stoichiometry of the balanced equation:
- From the balanced equation, 4 moles of NH₃ react with 5 moles of O₂.
- Therefore, the mole ratio of NH₃ to O₂ is 4 : 5.
- Moles of O₂ required = (Moles of NH₃) × (5 moles of O₂ / 4 moles of NH₃)
- Moles of O₂ required = 3.333 mol × (5/4) ≈ 4.167 mol

4. Calculate the mass of O₂:
- Molar mass of O₂ = 2 × Atomic mass of O = 2 × 16.0 g/mol = 32.0 g/mol
- Mass of O₂ = Moles of O₂ × Molar mass of O₂
- Mass of O₂ = 4.167 mol × 32.0 g/mol ≈ 133.3 g

Answer:
$$
\boxed{4.17 \text{ moles}, 133 \text{ grams}}
$$

---

Problem 4:

Using the following equation:
$$
\text{NaIO}_3 + 6 \text{HI} \rightarrow 3 \text{I}_2 + \text{NaI} + 3 \text{H}_2\text{O}
$$
Calculate the number of moles and the number of grams of iodine (I₂) that can be made this way from 16.4 grams of NaIO₃.

#### Solution:
1. Determine the molar mass of NaIO₃:
- Molar mass of NaIO₃ = Atomic mass of Na + Atomic mass of I + 3 × Atomic mass of O
- Atomic mass of Na = 22.99 g/mol
- Atomic mass of I = 126.9 g/mol
- Atomic mass of O = 16.0 g/mol
- Molar mass of NaIO₃ = 22.99 + 126.9 + 3 × 16.0 = 198.89 g/mol

2. Calculate the moles of NaIO₃:
- Moles of NaIO₃ = Mass of NaIO₃ / Molar mass of NaIO₃
- Moles of NaIO₃ = 16.4 g / 198.89 g/mol ≈ 0.0824 mol

3. Use the stoichiometry of the balanced equation:
- From the balanced equation, 1 mole of NaIO₃ produces 3 moles of I₂.
- Therefore, the mole ratio of NaIO₃ to I₂ is 1 : 3.
- Moles of I₂ produced = (Moles of NaIO₃) × (3 moles of I₂ / 1 mole of NaIO₃)
- Moles of I₂ produced = 0.0824 mol × 3 ≈ 0.2472 mol

4. Calculate the mass of I₂:
- Molar mass of I₂ = 2 × Atomic mass of I = 2 × 126.9 g/mol = 253.8 g/mol
- Mass of I₂ = Moles of I₂ × Molar mass of I₂
- Mass of I₂ = 0.2472 mol × 253.8 g/mol ≈ 62.7 g

Answer:
$$
\boxed{0.247 \text{ moles}, 62.7 \text{ grams}}
$$

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Final Answers:

1. Problem 1: $\boxed{17.87 \text{ grams}}$
2. Problem 2: $\boxed{0.0607 \text{ moles}, 9.56 \text{ grams}}$
3. Problem 3: $\boxed{4.17 \text{ moles}, 133 \text{ grams}}$
4. Problem 4: $\boxed{0.247 \text{ moles}, 62.7 \text{ grams}}$