Stoichiometry Practice Worksheet featuring four chemistry problems with balanced equations and calculations.

Let’s work through each problem step by step.

---

Problem 1:

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

We are given: 28.6 grams of ICl produced.
We need to find: how many grams of iodine (I₂) were needed.

Step 1: Find molar mass of ICl.
Iodine (I) = 126.90 g/mol
Chlorine (Cl) = 35.45 g/mol
→ Molar mass of ICl = 126.90 + 35.45 = 162.35 g/mol

Step 2: Convert grams of ICl to moles.
moles of ICl = 28.6 g / 162.35 g/mol ≈ 0.1762 mol

Step 3: Use mole ratio from equation.
From the balanced equation: 2 mol I₂ produce 5 mol ICl
So, moles of I₂ needed = (2/5) × moles of ICl
= (2/5) × 0.1762 ≈ 0.07048 mol

Step 4: Convert moles of I₂ to grams.
Molar mass of I₂ = 2 × 126.90 = 253.80 g/mol
Grams of I₂ = 0.07048 mol × 253.80 g/mol ≈ 17.89 g

✔ Final Answer for Problem 1: 17.9 grams of iodine

---

Problem 2:

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

Given: 11.4 grams of KNO₂
Find: moles and grams of KMnO₄ needed.

Step 1: Molar mass of KNO₂
K = 39.10, N = 14.01, O₂ = 32.00 → Total = 39.10 + 14.01 + 32.00 = 85.11 g/mol

Step 2: Moles of KNO₂ = 11.4 g / 85.11 g/mol ≈ 0.1339 mol

Step 3: Mole ratio — from equation: 5 mol KNO₂ require 2 mol KMnO₄
So, moles of KMnO₄ = (2/5) × 0.1339 ≈ 0.05356 mol

Step 4: Molar mass of KMnO₄
K = 39.10, Mn = 54.94, O₄ = 64.00 → Total = 39.10 + 54.94 + 64.00 = 158.04 g/mol

Grams of KMnO₄ = 0.05356 mol × 158.04 g/mol ≈ 8.465 g

✔ Final Answer for Problem 2: 0.0536 moles and 8.47 grams of KMnO₄

---

Problem 3:

Equation:
4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

Given: 56.8 grams of ammonia (NH₃)
Find: moles and grams of oxygen (O₂) needed.

Step 1: Molar mass of NH₃
N = 14.01, H₃ = 3.03 → Total = 17.04 g/mol

Step 2: Moles of NH₃ = 56.8 g / 17.04 g/mol ≈ 3.333 mol

Step 3: Mole ratio — 4 mol NH₃ require 5 mol O₂
So, moles of O₂ = (5/4) × 3.333 ≈ 4.166 mol

Step 4: Molar mass of O₂ = 32.00 g/mol
Grams of O₂ = 4.166 mol × 32.00 g/mol ≈ 133.3 g

✔ Final Answer for Problem 3: 4.17 moles and 133 grams of oxygen

---

Problem 4:

Equation:
NaIO₃ + 6 HI → 3 I₂ + NaI + 3 H₂O

Given: 16.4 grams of NaIO₃
Find: moles and grams of I₂ produced.

Step 1: Molar mass of NaIO₃
Na = 22.99, I = 126.90, O₃ = 48.00 → Total = 22.99 + 126.90 + 48.00 = 197.89 g/mol

Step 2: Moles of NaIO₃ = 16.4 g / 197.89 g/mol ≈ 0.08287 mol

Step 3: Mole ratio — 1 mol NaIO₃ produces 3 mol I₂
So, moles of I₂ = 3 × 0.08287 ≈ 0.2486 mol

Step 4: Molar mass of I₂ = 253.80 g/mol (from Problem 1)
Grams of I₂ = 0.2486 mol × 253.80 g/mol ≈ 63.10 g

✔ Final Answer for Problem 4: 0.249 moles and 63.1 grams of iodine

---

Final Answer:
1. 17.9 grams of iodine
2. 0.0536 moles and 8.47 grams of KMnO₄
3. 4.17 moles and 133 grams of oxygen
4. 0.249 moles and 63.1 grams of iodine