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

I'll solve each stoichiometry problem step by step, showing all calculations.

Problem 1:
Equation: 2 I₂ + KIO₃ + 6 HCl → 5 ICl + KCl + 3 H₂O

Given: 28.6 grams of ICl
Find: grams of I₂ needed

Step 1: Calculate molar mass of ICl
- I = 126.90 g/mol
- 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 equation: 2 mol I₂ produces 5 mol ICl
So: moles of I₂ = (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
Mass of I₂ = 0.07048 mol × 253.80 g/mol = 17.89 g ≈ 17.9 g

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: Calculate molar mass of KNO₂
- K = 39.10 g/mol
- N = 14.01 g/mol
- O = 16.00 g/mol (×2 = 32.00)
- Molar mass of KNO₂ = 39.10 + 14.01 + 32.00 = 85.11 g/mol

Step 2: Convert grams of KNO₂ to moles
Moles of KNO₂ = 11.4 g ÷ 85.11 g/mol = 0.1339 mol

Step 3: Use mole ratio from equation
From the equation: 5 mol KNO₂ reacts with 2 mol KMnO₄
So: moles of KMnO₄ = (2/5) × 0.1339 = 0.05356 mol

Step 4: Calculate molar mass of KMnO₄
- K = 39.10 g/mol
- Mn = 54.94 g/mol
- O = 16.00 g/mol (×4 = 64.00)
- Molar mass of KMnO₄ = 39.10 + 54.94 + 64.00 = 158.04 g/mol

Step 5: Convert moles of KMnO₄ to grams
Mass of KMnO₄ = 0.05356 mol × 158.04 g/mol = 8.465 g ≈ 8.47 g

Problem 3:
Equation: 4 NH + 5 O₂ → 4 NO + 6 H₂O

Given: 56.8 grams of NH₃
Find: moles and grams of O₂ needed

Step 1: Calculate molar mass of NH₃
- N = 14.01 g/mol
- H = 1.01 g/mol (×3 = 3.03)
- Molar mass of NH = 14.01 + 3.03 = 17.04 g/mol

Step 2: Convert grams of NH₃ to moles
Moles of NH₃ = 56.8 g ÷ 17.04 g/mol = 3.333 mol

Step 3: Use mole ratio from equation
From the equation: 4 mol NH₃ reacts with 5 mol O₂
So: moles of O₂ = (5/4) × 3.333 = 4.166 mol

Step 4: Calculate molar mass of O₂
Molar mass of O₂ = 2 × 16.00 = 32.00 g/mol

Step 5: Convert moles of O₂ to grams
Mass of O₂ = 4.166 mol × 32.00 g/mol = 133.3 g ≈ 133 g

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: Calculate molar mass of NaIO₃
- Na = 22.99 g/mol
- I = 126.90 g/mol
- O = 16.00 g/mol (×3 = 48.00)
- Molar mass of NaIO₃ = 22.99 + 126.90 + 48.00 = 197.89 g/mol

Step 2: Convert grams of NaIO₃ to moles
Moles of NaIO₃ = 16.4 g 197.89 g/mol = 0.08287 mol

Step 3: Use mole ratio from equation
From the equation: 1 mol NaIO₃ produces 3 mol I₂
So: moles of I₂ = 3 × 0.08287 = 0.2486 mol

Step 4: Calculate molar mass of I₂
Molar mass of I₂ = 2 × 126.90 = 253.80 g/mol

Step 5: Convert moles of I₂ to grams
Mass of I₂ = 0.2486 mol × 253.80 g/mol = 63.09 g ≈ 63.1 g

Final Answer:
1. 17.9 grams of I₂
2. 0.0536 moles and 8.47 grams of KMnO₄
3. 4.17 moles and 133 grams of O₂
4. 0.249 moles and 63.1 grams of I₂