Let’s solve each problem step by step.
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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:
grams of I₂ needed.
Step 1: Find molar mass of ICl.
Iodine (I) = 126.90 g/mol
Chlorine (Cl) = 35.45 g/mol
→ 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 equation:
2 mol I₂ → 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
Mass of I₂ = 0.07048 mol × 253.80 g/mol ≈
17.89 grams
✔ Check:
28.6 g ICl → 0.1762 mol → requires 0.07048 mol I₂ → 17.89 g → makes sense.
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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 =
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 : 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 =
158.04 g/mol
Grams of KMnO₄ = 0.05356 mol × 158.04 g/mol ≈
8.465 grams
✔ Check:
11.4 g KNO₂ → ~0.134 mol → needs ~0.0536 mol KMnO₄ → ~8.47 g → reasonable.
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Problem 3:
Equation:
4 NH₃ + 5 O₂ → 4 NO + 6 H₂O
Given:
56.8 grams of NH₃
Find:
moles and grams of 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 — from equation:
4 mol NH₃ : 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 grams
✔ Check:
56.8 g NH₃ → ~3.33 mol → needs ~4.17 mol O₂ → ~133 g → correct.
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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 =
197.89 g/mol
Step 2: Moles of NaIO₃ = 16.4 g / 197.89 g/mol ≈
0.08287 mol
Step 3: Mole ratio — from equation:
1 mol NaIO₃ → 3 mol I₂
So, moles of I₂ = 3 × 0.08287 ≈
0.2486 mol
Step 4: Molar mass of I₂ = 253.80 g/mol
Grams of I₂ = 0.2486 mol × 253.80 g/mol ≈
63.10 grams
✔ Check:
16.4 g NaIO₃ → ~0.083 mol → produces ~0.249 mol I₂ → ~63.1 g → correct.
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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
Parent Tip: Review the logic above to help your child master the concept of stoichiometry problems worksheet.
