15.6 Limiting reactant · Stoichiometry · @everexams
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15.6 Limiting Reactant · Stoichiometry

with quick check 15.8, 10 MCQ quiz & lesson planner · @everexams

📘 Lesson plan & activities

In a chemical reaction, reactants are not always mixed according to stoichiometric ratio. Often one reactant is used in large excess to ensure expensive reactant is completely used.

🧪 STEPS TO FIND LIMITING REACTANT
  1. 1. Calculate moles of each reactant from given amounts.
  2. 2. Calculate moles of product (e.g., CO₂) that would form from each reactant, using reaction stoichiometry.
  3. 3. The reactant that yields the least amount of product is the limiting reactant – it controls product amount.

⚗️ Example 15.16 (glucose + oxygen)

C₆H₁₂O₆ (s) + 6O₂ (g) → 6CO₂ (g) + 6H₂O (l)

Given: 25 g glucose (molar mass 180) & 40 g oxygen (32).

Moles CO₂ from glucose: 1 mol glucose → 6 mol CO₂ → 0.139 × 6 = 0.834 mol
Moles CO₂ from oxygen: 6 mol O₂ → 6 mol CO₂ → 1.25 mol O₂ → 1.25 mol CO₂ (since 6/6 ratio).

✔ Glucose gives 0.834 mol CO₂ (less than 1.25 from O₂).
➡ limiting reactant = glucose

Amount CO₂ produced = 0.834 × 44 = 36.7 g (approx).

🧠 Memorization tip: “Limiting = Least product moles”. Compare product moles from each reactant; smallest is the boss. Use BCA table: Before, Change, After.

🔁 Quick Check 15.8 (from text)

If you react 100 g of chlorine gas with 500 g of KI, which reactant will be the limiting reactant?

Cl₂ (g) + 2KI (aq) → 2KCl (aq) + I₂ (g) [atomic masses: Cl=35.5, K=39, I=127]

✅ Solution: Molar mass Cl₂ = 71 g/mol → moles Cl₂ = 100/71 = 1.41 mol.
Molar mass KI = 166 g/mol → moles KI = 500/166 = 3.01 mol.
Reaction: 1 mol Cl₂ requires 2 mol KI. For 1.41 mol Cl₂ need 2.82 mol KI (available 3.01, so KI excess).
Limiting reactant = Cl₂

⭐ Tip: always compare required vs available.

📝 10 MCQs quiz · limiting reactant

Instructions: click on answer. Correct turns ✔green, wrong turns ✘ red. Use submit to see score & key.

Score: — / 10
🔑 Answer key will appear after submit.

📅 Lesson Planner & Activity

Track your study & apply with a challenge problem.

✅ Today’s stoichiometry goals

⚡ Activity: Find limiting reactant

2 Na (s) + Cl₂ (g) → 2 NaCl (s)
Given: 46 g sodium (Na = 23) and 71 g chlorine (Cl₂ = 71).

Which reactant limits? How many g NaCl formed?

⭐ After planner, review your quiz errors. Use notebook for calculations.

📚 Good guidelines for students

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