Stoichiometry and Limiting Reactants

⚖️ Mole Ratios

Part 1 of 7 — The Foundation of Stoichiometry

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 1

• Understanding the core concepts covered in Part 1
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

📌 Coefficients Tell the Story

In the balanced equation:

$2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$

The coefficients tell us that:

• 2 molecules of $H_{2}$ react with 1 molecule of $O_{2}$ to produce 2 molecules of $H_{2}O$
• 2 moles of $H_{2}$ react with 1 mole of $O_{2}$ to produce 2 moles of $H_{2}O$

🔑 Key Concept: Coefficients give mole ratios, not mass ratios. The ratio 2:1:2 means 2 mol $H_{2}$ : 1 mol $O_{2}$ : 2 mol $H_{2}O$. This ratio is the conversion factor for all stoichiometric calculations.

✍️ Writing Mole Ratios

From any balanced equation, you can write a mole ratio between any two substances.

Example: $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$

All possible mole ratios:

💡 Tip: Pick the ratio that cancels the given unit and introduces the desired unit.

If you know moles of $N_{2}$ and want moles of $NH_{3}$:

$\text{mol NH}_3 = \text{mol N}_2 \times \frac{2 \text{ mol NH}_3}{1 \text{ mol N}_2}$

⚖️ Mole-to-Mole Conversions

The Simplest Stoichiometry Problem

Given moles of one substance, find moles of another using the mole ratio.

Problem: Given $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$, how many moles of $NH_{3}$ are produced from 5.0 mol $N_{2}$?

Solution:

$\text{mol NH}_3 = 5.0 \text{ mol N}_2 \times \frac{2 \text{ mol NH}_3}{1 \text{ mol N}_2} = 10.0 \text{ mol NH}_3$

Problem: How many moles of $H_{2}$ are needed to react with 4.0 mol $N_{2}$?

Solution:

$\text{mol H}_2 = 4.0 \text{ mol N}_2 \times \frac{3 \text{ mol H}_2}{1 \text{ mol N}_2} = 12.0 \text{ mol H}_2$

General Formula

$\boxed{\text{mol of B} = \text{mol of A} \times \frac{\text{coefficient of B}}{\text{coefficient of A}}}$

Mole Ratio Concept Quiz 🎯

Mole-to-Mole Calculation Drill 🧮

Use the equation: $2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O}$

1) How many moles of $O_{2}$ are needed to react with 5.0 mol $C_{2}H_{6}$? (to 3 significant figures)

2) How many moles of $CO_{2}$ are produced from 5.0 mol $C_{2}H_{6}$? (to 3 significant figures)

3) How many moles of $H_{2}O$ are produced from 3.5 mol $O_{2}$? (to 3 significant figures)

Mole Ratio Concepts — Fill in the Blanks 🔽

Exit Quiz — Mole Ratios ✅

🔬 Mass-to-Mass Stoichiometry

Part 2 of 7 — Converting Between Grams

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 2

• Understanding the core concepts covered in Part 2
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

⚖️ The Stoichiometry Roadmap

🔑 Key Concept: The Mole Road Map — every stoichiometry problem follows this path:

$\text{grams A} \xrightarrow{\div M_A} \text{moles A} \xrightarrow{\text{mole ratio}} \text{moles B} \xrightarrow{\times M_B} \text{grams B}$

The Three Steps

Combined Formula

$\boxed{\text{grams B} = \text{grams A} \times \frac{1}{M_A} \times \frac{\text{coeff B}}{\text{coeff A}} \times M_B}$

⚠️ Warning: You cannot skip steps! You must go through moles — there is no direct grams-to-grams conversion factor from the balanced equation.

💡 Tip: Set up the entire calculation as one line of dimensional analysis before computing — verify that all units cancel before reaching for your calculator.

🧪 Worked Example 1

Problem: How many grams of water are produced from burning 32.0 g of methane? Solution: $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$

Molar masses: $CH_{4}$ = 16.04 g/mol, $H_{2}O$ = 18.02 g/mol

Step 1: Grams $CH_{4}$ → Moles $CH_{4}$

$n_{\text{CH}_4} = 32.0 ; \cancel{\text{g CH}_4} \times \frac{1 \text{ mol CH}_4}{16.04 ; \cancel{\text{g CH}_4}} = 1.995 \text{ mol CH}_4$

Step 2: Moles $CH_{4}$ → Moles $H_{2}O$

$n_{\text{H}_2\text{O}} = 1.995 ; \cancel{\text{mol CH}_4} \times \frac{2 \text{ mol H}_2\text{O}}{1 ; \cancel{\text{mol CH}_4}} = 3.990 \text{ mol H}_2\text{O}$

Step 3: Moles $H_{2}O$ → Grams $H_{2}O$

$m_{\text{H}_2\text{O}} = 3.990 ; \cancel{\text{mol H}_2\text{O}} \times \frac{18.02 \text{ g H}_2\text{O}}{1 ; \cancel{\text{mol H}_2\text{O}}} = 71.9 \text{ g H}_2\text{O}$

Answer: 71.9 g of $H_{2}O$

🧪 Worked Example 2

Problem: How many grams of aluminum are needed to produce 51.0 g of aluminum oxide?

Solution:

$4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3$

Molar masses: Al = 26.98 g/mol, $Al_{2}O_{3}$ = 101.96 g/mol

Step 1: Grams $Al_{2}O_{3}$ → Moles $Al_{2}O_{3}$

$n_{\text{Al}_2\text{O}_3} = 51.0 ; \cancel{\text{g Al}_2\text{O}_3} \times \frac{1 \text{ mol Al}_2\text{O}_3}{101.96 ; \cancel{\text{g Al}_2\text{O}_3}} = 0.5002 \text{ mol Al}_2\text{O}_3$

Step 2: Moles $Al_{2}O_{3}$ → Moles Al

$n_{\text{Al}} = 0.5002 ; \cancel{\text{mol Al}_2\text{O}_3} \times \frac{4 \text{ mol Al}}{2 ; \cancel{\text{mol Al}_2\text{O}_3}} = 1.000 \text{ mol Al}$

Step 3: Moles Al → Grams Al

$m_{\text{Al}} = 1.000 ; \cancel{\text{mol Al}} \times \frac{26.98 \text{ g Al}}{1 ; \cancel{\text{mol Al}}} = 27.0 \text{ g Al}$

Answer: 27.0 g of Al

One-Line Setup

$51.0 \; \cancel{\text{g Al}_2\text{O}_3} \times \frac{1 \; \cancel{\text{mol Al}_2\text{O}_3}}{101.96 \; \cancel{\text{g Al}_2\text{O}_3}} \times \frac{4 \; \cancel{\text{mol Al}}}{2 \; \cancel{\text{mol Al}_2\text{O}_3}} \times \frac{26.98 \text{ g Al}}{1 \; \cancel{\text{mol Al}}} = 27.0 \text{ g Al}$

Mass-to-Mass Stoichiometry Quiz 🎯

Mass-to-Mass Calculation Drill 🧮

Use the equation: $\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$

Molar masses: $Fe_{2}O_{3}$ = 159.7 g/mol, CO = 28.01 g/mol, Fe = 55.85 g/mol, $CO_{2}$ = 44.01 g/mol

1) How many grams of Fe are produced from 159.7 g of $Fe_{2}O_{3}$? (to 3 significant figures)

2) How many grams of CO are needed to react with 79.85 g of $Fe_{2}O_{3}$? (to 3 significant figures)

3) How many grams of $CO_{2}$ are produced from 84.03 g of CO? (to 3 significant figures)

Mass-to-Mass Concepts — Fill in the Blanks 🔽

Exit Quiz — Mass-to-Mass Stoichiometry ✅

🧪 Limiting Reactant

Part 3 of 7 — Which Reactant Runs Out First?

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 3

• Understanding the core concepts covered in Part 3
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

📌 The Sandwich Analogy

To make 1 sandwich, you need: 2 slices of bread + 1 slice of cheese

$2\text{B} + \text{C} \rightarrow \text{B}_2\text{C}$

If you have 10 slices of bread and 7 slices of cheese:

• Bread can make: $10/2 = 5$ sandwiches
• Cheese can make: $7/1 = 7$ sandwiches
• You run out of bread first → bread is the limiting reactant
• You can only make 5 sandwiches
• Leftover cheese: $7 - 5 = 2$ slices → cheese is in excess

🔑 Key Concept: The limiting reactant is the one that runs out first — it determines the maximum amount of product. The excess reactant is left over.

⚖️ Finding the Limiting Reactant

Method: Compare Moles of Product Each Reactant Can Produce

1. Convert each reactant's given amount to moles
2. Use the mole ratio to calculate how much product each reactant could produce
3. The reactant that produces the lesser amount of product is the limiting reactant

Worked Example

$2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$

Given: 3.0 mol $H_{2}$ and 2.0 mol $O_{2}$. Which is limiting?

From $H_{2}$: $3.0 \text{ mol H}_2 \times \frac{2 \text{ mol H}_2\text{O}}{2 \text{ mol H}_2} = 3.0 \text{ mol H}_2\text{O}$

From $O_{2}$: $2.0 \text{ mol O}_2 \times \frac{2 \text{ mol H}_2\text{O}}{1 \text{ mol O}_2} = 4.0 \text{ mol H}_2\text{O}$

$H_{2}$ produces less → $H_{2}$ is the limiting reactant

Maximum $H_{2}O$ produced = 3.0 mol (from the limiting reactant)

💡 Tip: Always calculate how much product each reactant could produce separately. The one that gives less product is the limiting reactant — it’s not necessarily the one with fewer moles!

🔍 Finding the Excess Amount

💡 Tip: To find leftover excess, use the limiting reactant and the mole ratio to calculate how much excess was consumed, then subtract from the initial amount.

Continuing the Example

$2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$

$H_{2}$ is limiting (3.0 mol). How much $O_{2}$ is left over?

Step 1: How much $O_{2}$ is consumed?

$\text{mol O}_2 \text{ consumed} = 3.0 \text{ mol H}_2 \times \frac{1 \text{ mol O}_2}{2 \text{ mol H}_2} = 1.5 \text{ mol O}_2$

Step 2: How much $O_{2}$ remains?

$\boxed{\text{excess O}_2 = 2.0 - 1.5 = 0.5 \text{ mol O}_2}$

Summary Table

Limiting Reactant Quiz 🎯

Limiting Reactant Calculations 🧮

Given: $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$

You start with 2.0 mol $N_{2}$ and 5.0 mol $H_{2}$.

1) How many moles of $NH_{3}$ could $N_{2}$ produce? (to 3 significant figures)

2) How many moles of $NH_{3}$ could $H_{2}$ produce? (to 3 significant figures)

3) Which is limiting? Type N2 or H2.

Limiting Reactant Concepts — Fill in the Blanks 🔽

Exit Quiz — Limiting Reactant ✅

📊 Theoretical, Actual, and Percent Yield

Part 4 of 7 — How Efficient Is Your Reaction?

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 4

• Understanding the core concepts covered in Part 4
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

📂 Three Types of Yield

Theoretical Yield

The maximum amount of product that could form based on stoichiometric calculations (assuming the limiting reactant is completely converted).

Actual Yield

The amount of product actually obtained in the lab (measured experimentally).

Percent Yield

The ratio of actual to theoretical yield, expressed as a percentage:

🔑 Key Concept: Percent yield measures reaction efficiency:

$\boxed{\% \text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100}$

Key Facts

🤔 Why Is Actual Yield Less Than Theoretical?

Common Reasons for Reduced Yield

1. Incomplete reactions — not all reactant converts to product
2. Side reactions — reactants form unintended products
3. Loss during transfer — product sticks to glassware, spatulas, filter paper
4. Impure reactants — some of the starting material isn't what you think
5. Equilibrium — reversible reactions don't go to completion
6. Evaporation — volatile products may escape

⚠️ Warning: A percent yield greater than 100% is physically impossible for pure product — it signals contamination, impurities, or measurement error.

In Practice

• Industrial processes aim for yields of 60–90%
• Pharmaceutical synthesis may involve many steps, each with <100% yield
• Multi-step synthesis: overall yield = product of individual yields
  • Example: 3 steps at 80% each → $0.80^3 = 0.512 = 51.2\%$ overall

💡 Tip: To find the required theoretical yield when you know the desired actual yield: Theoretical = Actual ÷ (% yield / 100).

🧪 Worked Example

Problem: In the reaction $2\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3$, a student starts with 54.0 g of Al ($M = 26.98$) and excess $Cl_{2}$. The student obtains 200.0 g of $AlCl_{3}$ ($M = 133.34$). What is the percent yield?

Step 1: Find Theoretical Yield

Moles Al: $54.0 / 26.98 = 2.001$ mol

Moles $AlCl_{3}$ (theoretical): $2.001 \times \frac{2}{2} = 2.001$ mol

Grams $AlCl_{3}$ (theoretical): $2.001 \times 133.34 = 266.8$ g

Step 2: Calculate Percent Yield

$\boxed{\% \text{ yield} = \frac{200.0}{266.8} \times 100 = 75.0\%}$

Answer: 75.0%

This means 75% of the theoretical product was actually recovered. The remaining 25% was lost to various factors.

Percent Yield Quiz 🎯

Percent Yield Calculations 🧮

1) Theoretical yield = 80.0 g, actual yield = 68.0 g. Percent yield = ? (to 3 significant figures)

2) A student calculates a theoretical yield of 120.0 g and achieves 92% yield. What mass of product was obtained? (to 3 significant figures)

3) A reaction produces 35.0 g of product. If the percent yield is 70.0%, what was the theoretical yield? (to 3 significant figures)

Yield Concepts — Fill in the Blanks 🔽

Exit Quiz — Percent Yield ✅

🧫 Solution Stoichiometry

Part 5 of 7 — Using Molarity in Stoichiometry

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 5

• Understanding the core concepts covered in Part 5
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

🔄 Molarity Review

Molarity ($M$) is the concentration of a solution in moles per liter:

$M = \frac{n}{V} \quad \text{or} \quad n = M \times V$

📌 Variable Reference

⚠️ Always convert mL to L before using the formula! (÷ 1000)

🧪 Worked Example

💡 Once you have moles from $n = M \times V$, follow the same mole ratio logic as any other stoichiometry problem.

🧪 Solution Stoichiometry Roadmap

$M_A \times V_A \xrightarrow{= n_A} \text{moles A} \xrightarrow{\text{mole ratio}} \text{moles B} \xrightarrow{\div M_B \text{ or } \times M_B} V_B \text{ or grams B}$

The Steps

1. Find moles of the known substance: $n = M \times V$
2. Use the mole ratio to find moles of the unknown
3. Convert to the desired unit (volume, grams, or molarity)

Worked Example

How many mL of 0.100 M $AgNO_{3}$ are needed to react completely with 25.0 mL of 0.200 M NaCl?

$\text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3$

Step 1: Moles NaCl = $0.200 \times 0.0250 = 0.00500$ mol

Step 2: Mole ratio is 1:1, so moles $AgNO_{3}$ = 0.00500 mol

Step 3: Volume = $n/M = 0.00500/0.100 = 0.0500$ L $= 50.0$ mL

Answer: 50.0 mL of 0.100 M $AgNO_{3}$

🧪 Titration — A Key Application

A titration is a lab technique where you add a solution of known concentration (the titrant) to a solution of unknown concentration until the reaction is complete (the equivalence point).

At the Equivalence Point

$n_{\text{acid}} \times \text{(acid-to-base ratio)} = n_{\text{base}}$

🔑 Key Concept: For a 1:1 acid-base reaction at the equivalence point:

$\boxed{M_A \times V_A = M_B \times V_B}$

Worked Example

A student titrates 25.0 mL of HCl of unknown concentration with 0.150 M NaOH. It takes 32.0 mL of NaOH to reach the equivalence point. Find $M_{\text{HCl}}$.

$\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \quad (1:1 \text{ ratio})$

$M_{\text{HCl}} \times 0.0250 = 0.150 \times 0.0320$

$M_{\text{HCl}} = \frac{0.150 \times 0.0320}{0.0250} = 0.192 \text{ M}$

For Non-1:1 Ratios

$\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$

Here: $M_A \times V_A \times 1 = M_B \times V_B \times \frac{1}{2}$, or simply find moles and use the ratio.

Solution Stoichiometry Quiz 🎯

Solution Stoichiometry Calculations 🧮

1) How many moles of KOH are in 250 mL of 0.400 M KOH? (to 3 significant figures)

2) In the reaction $\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$, how many mL of 0.250 M NaOH are needed to neutralize 50.0 mL of 0.100 M HCl? (to 3 significant figures)

3) A titration requires 28.5 mL of 0.200 M KOH to neutralize 25.0 mL of $HNO_{3}$ (1:1 ratio). What is the molarity of $HNO_{3}$? (to 3 significant figures)

Solution Stoichiometry Concepts — Fill in the Blanks 🔽

Exit Quiz — Solution Stoichiometry ✅

🛠️ Problem-Solving Workshop

Part 6 of 7 — Multi-Step Stoichiometry with Limiting Reactants and Yield

Practice Makes Perfect

This workshop features multi-step problems that mirror the AP Chemistry exam format. Each problem requires you to combine concepts from previous parts and show your work clearly.

🔑 Why this matters: The AP Chemistry exam rewards students who can apply concepts to unfamiliar problems — structured practice is the best preparation.

What You'll Master in Part 6

• Working through complete multi-step problems from start to finish
• Building problem-solving strategies you can apply on the AP exam
• Identifying which concepts to apply and in what order

🛠️ The Complete Problem-Solving Strategy

For Multi-Step Stoichiometry Problems

1. Write and balance the equation
2. Convert all given amounts to moles
3. Identify the limiting reactant (compare moles of product each can produce)
4. Calculate theoretical yield using the limiting reactant
5. Find excess remaining (if asked)
6. Apply percent yield (if given or asked)

🔑 Key Concept: Every stoichiometry problem is a series of conversions — grams ↔ moles ↔ moles ↔ grams — always chained through the mole ratio from a balanced equation.

Master Formula Chain

$\text{grams A} \xrightarrow{\div M_A} \text{mol A} \xrightarrow{\text{ratio}} \text{mol product} \xrightarrow{\times M_P} \text{theoretical yield (g)} \xrightarrow{\times \%/100} \text{actual yield (g)}$

⚠️ Warning: Never apply percent yield to the excess reactant — only to the product calculated from the limiting reactant.

🧪 Comprehensive Worked Example

Problem: In the reaction below, 50.0 g of $Fe_{2}O_{3}$ ($M = 159.7$) reacts with 30.0 g of Al ($M = 26.98$). The percent yield is 78%. Find: a) the limiting reactant b) the theoretical yield of Fe ($M = 55.85$) c) the actual yield of Fe d) the mass of excess reactant remaining

Solution:

$\text{Fe}_2\text{O}_3 + 2\text{Al} \rightarrow \text{Al}_2\text{O}_3 + 2\text{Fe}$

Step 1: Convert to Moles

• Moles $Fe_{2}O_{3}$: $50.0 \; \cancel{\text{g Fe}_2\text{O}_3} \times \frac{1 \text{ mol Fe}_2\text{O}_3}{159.7 \; \cancel{\text{g Fe}_2\text{O}_3}} = 0.3131$ mol $Fe_{2}O_{3}$
• Moles Al: $30.0 \; \cancel{\text{g Al}} \times \frac{1 \text{ mol Al}}{26.98 \; \cancel{\text{g Al}}} = 1.112$ mol Al

Step 2: Find Limiting Reactant

• From $Fe_{2}O_{3}$: $0.3131 \; \cancel{\text{mol Fe}_2\text{O}_3} \times \frac{2 \text{ mol Fe}}{1 \; \cancel{\text{mol Fe}_2\text{O}_3}} = 0.6262$ mol Fe
• From Al: $1.112 \; \cancel{\text{mol Al}} \times \frac{2 \text{ mol Fe}}{2 \; \cancel{\text{mol Al}}} = 1.112$ mol Fe
• $Fe_{2}O_{3}$ produces less → $Fe_{2}O_{3}$ is limiting

Step 3: Theoretical Yield

$m_{\text{Fe}} = 0.6262 \; \cancel{\text{mol Fe}} \times \frac{55.85 \text{ g Fe}}{1 \; \cancel{\text{mol Fe}}} = 35.0 \text{ g Fe}$

Step 4: Actual Yield

$\text{actual} = 35.0 \times 0.78 = 27.3 \text{ g Fe}$

Step 5: Excess Al Remaining

Al consumed: $0.3131 \; \cancel{\text{mol Fe}_2\text{O}_3} \times \frac{2 \text{ mol Al}}{1 \; \cancel{\text{mol Fe}_2\text{O}_3}} = 0.6262$ mol Al

Al remaining: $(1.112 - 0.6262) \; \cancel{\text{mol Al}} \times \frac{26.98 \text{ g Al}}{1 \; \cancel{\text{mol Al}}} = 13.1$ g Al

💡 Tip: Use an ICE-style table (Initial → Consumed → End) to organize your work and verify that the limiting reactant reaches exactly zero.

Multi-Step Stoichiometry Quiz 🎯

Multi-Step Calculation Drill 🧮

Given: $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$

$M_{\text{N}_2} = 28.02$ g/mol, $M_{\text{H}_2} = 2.016$ g/mol, $M_{\text{NH}_3} = 17.03$ g/mol

A reaction starts with 28.02 g of $N_{2}$ and 8.064 g of $H_{2}$. Percent yield = 85%.

1) Which is limiting? Type N2 or H2. (Hint: calculate mol product from each)

2) What is the theoretical yield of $NH_{3}$ in grams? (to 3 significant figures)

3) What is the actual yield of $NH_{3}$ in grams? (to 3 significant figures)

Workshop Review — Fill in the Blanks 🔽

Exit Quiz — Multi-Step Stoichiometry ✅

🎓 Synthesis & AP Review

Part 7 of 7 — Comprehensive Stoichiometry Problems & AP Exam Strategies

Bringing It All Together

This comprehensive review connects every concept from Parts 1–6 with AP-style problems. The questions are designed to mirror what you'll see on the actual exam — multi-step, multi-concept, and requiring clear written explanations.

🔑 Why this matters: AP Chemistry exam questions rarely test one concept in isolation — success requires connecting ideas across topics.

What You'll Master in Part 7

• Solving AP-style questions that integrate multiple concepts from this unit
• Writing clear, concise explanations using proper chemistry terminology
• Identifying and avoiding common AP exam traps and mistakes

⚖️ Complete Stoichiometry Toolkit

📌 AP Exam Tips

1. Show your work — AP graders want to see the setup, not just the answer
2. Use dimensional analysis — set up conversion factors so units cancel
3. Significant figures — match the least precise measurement
4. Label everything — always include units and chemical formulas
5. Check your answer — does the magnitude make sense?
6. Common mistake: forgetting to balance the equation before using mole ratios

🔑 Key Concept: Dimensional analysis is the universal approach — set up conversion factors so units cancel, whether working with mass, moles, volume, or molarity.

⚠️ Warning: On the AP exam, a common mistake is using subscripts instead of coefficients for mole ratios. Subscripts describe the formula; coefficients describe the reaction.

💡 Tip: On free-response questions, show every conversion factor and cancel units explicitly — this earns partial credit even if your arithmetic has an error.

📌 AP-Style Problem Walkthrough

Problem: A student reacts 25.0 mL of 0.400 M $Pb(NO_{3})_{2}$ with 35.0 mL of 0.300 M KI. Find the mass of $PbI_{2}$ precipitate formed.

Solution:

$\text{Pb(NO}_3)_2(\text{aq}) + 2\text{KI}(\text{aq}) \rightarrow \text{PbI}_2(\text{s}) + 2\text{KNO}_3(\text{aq})$

$M_{\text{PbI}_2} = 461.0$ g/mol

Step 1: Find Moles

• mol $Pb(NO_{3})_{2}$ = $0.400 \times 0.0250 = 0.0100$ mol
• mol KI = $0.300 \times 0.0350 = 0.0105$ mol

Step 2: Limiting Reactant

• From $Pb(NO_{3})_{2}$: $0.0100 \; \cancel{\text{mol Pb(NO}_3)_2} \times \frac{1 \text{ mol PbI}_2}{1 \; \cancel{\text{mol Pb(NO}_3)_2}} = 0.0100$ mol $PbI_{2}$
• From KI: $0.0105 \; \cancel{\text{mol KI}} \times \frac{1 \text{ mol PbI}_2}{2 \; \cancel{\text{mol KI}}} = 0.00525$ mol $PbI_{2}$
• KI produces less → KI is limiting

Step 3: Mass of $PbI_{2}$

$m = 0.00525 \; \cancel{\text{mol PbI}_2} \times \frac{461.0 \text{ g PbI}_2}{1 \; \cancel{\text{mol PbI}_2}} = 2.42 \text{ g PbI}_2$

Key Insight

Even though we had fewer moles of $Pb(NO_{3})_{2}$ initially (0.0100 vs 0.0105), KI was limiting because the reaction requires 2 mol KI per 1 mol $Pb(NO_{3})_{2}$.

AP-Style Questions — Set 1 🎯

Comprehensive Stoichiometry Drill 🧮

Given: $2\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2$

$M_{\text{Al}} = 26.98$, $M_{\text{HCl}} = 36.46$, $M_{\text{AlCl}_3} = 133.34$, $M_{\text{H}_2} = 2.016$

A student reacts 13.49 g of Al with 109.4 g of HCl. Percent yield = 90%.

1) The limiting reactant is which? Type Al or HCl.

2) What is the theoretical yield of $AlCl_{3}$ in grams? (to 3 significant figures)

3) What is the actual yield of $AlCl_{3}$ in grams? (to 3 significant figures)

AP-Style Questions — Set 2 🔬

Comprehensive Review — Fill in the Blanks 🔽

Final Exit Quiz — Stoichiometry Mastery ✅