🎯⭐ INTERACTIVE LESSON

Stoichiometry and Limiting Reactants

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Stoichiometry and Limiting Reactants - Complete Interactive Lesson

Part 1: Mole Ratios

⚖️ Mole Ratios

Part 1 of 7 — The Foundation of Stoichiometry

Stoichiometry is the math of chemistry — it lets you calculate how much of each substance is involved in a reaction. The foundation of all stoichiometric calculations is the mole ratio, which comes directly from the coefficients in a balanced equation.

📌 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 H₂ react with 1 molecule of O₂ to produce 2 molecules of H₂O
2 moles of H₂ react with 1 mole of O₂ to produce 2 moles of H₂O

Key Insight

Coefficients give mole ratios, not mass ratios. The ratio 2:1:2 means:

2 mol H₂ : 1 mol O₂ : 2 mol H₂O

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

⚖️ Mole-to-Mole Conversions

The Simplest Stoichiometry Problem

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

Worked Example

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

Mole Ratio Concept Quiz 🎯

Mole-to-Mole Calculation Drill 🧮

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

Mole Ratio Concepts — Fill in the Blanks 🔽

Exit Quiz — Mole Ratios ✅

Part 2: Mass-to-Mass Calculations

🔬 Mass-to-Mass Stoichiometry

Part 2 of 7 — Converting Between Grams

In the lab, you weigh substances in grams, not moles. Mass-to-mass stoichiometry lets you convert grams of one substance to grams of another using the balanced equation. The key is the three-step bridge: grams → moles → moles → grams.

⚖️ The Stoichiometry Roadmap

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

Part 3: Limiting Reactant

🧪 Limiting Reactant

Part 3 of 7 — Which Reactant Runs Out First?

In real chemistry, reactants are rarely present in perfect stoichiometric proportions. One reactant will be used up first, stopping the reaction. That reactant is the limiting reactant — it limits how much product can form. The other reactant is in excess.

📌 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 :

Part 4: Excess Reactant & Theoretical Yield

📊 Theoretical, Actual, and Percent Yield

Part 4 of 7 — How Efficient Is Your Reaction?

In the real world, reactions rarely produce as much product as calculations predict. The percent yield measures how efficient a reaction actually is compared to the theoretical maximum.

📂 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:

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

Part 5: Percent Yield

🧫 Solution Stoichiometry

Part 5 of 7 — Using Molarity in Stoichiometry

Many reactions take place in aqueous solution. Instead of weighing solids, you measure volumes of solutions with known molarities. The key relationship is:

$M \times V = n \quad \text{(moles = molarity × volume in liters)}$

🔄 Molarity Review

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

$M$

Part 6: Problem-Solving Workshop

🛠️ Problem-Solving Workshop

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

This workshop brings together everything: mass-to-mass conversions, limiting reactants, excess calculations, and percent yield — all in one problem. These are the types of problems you'll see on the AP exam.

🛠️ The Complete Problem-Solving Strategy

For Multi-Step Stoichiometry Problems

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

Master Formula Chain

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

Part 7: Synthesis & AP Review

🎓 Synthesis & AP Review

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

This final part challenges you with AP-level stoichiometry problems that combine every concept: mole ratios, mass-to-mass conversions, limiting reactants, percent yield, and solution stoichiometry. Master these, and you're ready for the AP exam.

⚖️ Complete Stoichiometry Toolkit

Concept	Key Formula
Moles from grams	$n = m / M$
Moles from solution	$n = M_{\text{soln}} \times V$

All possible mole ratios:

Ratio	Value
N₂ to H₂	$\frac{1 \text{ mol N}_2}{3 \text{ mol H}_2}$ or $\frac{3 \text{ mol H}_2}{1 \text{ mol N}_2}$
N₂ to NH₃	$\frac{1 \text{ mol N}_2}{2 \text{ mol NH}_3}$ or
H₂ to NH₃	$\frac{3 \text{ mol H}_2}{2 \text{ mol NH}_3}$ or

Pick the ratio that cancels the given unit and introduces the desired unit.

If you know moles of N₂ and want moles of NH₃:

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

How many moles of NH₃ are produced from 5.0 mol N₂?

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

How many moles of H₂ are needed to react with 4.0 mol N₂?

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

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

1) How many moles of O₂ are needed to react with 5.0 mol C₂H₆? (to 3 significant figures)

2) How many moles of CO₂ are produced from 5.0 mol C₂H₆? (to 3 significant figures)

3) How many moles of H₂O are produced from 3.5 mol O₂? (to 3 significant figures)

Step	Conversion	Tool Used
1	Grams A → Moles A	Divide by molar mass of A
2	Moles A → Moles B	Multiply by mole ratio
3	Moles B → Grams B	Multiply by molar mass of B

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

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

🧪 Worked Example 1

Problem: How many grams of water are produced from burning 32.0 g of methane?

$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$

Molar masses: CH₄ = 16.04 g/mol, H₂O = 18.02 g/mol

Step 1: Grams CH₄ → Moles CH₄

$n_{ ext{CH}_4} = 32.0 ; cancel{ ext{g CH}_4} imes rac{1 ext{ mol CH}_4}{16.04 ; cancel{ ext{g CH}_4}} = 1.995 ext{ mol CH}_4$

Step 2: Moles CH₄ → Moles H₂O

$n_{ ext{H}_2 ext{O}} = 1.995 ; cancel{ ext{mol CH}_4} imes rac{2 ext{ mol H}_2 ext{O}}{1 ; cancel{ ext{mol CH}_4}} = 3.990 ext{ mol H}_2 ext{O}$

Step 3: Moles H₂O → Grams H₂O

$m_{ ext{H}_2 ext{O}} = 3.990 ; cancel{ ext{mol H}_2 ext{O}} imes rac{18.02 ext{ g H}_2 ext{O}}{1 ; cancel{ ext{mol H}_2 ext{O}}} = 71.9 ext{ g H}_2 ext{O}$

Answer: 71.9 g of H₂O

🧪 Worked Example 2

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

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

Molar masses: Al = 26.98 g/mol, Al₂O₃ = 101.96 g/mol

Step 1: Grams Al₂O₃ → Moles Al₂O₃

$n_{ ext{Al}_2 ext{O}_3} = 51.0 ; cancel{ ext{g Al}_2 ext{O}_3} imes rac{1 ext{ mol Al}_2 ext{O}_3}{101.96 ; cancel{ ext{g Al}_2 ext{O}_3}} = 0.5002 ext{ mol Al}_2 ext{O}_3$

Step 2: Moles Al₂O₃ → Moles Al

$n_{ ext{Al}} = 0.5002 ; cancel{ ext{mol Al}_2 ext{O}_3} imes rac{4 ext{ mol Al}}{2 ; cancel{ ext{mol Al}_2 ext{O}_3}} = 1.000 ext{ mol Al}$

Step 3: Moles Al → Grams Al

$m_{ ext{Al}} = 1.000 ; cancel{ ext{mol Al}} imes rac{26.98 ext{ g Al}}{1 ; cancel{ ext{mol Al}}} = 27.0 ext{ g Al}$

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: Fe₂O₃ = 159.7 g/mol, CO = 28.01 g/mol, Fe = 55.85 g/mol, CO₂ = 44.01 g/mol

1) How many grams of Fe are produced from 159.7 g of Fe₂O₃? (to 3 significant figures)

2) How many grams of CO are needed to react with 79.85 g of Fe₂O₃? (to 3 significant figures)

3) How many grams of CO₂ 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 ✅

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

The limiting reactant 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

Convert each reactant's given amount to moles
Use the mole ratio to calculate how much product each reactant could produce
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 H₂ and 2.0 mol O₂. Which is limiting?

From H₂: $3.0 \text{ mol H}_2 \times \frac{2 \text{ mol H}_2\text{O}}{2 \text{ mol H}_2} = 3.0 \text{ mol H}_2\text{O}$

From O₂: $2.0 \text{ mol O}_2 \times \frac{2 \text{ mol H}_2\text{O}}{1 \text{ mol O}_2} = 4.0 \text{ mol H}_2\text{O}$

H₂ produces less → H₂ is the limiting reactant

Maximum H₂O produced = 3.0 mol (from the limiting reactant)

🔍 Finding the Excess Amount

After identifying the limiting reactant, you can find how much excess reactant remains.

Continuing the Example

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

H₂ is limiting (3.0 mol). How much O₂ is left over?

Step 1: How much O₂ 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 O₂ remains?

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

Summary Table

Substance	Initial	Consumed	Remaining
H₂ (limiting)	3.0 mol	3.0 mol	0 mol
O₂ (excess)	2.0 mol	1.5 mol	0.5 mol
H₂O (product)	0 mol	—	3.0 mol

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 N₂ and 5.0 mol H₂.

1) How many moles of NH₃ could N₂ produce? (to 3 significant figures)

2) How many moles of NH₃ could H₂ produce? (to 3 significant figures)

3) Which is limiting? Type N2 or H2.

Limiting Reactant Concepts — Fill in the Blanks 🔽

Exit Quiz — Limiting Reactant ✅

Fact	Detail
Percent yield is always ≤ 100%	You can't create more product than the theoretical maximum
100% yield is ideal but rare	Side reactions, incomplete reactions, and losses reduce yield
Percent yield is unitless	It's a ratio — grams/grams cancels out

🤔 Why Is Actual Yield Less Than Theoretical?

Common Reasons for Reduced Yield

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

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

🧪 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 Cl₂. The student obtains 200.0 g of AlCl₃ ( $M = 133.34$ ). What is the percent yield?

Step 1: Find Theoretical Yield

Moles Al: $54.0 / 26.98 = 2.001$ mol

Moles AlCl₃ (theoretical): $2.001 \times \frac{2}{2} = 2.001$ mol

Grams AlCl₃ (theoretical): $2.001 \times 133.34 = 266.8$ g

Step 2: Calculate Percent Yield

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

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 ✅

M = \frac{n}{V} or n = M \times V

$M$ = molarity (mol/L)
$n$ = moles of solute
$V$ = volume of solution in liters

Unit Conversion Reminder

1 L = 1000 mL
Always convert mL to L before using the formula!
Example: 250 mL = 0.250 L

Problem: How many moles of NaOH are in 500 mL of 0.200 M NaOH?

$n = M \times V = 0.200 \times 0.500 = 0.100 \text{ mol NaOH}$

🧪 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

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

Worked Example

How many mL of 0.100 M AgNO₃ 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 AgNO₃ = 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 AgNO₃

🧪 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

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

For a 1:1 acid-base reaction:

$M_A \times V_A = M_B \times V_B$

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 HNO₃ (1:1 ratio). What is the molarity of HNO₃? (to 3 significant figures)

Solution Stoichiometry Concepts — Fill in the Blanks 🔽

Exit Quiz — Solution Stoichiometry ✅

grams A \div M_{A} mol A ratio mol product \times M_{P} theoretical yield (g) \times %/100 actual yield (g)

🧪 Comprehensive Worked Example

Problem: In the reaction below, 50.0 g of Fe₂O₃ ( $M = 159.7$ ) 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

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

Step 1: Convert to Moles

Moles Fe₂O₃: $50.0 \; \cancel{\text{g Fe}_2\text{O}_3} \times \frac{1 \text{ mol Fe}_2\text{O}_3}{159.7 \; \cancel{\text{g Fe}_2\text{O}_3}} = 0.3131$ mol Fe₂O₃

Step 2: Find Limiting Reactant

From Fe₂O₃: $0.3131 \; \cancel{\text{mol Fe}_2\text{O}_3} \times \frac{2 \text{ mol Fe}}{1 \; \cancel{\text{mol Fe}_2\text{O}_3}} = 0.6262$ mol Fe

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

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, g/mol

A reaction starts with 28.02 g of N₂ and 8.064 g of H₂. Percent yield = 85%.

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

2) What is the theoretical yield of NH₃ in grams? (to 3 significant figures)

3) What is the actual yield of NH₃ in grams? (to 3 significant figures)

Workshop Review — Fill in the Blanks 🔽

Exit Quiz — Multi-Step Stoichiometry ✅

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

📌 AP-Style Problem Walkthrough

Problem: A student reacts 25.0 mL of 0.400 M Pb(NO₃)₂ with 35.0 mL of 0.300 M KI. Find the mass of PbI₂ precipitate formed.

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

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

Step 1: Find Moles

mol Pb(NO₃)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(NO₃)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 PbI₂

Step 3: Mass of PbI₂

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

Key Insight

Even though we had fewer moles of Pb(NO₃)₂ initially (0.0100 vs 0.0105), KI was limiting because the reaction requires 2 mol KI per 1 mol Pb(NO₃)₂.

AP-Style Questions — Set 1 🎯

Comprehensive Stoichiometry Drill 🧮

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

$M_{\text{Al}} = 26.98$ , $M_{\text{HCl}} = 36.46$ , ,

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 AlCl₃ in grams? (to 3 significant figures)

3) What is the actual yield of AlCl₃ in grams? (to 3 significant figures)

AP-Style Questions — Set 2 🔬

Comprehensive Review — Fill in the Blanks 🔽

Final Exit Quiz — Stoichiometry Mastery ✅

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

\frac{2 \text{ mol NH}_3}{3 \text{ mol H}_2}

101.96g Al2O31mol Al2O3×

159.7g Fe2O31 mol Fe2O3=

Moles Al:

30.0 \; \cancel{\text{g Al}} \times \frac{1 \text{ mol Al}}{26.98 \; \cancel{\text{g Al}}} = 1.112

mol Al

From Al:

1.112 \; \cancel{\text{mol Al}} \times \frac{2 \text{ mol Fe}}{2 \; \cancel{\text{mol Al}}} = 1.112

mol Fe

Fe₂O₃ produces less → Fe₂O₃ is limiting

M_{\text{NH}_3} = 17.03

1mol Pb(NO3)21 mol PbI2=

From KI:

0.0105 \; \cancel{\text{mol KI}} \times \frac{1 \text{ mol PbI}_2}{2 \; \cancel{\text{mol KI}}} = 0.00525

mol PbI₂

KI produces less → KI is limiting

1mol PbI2461.0 g PbI2

M_{\text{AlCl}_3} = 133.34

M_{\text{H}_2} = 2.016

Stoichiometry and Limiting Reactants

Stoichiometry and Limiting Reactants - Complete Interactive Lesson

Part 1: Mole Ratios

⚖️ Mole Ratios

📌 Coefficients Tell the Story

Key Insight

✍️ Writing Mole Ratios

Example: N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3N2​+3H2​→2NH

⚖️ Mole-to-Mole Conversions

The Simplest Stoichiometry Problem

Worked Example

Part 2: Mass-to-Mass Calculations

🔬 Mass-to-Mass Stoichiometry

⚖️ The Stoichiometry Roadmap

Part 3: Limiting Reactant

🧪 Limiting Reactant

📌 The Sandwich Analogy

Part 4: Excess Reactant & Theoretical Yield

📊 Theoretical, Actual, and Percent Yield

📂 Three Types of Yield

Theoretical Yield

Actual Yield

Percent Yield

Part 5: Percent Yield

🧫 Solution Stoichiometry

🔄 Molarity Review

Part 6: Problem-Solving Workshop

🛠️ Problem-Solving Workshop

🛠️ The Complete Problem-Solving Strategy

For Multi-Step Stoichiometry Problems

Master Formula Chain

Part 7: Synthesis & AP Review

🎓 Synthesis & AP Review

⚖️ Complete Stoichiometry Toolkit

How to Choose

Another Example

General Formula

The Three Steps

Combined Formula

Key Point

🧪 Worked Example 1

Step 1: Grams CH₄ → Moles CH₄

Step 2: Moles CH₄ → Moles H₂O

Step 3: Moles H₂O → Grams H₂O

Answer: 71.9 g of H₂O

🧪 Worked Example 2

Step 1: Grams Al₂O₃ → Moles Al₂O₃

Step 2: Moles Al₂O₃ → Moles Al

Step 3: Moles Al → Grams Al

Answer: 27.0 g of Al

One-Line Setup

The Principle

⚖️ Finding the Limiting Reactant

Method: Compare Moles of Product Each Reactant Can Produce

Worked Example

🔍 Finding the Excess Amount

Continuing the Example

Summary Table

Key Facts

🤔 Why Is Actual Yield Less Than Theoretical?

Common Reasons for Reduced Yield

In Practice

🧪 Worked Example

Step 1: Find Theoretical Yield

Step 2: Calculate Percent Yield

Answer: 75.0%

Unit Conversion Reminder

Example

🧪 Solution Stoichiometry Roadmap

The Steps

Worked Example

🧪 Titration — A Key Application

At the Equivalence Point

Worked Example

For Non-1:1 Ratios

🧪 Comprehensive Worked Example

Step 1: Convert to Moles

Step 2: Find Limiting Reactant

Step 3: Theoretical Yield

Step 4: Actual Yield

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