Stoichiometry and Limiting Reactants

Q: How can I study Stoichiometry and Limiting Reactants effectively?

Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 5 problems provided, checking solutions as you go. Regular review and active practice are key to retention.

Q: Is this Stoichiometry and Limiting Reactants study guide free?

Yes — all study notes, flashcards, and practice problems for Stoichiometry and Limiting Reactants on Study Mondo are free to access. No account is needed.

Q: What course covers Stoichiometry and Limiting Reactants?

Stoichiometry and Limiting Reactants is part of the AP Chemistry course on Study Mondo, specifically in the Chemical Reactions section. You can explore the full course for more related topics and practice resources.

Q: Are there practice problems for Stoichiometry and Limiting Reactants?

Yes, this page includes 5 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.

Master stoichiometric calculations, mole ratios, limiting reactants, theoretical yield, percent yield, and solution stoichiometry.

Written and reviewed by the Study Mondo Education TeamLast updated April 7, 2026

🎯⭐ INTERACTIVE LESSON

Try the Interactive Version!

Learn step-by-step with practice exercises built right in.

Start Interactive Lesson →

Stoichiometry and Limiting Reactants

Introduction to Stoichiometry

Stoichiometry: The quantitative study of reactants and products in chemical reactions

From Greek:

stoicheion = "element"
metron = "measure"

What stoichiometry tells us:

How much product forms from given reactants
How much reactant needed to make desired product
Which reactant limits the reaction
Theoretical vs. actual yield

Foundation: Balanced chemical equations

Mole Ratios from Balanced Equations

Balanced equation provides mole ratios

Example:

$N2\text{(g)} + 3H2\text{(g)} \rightarrow 2NH3\text{(g)}$

Mole ratios:

1 mol N₂ : 3 mol H₂
1 mol N₂ : 2 mol NH₃
3 mol H₂ : 2 mol NH₃

These ratios are exact (not measured)

Can have infinite sig figs
Used as conversion factors

Interpretation:

1 mole of N₂ reacts with 3 moles of H₂
1 mole of N₂ produces 2 moles of NH₃
3 moles of H₂ produce 2 moles of NH₃

Key concept: Coefficients give mole-to-mole relationships

Basic Stoichiometry Calculations

Mass-to-Mass Calculations

Problem type: Given mass of reactant, find mass of product

General strategy:

$\text{Mass A} \xrightarrow{\text{÷ Molar mass}} \text{Moles A} \xrightarrow{\text{Mole ratio}} \text{Moles B} \xrightarrow{\text{× Molar mass}} \text{Mass B}$

Example:

How many grams of NH₃ can be produced from 28.0 g of N₂?

$N2 + 3H2 \rightarrow 2NH3$

Step 1: Convert mass N₂ to moles

$n_{N_2} = \frac{28.0 \text{ g}}{28.0 \text{ g/mol}} = 1.00 \text{ mol N}_2$

Step 2: Use mole ratio to find moles NH₃

$n_{NH_3} = 1.00 \text{ mol N}_2 \times \frac{2 \text{ mol NH}_3}{1 \text{ mol N}_2} = 2.00 \text{ mol NH}_3$

Step 3: Convert moles NH₃ to mass

$\text{mass}_{NH_3} = 2.00 \text{ mol} \times 17.0 \text{ g/mol} = 34.0 \text{ g}$

Answer: 34.0 g NH₃

Mole-to-Mole Calculations

Simplest type: Already have moles

Example:

How many moles of O₂ needed to react with 5.0 mol glucose?

$C6H12O6 + 6O2 \rightarrow 6CO2 + 6H2O$

Solution:

$n_{O_2} = 5.0 \text{ mol C}_6\text{H}_{12}\text{O}_6 \times \frac{6 \text{ mol O}_2}{1 \text{ mol C}_6\text{H}_{12}\text{O}_6} = 30 \text{ mol O}_2$

Answer: 30 mol O₂

Mass-to-Particles Calculations

Use Avogadro's number: $N_A = 6.022 \times 10^{23}$ particles/mol

Example:

How many molecules of H₂O form from 9.0 g H₂?

$2H2 + O2 \rightarrow 2H2O$

Step 1: Moles H₂

$n_{H_2} = \frac{9.0 \text{ g}}{2.0 \text{ g/mol}} = 4.5 \text{ mol}$

Step 2: Moles H₂O

$n_{H_2O} = 4.5 \text{ mol H}_2 \times \frac{2 \text{ mol H}_2\text{O}}{2 \text{ mol H}_2} = 4.5 \text{ mol}$

Step 3: Convert to molecules

$N = 4.5 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} = 2.7 \times 10^{24} \text{ molecules}$

Answer: $2.7 \times 10^{24}$ molecules H₂O

Limiting Reactants

Limiting reactant: Reactant that is completely consumed first, limiting the amount of product

Excess reactant: Reactant that remains after the reaction stops

Analogy: Making sandwiches

10 slices bread + 6 slices cheese
1 sandwich needs 2 bread + 1 cheese
Can make only 5 sandwiches (limited by bread)
Bread = limiting reactant
Cheese = excess (1 slice left over)

Identifying Limiting Reactant

Method 1: Calculate product from each reactant

Compare: Which reactant produces less product?

That reactant is limiting

Example:

$2Al + 3Cl2 \rightarrow 2AlCl3$

Given: 5.0 mol Al and 6.0 mol Cl₂

From Al:

$n_{AlCl_3} = 5.0 \text{ mol Al} \times \frac{2 \text{ mol AlCl}_3}{2 \text{ mol Al}} = 5.0 \text{ mol AlCl}_3$

From Cl₂:

$n_{AlCl_3} = 6.0 \text{ mol Cl}_2 \times \frac{2 \text{ mol AlCl}_3}{3 \text{ mol Cl}_2} = 4.0 \text{ mol AlCl}_3$

Cl₂ produces less product → Cl₂ is limiting reactant

Al is in excess

Actual product formed: 4.0 mol AlCl₃ (limited by Cl₂)

Method 2: Compare mole ratios

Calculate: $\frac{\text{moles available}}{\text{coefficient}}$ for each reactant

Smallest ratio → limiting reactant

Example (same reaction):

$2Al + 3Cl2 \rightarrow 2AlCl3$

Given: 5.0 mol Al and 6.0 mol Cl₂

For Al:

$\frac{5.0 \text{ mol}}{2} = 2.5$

For Cl₂:

$\frac{6.0 \text{ mol}}{3} = 2.0$

Cl₂ has smaller ratio → Cl₂ is limiting ✓

Method 3: Calculate how much of one reactant is needed

Calculate: How much of reactant B is needed to react with all of reactant A?

Compare: Available B vs. needed B

If available < needed → B is limiting
If available > needed → A is limiting

Example (same reaction):

Given: 5.0 mol Al and 6.0 mol Cl₂

How much Cl₂ needed for all 5.0 mol Al?

$n_{Cl_2 \text{ needed}} = 5.0 \text{ mol Al} \times \frac{3 \text{ mol Cl}_2}{2 \text{ mol Al}} = 7.5 \text{ mol Cl}_2$

Available Cl₂: 6.0 mol Needed Cl₂: 7.5 mol

6.0 < 7.5 → Not enough Cl₂ → Cl₂ is limiting ✓

Calculating Excess Reactant Remaining

After reaction, how much excess reactant left?

Strategy:

Use limiting reactant to find how much excess reactant consumed
Subtract consumed from initial amount

Example:

$2Al + 3Cl2 \rightarrow 2AlCl3$

Given: 5.0 mol Al, 6.0 mol Cl₂ (Cl₂ is limiting)

How much Al remains?

Step 1: How much Al consumed by 6.0 mol Cl₂?

$n_{Al \text{ used}} = 6.0 \text{ mol Cl}_2 \times \frac{2 \text{ mol Al}}{3 \text{ mol Cl}_2} = 4.0 \text{ mol Al}$

Step 2: Calculate Al remaining

$n_{Al \text{ remaining}} = 5.0 \text{ mol initial} - 4.0 \text{ mol used} = 1.0 \text{ mol Al}$

Answer: 1.0 mol Al remains (excess)

Theoretical, Actual, and Percent Yield

Theoretical Yield

Theoretical yield: Maximum amount of product that can form from given reactants

Calculated from stoichiometry
Assumes 100% efficiency
Based on limiting reactant

Always use limiting reactant for theoretical yield!

Actual Yield

Actual yield: Amount of product actually obtained in lab

Measured experimentally
Always less than theoretical yield
Due to side reactions, losses, incomplete reactions

Percent Yield

Percent yield: Efficiency of reaction

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

Example:

Theoretical yield: 50.0 g product Actual yield: 42.5 g product

$\text{Percent yield} = \frac{42.5 \text{ g}}{50.0 \text{ g}} \times 100\% = 85.0\%$

Interpretation: Reaction is 85% efficient

Why actual < theoretical?

Side reactions: Reactants form unwanted products
Incomplete reactions: Not all reactants convert to products
Losses during purification: Some product lost in transfer, filtering, etc.
Impure reactants: Starting materials not 100% pure

Percent yield range:

0-100% (cannot exceed 100% with pure products)
Good lab technique: 80-95%
Industrial processes often optimize for high percent yield

Using Percent Yield

If percent yield known, can calculate actual yield:

$\text{Actual yield} = \frac{\text{percent yield}}{100} \times \text{theoretical yield}$

Example:

If theoretical yield = 30.0 g and reaction is 75% efficient:

$\text{Actual yield} = \frac{75}{100} \times 30.0 \text{ g} = 22.5 \text{ g}$

Solution Stoichiometry

For reactions in solution, use molarity

Molarity (M):

$M = \frac{\text{moles solute}}{\text{liters solution}}$

Rearranged:

$\text{moles} = M \times V_{\text{liters}}$

Strategy for Solution Stoichiometry

Given volumes and molarities, find product:

Calculate moles of each reactant: $n = M \times V$
Identify limiting reactant
Use stoichiometry to find product moles
Convert to mass (if needed) or molarity (if in solution)

Example:

$2AgNO3\text{(aq)} + CaCl2\text{(aq)} \rightarrow 2AgCl\text{(s)} + Ca(NO3)2\text{(aq)}$

Mix 50.0 mL of 0.200 M AgNO₃ with 30.0 mL of 0.100 M CaCl₂

How many grams of AgCl precipitate?

Step 1: Calculate moles of reactants

$n_{AgNO_3} = 0.200 \text{ M} \times 0.0500 \text{ L} = 0.0100 \text{ mol}$

$n_{CaCl_2} = 0.100 \text{ M} \times 0.0300 \text{ L} = 0.00300 \text{ mol}$

Step 2: Identify limiting reactant

From AgNO₃:

$n_{AgCl} = 0.0100 \text{ mol AgNO}_3 \times \frac{2 \text{ mol AgCl}}{2 \text{ mol AgNO}_3} = 0.0100 \text{ mol}$

From CaCl₂:

$n_{AgCl} = 0.00300 \text{ mol CaCl}_2 \times \frac{2 \text{ mol AgCl}}{1 \text{ mol CaCl}_2} = 0.00600 \text{ mol}$

CaCl₂ produces less → CaCl₂ is limiting

Step 3: Calculate mass of AgCl

$\text{mass} = 0.00600 \text{ mol} \times 143.5 \text{ g/mol} = 0.861 \text{ g AgCl}$

Answer: 0.861 g AgCl precipitates

Consecutive Reactions

Sometimes reactions occur in sequence

Product of first reaction → reactant for second reaction

Example:

$S\text{(s)} + O2\text{(g)} \rightarrow SO2\text{(g)}$ (Reaction 1)

$2SO2\text{(g)} + O2\text{(g)} \rightarrow 2SO3\text{(g)}$ (Reaction 2)

If given amount of S, find SO₃:

Strategy: Chain the stoichiometry

$\text{S} \xrightarrow{\text{Reaction 1}} \text{SO}_2 \xrightarrow{\text{Reaction 2}} \text{SO}_3$

Use mole ratios from both equations

Summary of Stoichiometry Steps

General problem-solving strategy:

Write balanced equation
- Double-check coefficients
Convert given to moles
- Use molar mass (g → mol)
- Use molarity × volume (M × L → mol)
- Use Avogadro's number (particles → mol)
Identify limiting reactant (if multiple reactants given)
- Calculate product from each reactant
- Smallest product → that reactant is limiting
- OR use ratio method
Use mole ratios (from balanced equation)
- Convert moles of limiting reactant to moles of product
Convert to desired units
- Mass (mol → g using molar mass)
- Particles (mol → particles using $N_A$ )
- Volume (if gas at STP: 22.4 L/mol)
- Molarity (if in solution)
Calculate percent yield (if actual yield given)
- Compare actual to theoretical

Common Stoichiometry Errors

❌ Mistake 1: Using wrong mole ratio

Check coefficients carefully
Write ratio as fraction

❌ Mistake 2: Forgetting to identify limiting reactant

If given multiple reactants, MUST determine limiting
Calculate product from each reactant

❌ Mistake 3: Using excess reactant for calculations

Always use limiting reactant for product calculations
Excess reactant doesn't limit product

❌ Mistake 4: Rounding too early

Keep extra sig figs during calculation
Round at final answer

❌ Mistake 5: Unit errors

Track units throughout
Cancel units properly
Convert volumes to liters for molarity

Tips for Success

✓ Always start with balanced equation

✓ Write out conversion factors explicitly

✓ Track units through calculation

✓ For limiting reactant: Calculate product from EACH reactant

✓ Check reasonableness of answer

Product mass shouldn't exceed reactant mass (usually)
Percent yield can't exceed 100%

✓ Use dimensional analysis

Set up so units cancel

✓ For solution problems: V must be in liters for molarity

📚 Practice Problems

1Problem 1easy

❓ Question:

Ammonia (NH₃) is produced by the Haber process: N₂(g) + 3H₂(g) → 2NH₃(g). How many grams of NH₃ can be produced from 56.0 g of N₂ and 12.0 g of H₂? (N = 14.0 g/mol, H = 1.0 g/mol)

💡 Show Solution

Solution:

Given:

Balanced equation: $N2\text{(g)} + 3H2\text{(g)} \rightarrow 2NH3\text{(g)}$
Mass of N₂ = 56.0 g
Mass of H₂ = 12.0 g
Molar masses: N = 14.0 g/mol, H = 1.0 g/mol

Find: Mass of NH₃ produced

This is a LIMITING REACTANT problem (two reactants given!)

Step 1: Calculate molar masses

$M_{N_2} = 2 \times 14.0 = 28.0 \text{ g/mol}$

$M_{H_2} = 2 \times 1.0 = 2.0 \text{ g/mol}$

$M_{NH_3} = 14.0 + 3(1.0) = 17.0 \text{ g/mol}$

Step 2: Convert masses to moles

Moles of N₂:

$n_{N_2} = \frac{56.0 \text{ g}}{28.0 \text{ g/mol}} = 2.00 \text{ mol N}_2$

Moles of H₂:

$n_{H_2} = \frac{12.0 \text{ g}}{2.0 \text{ g/mol}} = 6.0 \text{ mol H}_2$

Step 3: Identify limiting reactant

Method: Calculate NH₃ from each reactant, compare

From N₂:

Using mole ratio: $\frac{2 \text{ mol NH}_3}{1 \text{ mol N}_2}$

$n_{NH_3} = 2.00 \text{ mol N}_2 \times \frac{2 \text{ mol NH}_3}{1 \text{ mol N}_2} = 4.00 \text{ mol NH}_3$

From H₂:

Using mole ratio: $\frac{2 \text{ mol NH}_3}{3 \text{ mol H}_2}$

$n_{NH_3} = 6.0 \text{ mol H}_2 \times \frac{2 \text{ mol NH}_3}{3 \text{ mol H}_2}$

$n_{NH_3} = 6.0 \times \frac{2}{3} = 4.0 \text{ mol NH}_3$

Compare:

From N₂: 4.00 mol NH₃
From H₂: 4.0 mol NH₃

Both give same amount! → Both are completely consumed (neither in excess)

This is a special case: stoichiometric amounts

Limiting reactant: Both (or can say either)

Product formed: 4.0 mol NH₃

Alternative check - Ratio method:

$N2 + 3H2 \rightarrow 2NH3$

For N₂: $\frac{2.00 \text{ mol}}{1} = 2.00$

For H₂: $\frac{6.0 \text{ mol}}{3} = 2.0$

Ratios equal → Stoichiometric amounts (both limiting) ✓

Step 4: Convert moles NH₃ to mass

$\text{mass}_{NH_3} = n_{NH_3} \times M_{NH_3}$

$\text{mass}_{NH_3} = 4.0 \text{ mol} \times 17.0 \text{ g/mol}$

$\text{mass}_{NH_3} = 68 \text{ g}$

Answer:

$\boxed{68 \text{ g NH}_3}$

Summary:

Substance	Given Mass	Moles	NH₃ Produced
N₂	56.0 g	2.00 mol	4.00 mol
H₂	12.0 g	6.0 mol	4.0 mol
NH₃	?	4.0 mol	68 g

Stoichiometry flow:

$56.0 \text{ g N}_2 \xrightarrow{÷28.0} 2.00 \text{ mol N}_2 \xrightarrow{×\frac{2}{1}} 4.0 \text{ mol NH}_3 \xrightarrow{×17.0} 68 \text{ g NH}_3$

Or:

$12.0 \text{ g H}_2 \xrightarrow{÷2.0} 6.0 \text{ mol H}_2 \xrightarrow{×\frac{2}{3}} 4.0 \text{ mol NH}_3 \xrightarrow{×17.0} 68 \text{ g NH}_3$

Additional insights:

Why are these stoichiometric amounts?

From balanced equation: $N2 + 3H2 \rightarrow 2NH3$

Mole ratio: 1 mol N₂ : 3 mol H₂

Given amounts:

N₂: 2.00 mol
H₂: 6.0 mol

Check ratio: $\frac{6.0}{2.00} = 3$ ✓

Exactly 3:1 ratio → Perfect stoichiometric amounts!

No excess reactant remains

In real industrial process:

Usually use excess H₂ (cheaper than N₂)
Forces equilibrium toward products
Not run at stoichiometric ratio

Check conservation of mass:

Reactants: 56.0 g + 12.0 g = 68.0 g
Products: 68 g NH₃
Mass conserved! ✓

This makes sense because all reactants consumed

2Problem 2medium

❓ Question:

Consider the reaction: 2 Al + 3 Cl₂ → 2 AlCl₃. If 5.40 g of Al reacts with 8.00 g of Cl₂, (a) identify the limiting reactant, (b) calculate the mass of AlCl₃ produced, and (c) determine how much of the excess reactant remains.

💡 Show Solution

Solution:

Molar masses: Al = 26.98 g/mol, Cl₂ = 70.90 g/mol, AlCl₃ = 133.34 g/mol

(a) Identify limiting reactant:

Moles of Al = 5.40 g / 26.98 g/mol = 0.200 mol Moles of Cl₂ = 8.00 g / 70.90 g/mol = 0.113 mol

Using stoichiometry (2 Al : 3 Cl₂):

If Al is limiting: needs 0.200 mol Al × (3 Cl₂/2 Al) = 0.300 mol Cl₂ (we only have 0.113) ✗
If Cl₂ is limiting: needs 0.113 mol Cl₂ × (2 Al/3 Cl₂) = 0.0753 mol Al (we have 0.200) ✓

Limiting reactant: Cl₂

(b) Mass of AlCl₃ produced: Moles AlCl₃ = 0.113 mol Cl₂ × (2 AlCl₃/3 Cl₂) = 0.0753 mol Mass AlCl₃ = 0.0753 mol × 133.34 g/mol = 10.0 g

(c) Excess Al remaining: Al consumed = 0.113 mol Cl₂ × (2 Al/3 Cl₂) = 0.0753 mol Al remaining = 0.200 - 0.0753 = 0.125 mol = 3.37 g

3Problem 3medium

❓ Question:

When 15.0 g of aluminum reacts with excess hydrochloric acid according to the equation 2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g), 16.8 L of H₂ gas is collected at STP. (a) Calculate the theoretical yield of H₂ in liters at STP. (b) Calculate the percent yield. (Al = 27.0 g/mol)

💡 Show Solution

Solution:

Given:

Mass of Al = 15.0 g
Equation: $2Al\text{(s)} + 6HCl\text{(aq)} \rightarrow 2AlCl3\text{(aq)} + 3H2\text{(g)}$
HCl in excess (Al is limiting reactant)
Actual yield of H₂ = 16.8 L at STP
Molar mass Al = 27.0 g/mol
At STP: 1 mol gas = 22.4 L

Find: (a) Theoretical yield of H₂, (b) Percent yield

Part (a): Calculate theoretical yield

Step 1: Convert mass Al to moles

$n_{Al} = \frac{\text{mass}}{\text{molar mass}}$

$n_{Al} = \frac{15.0 \text{ g}}{27.0 \text{ g/mol}}$

$n_{Al} = 0.556 \text{ mol Al}$

Step 2: Use stoichiometry to find moles H₂

From balanced equation:

$2Al + 6HCl \rightarrow 2AlCl3 + 3H2$

Mole ratio: $\frac{3 \text{ mol H}_2}{2 \text{ mol Al}}$

$n_{H_2} = 0.556 \text{ mol Al} \times \frac{3 \text{ mol H}_2}{2 \text{ mol Al}}$

$n_{H_2} = 0.556 \times 1.5$

$n_{H_2} = 0.834 \text{ mol H}_2$

Step 3: Convert moles H₂ to volume at STP

At STP: 1 mol gas = 22.4 L (molar volume)

$V_{H_2} = n_{H_2} \times 22.4 \text{ L/mol}$

$V_{H_2} = 0.834 \text{ mol} \times 22.4 \text{ L/mol}$

$V_{H_2} = 18.7 \text{ L}$

Answer (a):

$\boxed{\text{Theoretical yield} = 18.7 \text{ L H}_2}$

This is the maximum H₂ that could be produced from 15.0 g Al

Part (b): Calculate percent yield

Step 4: Use percent yield formula

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

Given:

Actual yield = 16.8 L (measured in lab)
Theoretical yield = 18.7 L (calculated above)

Calculate:

$\text{Percent yield} = \frac{16.8 \text{ L}}{18.7 \text{ L}} \times 100\%$

$\text{Percent yield} = 0.898 \times 100\%$

$\text{Percent yield} = 89.8\%$

Answer (b):

$\boxed{\text{Percent yield} = 89.8\%}$

Or: $\boxed{90\%}$ (2 sig figs)

Interpretation:

What does 89.8% yield mean?

Reaction produced 89.8% of theoretical maximum
This is actually quite good efficiency for lab reaction
10.2% of potential H₂ not collected

Why is actual < theoretical?

Possible reasons:

Gas escaped during collection
- Not all H₂ captured in collection apparatus
- Small leaks in setup
Incomplete reaction
- Not all Al dissolved
- Oxide coating on Al surface prevents complete reaction
Impure aluminum
- Al sample may contain impurities
- Less than 15.0 g pure Al actually present
Measurement errors
- Volume measurement not exact
- Temperature or pressure not exactly STP
Side reactions
- Some Al reacts with O₂ in air instead of HCl

In industrial processes:

89.8% would be excellent yield
Industries optimize conditions for high percent yield
Economic motivation (more product per reactant)

Summary of calculations:

Stoichiometry path:

$15.0 \text{ g Al} \xrightarrow{÷27.0} 0.556 \text{ mol Al} \xrightarrow{×\frac{3}{2}} 0.834 \text{ mol H}_2 \xrightarrow{×22.4} 18.7 \text{ L H}_2$

Percent yield:

$\frac{16.8 \text{ L (actual)}}{18.7 \text{ L (theoretical)}} \times 100\% = 89.8\%$

Comparison table:

Quantity	Value
Mass Al (given)	15.0 g
Moles Al	0.556 mol
Moles H₂ (from stoichiometry)	0.834 mol
Theoretical yield H₂	18.7 L
Actual yield H₂ (given)	16.8 L
Percent yield	89.8%

Additional practice:

What if we wanted mass of H₂ instead of volume?

$\text{mass}_{H_2} = 0.834 \text{ mol} \times 2.0 \text{ g/mol} = 1.67 \text{ g H}_2$ (theoretical)

$\text{mass}_{H_2 \text{ actual}} = 1.67 \text{ g} \times 0.898 = 1.50 \text{ g}$ (actual)

Or from volume:

At STP: 22.4 L H₂ = 2.0 g H₂

$18.7 \text{ L} \times \frac{2.0 \text{ g}}{22.4 \text{ L}} = 1.67 \text{ g}$ ✓

Key concepts demonstrated:

Stoichiometry with limiting reactant
Gas volume at STP (molar volume = 22.4 L/mol)
Theoretical yield calculation
Percent yield formula and interpretation
Understanding why actual < theoretical

4Problem 4hard

❓ Question:

Aspirin (C₉H₈O₄) is synthesized from salicylic acid (C₇H₆O₃) and acetic anhydride (C₄H₆O₃): C₇H₆O₃ + C₄H₆O₃ → C₉H₈O₄ + C₂H₄O₂. If 20.0 g of salicylic acid reacts with 15.0 g of acetic anhydride and produces 18.5 g of aspirin, calculate the percent yield.

💡 Show Solution

Solution:

Molar masses:

C₇H₆O₃ (salicylic acid) = 138.12 g/mol
C₄H₆O₃ (acetic anhydride) = 102.09 g/mol
C₉H₈O₄ (aspirin) = 180.16 g/mol

Step 1: Find limiting reactant

Moles salicylic acid = 20.0 g / 138.12 g/mol = 0.145 mol Moles acetic anhydride = 15.0 g / 102.09 g/mol = 0.147 mol

Stoichiometry is 1:1, so we have nearly equal moles, but salicylic acid (0.145 mol) is slightly less.

Limiting reactant: Salicylic acid

Step 2: Theoretical yield Since stoichiometry is 1:1:1:1, moles of aspirin = 0.145 mol

Theoretical mass = 0.145 mol × 180.16 g/mol = 26.1 g

Step 3: Percent yield Actual yield = 18.5 g Percent yield = (actual/theoretical) × 100% Percent yield = (18.5 g / 26.1 g) × 100% = 70.9%

5Problem 5hard

❓ Question:

50.0 mL of 0.200 M Pb(NO₃)₂ is mixed with 30.0 mL of 0.300 M KI. The reaction is: Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq). (a) Identify the limiting reactant. (b) Calculate the mass of PbI₂ precipitate formed. (c) Calculate the molar concentration of K⁺ ions in the final solution (assume volumes are additive). (Pb = 207.2 g/mol, I = 126.9 g/mol, K = 39.1 g/mol)

💡 Show Solution

Solution:

Given:

Volume Pb(NO₃)₂ = 50.0 mL = 0.0500 L
Molarity Pb(NO₃)₂ = 0.200 M
Volume KI = 30.0 mL = 0.0300 L
Molarity KI = 0.300 M
Equation: $Pb(NO3)2\text{(aq)} + 2KI\text{(aq)} \rightarrow PbI2\text{(s)} + 2KNO3\text{(aq)}$
Molar masses: Pb = 207.2, I = 126.9, K = 39.1 g/mol

Find: (a) Limiting reactant, (b) Mass PbI₂, (c) [K⁺] in final solution

Part (a): Identify limiting reactant

Step 1: Calculate moles of each reactant

Moles of Pb(NO₃)₂:

$n_{Pb(NO_3)_2} = M \times V$

$n_{Pb(NO_3)_2} = 0.200 \text{ M} \times 0.0500 \text{ L}$

$n_{Pb(NO_3)_2} = 0.0100 \text{ mol}$

Moles of KI:

$n_{KI} = M \times V$

$n_{KI} = 0.300 \text{ M} \times 0.0300 \text{ L}$

$n_{KI} = 0.00900 \text{ mol}$

Step 2: Identify limiting reactant

Method 1: Calculate product from each reactant

From balanced equation:

$Pb(NO3)2 + 2KI \rightarrow PbI2 + 2KNO3$

Mole ratio: 1 Pb(NO₃)₂ : 2 KI : 1 PbI₂

From Pb(NO₃)₂:

$n_{PbI_2} = 0.0100 \text{ mol Pb(NO}_3)_2 \times \frac{1 \text{ mol PbI}_2}{1 \text{ mol Pb(NO}_3)_2}$

$n_{PbI_2} = 0.0100 \text{ mol}$

From KI:

$n_{PbI_2} = 0.00900 \text{ mol KI} \times \frac{1 \text{ mol PbI}_2}{2 \text{ mol KI}}$

$n_{PbI_2} = 0.00450 \text{ mol}$

Compare:

From Pb(NO₃)₂: 0.0100 mol PbI₂
From KI: 0.00450 mol PbI₂

KI produces less product → KI is limiting reactant ✓

Method 2: Ratio method (verification)

$\frac{n_{Pb(NO_3)_2}}{\text{coefficient}} = \frac{0.0100}{1} = 0.0100$

$\frac{n_{KI}}{\text{coefficient}} = \frac{0.00900}{2} = 0.00450$

KI has smaller ratio → KI is limiting ✓

Answer (a):

$\boxed{\text{KI is the limiting reactant}}$

Pb(NO₃)₂ is in excess

Part (b): Calculate mass of PbI₂

Step 3: Calculate moles of PbI₂ formed

Use limiting reactant (KI):

$n_{PbI_2} = 0.00900 \text{ mol KI} \times \frac{1 \text{ mol PbI}_2}{2 \text{ mol KI}}$

$n_{PbI_2} = 0.00450 \text{ mol}$

Step 4: Calculate molar mass of PbI₂

$M_{PbI_2} = 207.2 + 2(126.9)$

$M_{PbI_2} = 207.2 + 253.8$

$M_{PbI_2} = 461.0 \text{ g/mol}$

Step 5: Convert moles to mass

$\text{mass}_{PbI_2} = n_{PbI_2} \times M_{PbI_2}$

$\text{mass}_{PbI_2} = 0.00450 \text{ mol} \times 461.0 \text{ g/mol}$

$\text{mass}_{PbI_2} = 2.07 \text{ g}$

Answer (b):

$\boxed{2.07 \text{ g PbI}_2 \text{ precipitates}}$

This is bright yellow precipitate

Part (c): Calculate [K⁺] in final solution

Step 6: Find total moles of K⁺

K⁺ comes from KI:

Each KI provides 1 K⁺
Started with 0.00900 mol KI
Therefore: 0.00900 mol K⁺ total

Note: All KI dissolves initially (before precipitation)

KI → K⁺ + I⁻ (complete dissociation)
Then: Pb²⁺ + 2I⁻ → PbI₂(s)
K⁺ is spectator ion (stays dissolved)

All K⁺ remains in solution

Step 7: Calculate total volume

Assume volumes are additive:

$V_{total} = V_{Pb(NO_3)_2} + V_{KI}$

$V_{total} = 50.0 \text{ mL} + 30.0 \text{ mL} = 80.0 \text{ mL}$

$V_{total} = 0.0800 \text{ L}$

Step 8: Calculate molarity of K⁺

$[K^+] = \frac{n_{K^+}}{V_{total}}$

$[K^+] = \frac{0.00900 \text{ mol}}{0.0800 \text{ L}}$

$[K^+] = 0.1125 \text{ M}$

$[K^+] = 0.113 \text{ M}$

Answer (c):

$\boxed{[K^+] = 0.113 \text{ M}}$

Summary Table:

Species	Initial moles	Final status
Pb(NO₃)₂	0.0100 mol	Excess (some remains)
KI	0.00900 mol	Limiting (all consumed)
PbI₂	0	0.00450 mol formed (2.07 g solid)
K⁺	0.00900 mol	0.00900 mol (dissolved, 0.113 M)

Additional analysis:

How much Pb(NO₃)₂ remains?

Pb(NO₃)₂ consumed:

$n_{Pb(NO_3)_2 \text{ used}} = 0.00900 \text{ mol KI} \times \frac{1 \text{ mol Pb(NO}_3)_2}{2 \text{ mol KI}}$

$n_{Pb(NO_3)_2 \text{ used}} = 0.00450 \text{ mol}$

Pb(NO₃)₂ remaining:

$n_{Pb(NO_3)_2 \text{ remaining}} = 0.0100 - 0.00450 = 0.00550 \text{ mol}$

Concentration of Pb²⁺ in final solution:

$[Pb^{2+}] = \frac{0.00550 \text{ mol}}{0.0800 \text{ L}} = 0.0688 \text{ M}$

Wait! This ignores PbI₂ solubility equilibrium

PbI₂ is "insoluble" but has tiny solubility
Some Pb²⁺ and I⁻ in equilibrium with solid
For this problem, assume PbI₂ completely insoluble

What about NO₃⁻ concentration?

NO₃⁻ from Pb(NO₃)₂:

Each Pb(NO₃)₂ provides 2 NO₃⁻
Started with 0.0100 mol Pb(NO₃)₂
Total NO₃⁻ = 0.0100 × 2 = 0.0200 mol

NO₃⁻ is spectator ion (stays dissolved)

$[NO_3^-] = \frac{0.0200 \text{ mol}}{0.0800 \text{ L}} = 0.250 \text{ M}$

Complete ion inventory in final solution:

Ion	Concentration
K⁺	0.113 M
NO₃⁻	0.250 M
Pb²⁺	~0.069 M (excess)
I⁻	~0 M (consumed)

Plus: Solid PbI₂ precipitate (2.07 g)

Key concepts demonstrated:

Solution stoichiometry using M × V = n
Limiting reactant determination with multiple reactants
Precipitate formation (PbI₂ is insoluble)
Spectator ions (K⁺ and NO₃⁻ don't react)
Final concentration calculation with mixed volumes
Excess reactant remaining after reaction

Observable evidence:

Solutions mixed → immediately bright yellow precipitate forms
Yellow solid settles to bottom
Clear solution above contains K⁺, NO₃⁻, excess Pb²⁺

Explain using:

📋 AP Chemistry — Exam Format Guide

⏱ 3 hours 15 minutes📝 67 questions📊 3 sections

Section	Format	Questions	Time	Weight	Calculator
Multiple Choice	MCQ	60	90 min	50%	✅
Free Response (Long)	FRQ	3	69 min	30%	✅
Free Response (Short)	FRQ	4	36 min	20%	✅

📊 Scoring: 1-5

Extremely Qualified

~12%

Well Qualified

~16%

Qualified

~24%

Possibly Qualified

~24%

No Recommendation

~24%

💡 Key Test-Day Tips

✓Memorize common polyatomic ions
✓Practice dimensional analysis
✓Know your gas laws

⚠️ Common Mistakes: Stoichiometry and Limiting Reactants

Avoid these 3 frequent errors

🌍 Real-World Applications: Stoichiometry and Limiting Reactants

See how this math is used in the real world

📝 Worked Example: Stoichiometry — Limiting Reagent

Problem:

$2$ mol of $H_2$ reacts with $1$ mol of $O_2$ . How many grams of water are produced? Which is the limiting reagent? ( $2H_2 + O_2 \to 2H_2O$ )

2Determine the limiting reagent

3Calculate moles of product

4Convert moles to grams

← Previous Topic

Types of Chemical Reactions

Next Topic →

Net Ionic Equations and Spectator Ions

🎴

Practice with Flashcards

Rate this topic's cards with spaced repetition. Cards join your deck when you finish a topic's lesson and take its exit quiz.

📖

Browse All Topics

Explore more AP Chemistry topics

📌 Related Topics in Chemical Reactions

Types of Chemical Reactions Net Ionic Equations and Spectator Ions Oxidation-Reduction (Redox) Reactions

❓ Frequently Asked Questions

What is Stoichiometry and Limiting Reactants?▾

Master stoichiometric calculations, mole ratios, limiting reactants, theoretical yield, percent yield, and solution stoichiometry.

How can I study Stoichiometry and Limiting Reactants effectively?▾

Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 5 problems provided, checking solutions as you go. Regular review and active practice are key to retention.

Is this Stoichiometry and Limiting Reactants study guide free?▾

Yes — all study notes, flashcards, and practice problems for Stoichiometry and Limiting Reactants on Study Mondo are free to access. No account is needed.

What course covers Stoichiometry and Limiting Reactants?▾

Stoichiometry and Limiting Reactants is part of the AP Chemistry course on Study Mondo, specifically in the Chemical Reactions section. You can explore the full course for more related topics and practice resources.

Are there practice problems for Stoichiometry and Limiting Reactants?▾

Yes, this page includes 5 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.

Stoichiometry and Limiting Reactants

Try the Interactive Version!

Stoichiometry and Limiting Reactants

Introduction to Stoichiometry

Mole Ratios from Balanced Equations

Basic Stoichiometry Calculations

Mass-to-Mass Calculations

Mole-to-Mole Calculations

Mass-to-Particles Calculations

Limiting Reactants

Identifying Limiting Reactant

Calculating Excess Reactant Remaining

Theoretical, Actual, and Percent Yield

Theoretical Yield

Actual Yield

Percent Yield

Using Percent Yield

Solution Stoichiometry

Strategy for Solution Stoichiometry

Consecutive Reactions

Summary of Stoichiometry Steps

Common Stoichiometry Errors

Tips for Success

📚 Practice Problems

1Problem 1easy

❓ Question:

2Problem 2medium

❓ Question:

3Problem 3medium

❓ Question:

4Problem 4hard

❓ Question:

5Problem 5hard

❓ Question:

📋 AP Chemistry — Exam Format Guide

📊 Scoring: 1-5

💡 Key Test-Day Tips

⚠️ Common Mistakes: Stoichiometry and Limiting Reactants

🌍 Real-World Applications: Stoichiometry and Limiting Reactants

📝 Worked Example: Stoichiometry — Limiting Reagent

Practice Lab

Practice with Flashcards

Browse All Topics

📌 Related Topics in Chemical Reactions

❓ Frequently Asked Questions