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:

1. Use limiting reactant to find how much excess reactant consumed
2. 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?

1. Side reactions: Reactants form unwanted products
2. Incomplete reactions: Not all reactants convert to products
3. Losses during purification: Some product lost in transfer, filtering, etc.
4. 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:

1. Calculate moles of each reactant: $n = M \times V$
2. Identify limiting reactant
3. Use stoichiometry to find product moles
4. 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:

1. Write balanced equation

   • Double-check coefficients

2. Convert given to moles

   • Use molar mass (g → mol)
   • Use molarity × volume (M × L → mol)
   • Use Avogadro's number (particles → mol)

3. Identify limiting reactant (if multiple reactants given)

   • Calculate product from each reactant
   • Smallest product → that reactant is limiting
   • OR use ratio method

4. Use mole ratios (from balanced equation)

   • Convert moles of limiting reactant to moles of product

5. 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)

6. 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

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:

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