Stoichiometry Master Slides Plus Practice Stoichiometry Master
Unit 8 Review & Strategy Session
The Golden Rule: Conservation
Key Concept
Matter cannot be created or destroyed. In a chemical reaction:
Total Atoms (Reactants) = Total Atoms (Products)
Total Mass (Reactants) = Total Mass (Products)
Watch Out!
Misconception: "The number of moles must be equal."
The Truth: Moles can change because atoms rearrange! Only mass and atom counts are strictly conserved.
PRACTICE: Conservation
"In a lab, 20g of Hydrogen reacts with 160g of Oxygen to produce Water."
What is the total mass of the reactants ?
Based on the law, what must be the mass of the water produced?
Molar Mass Calculations
The Steps
List every element in the formula.
Count how many of each atom (check subscripts!).
Multiply by atomic mass from Periodic Table.
Add all totals together.
Example: \( \text{Fe(NO}_3)_3 \)
Fe: 1 \(\times\) 55.85 = 55.85
PRACTICE: Molar Mass
Level 1: Simple
H\(_2\)SO\(_4\)
H = 1.01 | S = 32.07 | O = 16.00
Level 2: Parentheses
Al\(_2\)(SO\(_4\))\(_3\)
Al = 26.98 | S = 32.07 | O = 16.00
The Mole Ratio Bridge
Coefficients in a balanced equation tell us the ratio of moles.
1
mol C\(_3\)H\(_8\)
:
5
mol O\(_2\)
:
3
mol CO\(_2\)
PRACTICE: Mole Ratio
2 H\(_2\) + O\(_2\) → 2 H\(_2\)O
Ratio 1
H\(_2\) to O\(_2\)
____ : ____
Ratio 2
O\(_2\) to H\(_2\)O
____ : ____
Ratio 3
H\(_2\) to H\(_2\)O
____ : ____
The Stoichiometry Roadmap
Step 1
Grams → Moles
(Divide by MM)
Step 2
Mole Ratio
Bridge Crossing
Step 3
Moles → Grams
(Multiply by MM)
PRACTICE: Full Setup
"How many grams of Ammonia (NH\(_3\)) are produced from 10.0g of Hydrogen (H\(_2\))?"
N\(_2\) + 3 H\(_2\) → 2 NH\(_3\)
1. Convert H\(_2\) to mol
10.0g / 2.02 g/mol
2. Use Ratio
(mol H\(_2\)) \(\times\) (2/3 ratio)
3. Convert to NH\(_3\) g
(mol NH\(_3\)) \(\times\) 17.03 g/mol
Limiting Reactants
The Limit
The reactant that runs out first . It stops the reaction and determines the amount of product formed.
The Sandwich Rule
🍞 + 🧀 + 🍞 → 🥪
If you have 10 slices of bread and 1 piece of cheese...
The cheese is limiting.1 sandwich .
PRACTICE: Finding the Limit
"A student has 4.0 moles of Nitrogen and 9.0 moles of Hydrogen."
N\(_2\) + 3 H\(_2\) → 2 NH\(_3\)
Question A
How many moles of Hydrogen are needed to react with all 4 moles of Nitrogen?
Question B
Which reactant runs out first? (Is it the N\(_2\) or the H\(_2\)?)
Percent Yield
Theoretical Yield
The maximum possible amount (The Math result).
Actual Yield
What you really got (The Lab result).
The Formula
Actual
Theoretical
\(\times\) 100
PRACTICE: Percent Yield
"A student calculated they should get 15.5g of product. After the lab, they weighed 12.8g."
Setup:
(12.8 / 15.5) \(\times\) 100
The Question
What is the final percent yield for this student?
Result %
Final Exam Strategy
Equation balanced? (Check coefficients)
Molar masses calculated correctly?
Units included in every final answer?
Final rounding matches Sig Figs?
Stoichiometry Quest Cards Slice Space Stoichiometry Quest
Classroom Scavenger Hunt Challenge
Print: Pages 2 through 7 (Challenge Cards).
Prep: Cut along the dashed lines. Tape the 12 cards around the room.
Play: Students solve the bottom problem, then hunt for the card with that answer at the top.
STATION 1
Previous Answer:
8.00 g
Current Challenge:
Calculate the molar mass of calcium hydroxide, Ca(OH)2.
Ca: 40.08 | O: 16.00 | H: 1.01
Cut Along Line
STATION 2
Previous Answer:
74.10 g/mol
Current Challenge:
Given: 2H2 + O2 → 2H2O
STATION 3
Previous Answer:
10.0 mol
Current Challenge:
How many moles are in 88.0 grams of carbon dioxide (CO2)?
Molar Mass of CO2 = 44.01 g/mol
Cut Along Line
STATION 4
Previous Answer:
2.00 mol
Current Challenge:
What is the mass in grams of 0.50 moles of sodium chloride (NaCl)?
Molar Mass of NaCl = 58.44 g/mol
STATION 5
Previous Answer:
29.22 g
Current Challenge:
Given: N2 + 3H2 → 2NH3
N2: 28.02 g/mol | NH3: 17.04 g/mol
Cut Along Line
STATION 6
Previous Answer:
34.08 g
Current Challenge:
Reaction: CH4 + 2O2 → CO2 + 2H2O
CH4: 16.05 g/mol | O2: 32.00 g/mol
STATION 7
Previous Answer:
O2
Current Challenge:
Based on the limiting reactant from Card 6, what is the theoretical yield of H2O in grams?
Ratio 2 O2 : 2 H2O | H2O = 18.02 g/mol
Cut Along Line
STATION 8
Previous Answer:
18.02 g
Current Challenge:
If you collect 15.50 g H2O in the lab, what is the percent yield?
Yield = (Actual / Theoretical) × 100
STATION 9
Previous Answer:
86.02%
Current Challenge:
Stoichiometry Skill Builder Worksheet Stoichiometry Skill Builder
Unit 8: Quantitative Chemistry Review & Practice
Name:
Date:
Phase 1: Foundations
1. Conservation of Mass: Complete the missing values.
Reaction A: \( \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \)
Reactants: 16.0g \( \text{CH}_4 \) + 64.0g \( \text{O}_2 \)
Products: 44.0g \( \text{CO}_2 \) + ________ g \( \text{H}_2\text{O} \)
Reaction B: \( 2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO} \)
Reactants: ________ g Mg + 16.0g \( \text{O}_2 \)
Products: 40.3g MgO
2. Molar Mass Mission: Calculate the molar mass (g/mol) for each compound.
a) Sodium Phosphate: \( \text{Na}_3\text{PO}_4 \)
b) Calcium Hydroxide: \( \text{Ca(OH)}_2 \)
Phase 2: The Mole Bridge
The 3-Step Stoichiometry Logic
Step 1
Grams to Moles
Step 2
Mole Bridge
Step 3
Moles to Grams
Reaction: \( \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \)
Ratio of \( \text{H}_2 \) to \( \text{NH}_3 \):
Ratio of \( \text{N}_2 \) to \( \text{H}_2 \):
3. Mole-to-Mole Walkthrough
How many moles of \( \text{NH}_3 \) can be produced from 4.5 moles of \( \text{H}_2 \)?
4.5 mol H\(_2\) ×
___ mol NH\(_3\)
___ mol H\(_2\)
= ___ mol NH\(_3\)
Phase 3: The Roadmap
4. Mass-to-Mass Road Map Sets
Master Strategy
Scenario 1: Propane Combustion
"How many grams of CO\(_2\) are produced if 50.0g of C\(_3\)H\(_8\) is burned?"
Given
50.0g C\(_3\)H\(_8\)
1. Moles A
1 mol C\(_3\)H\(_8\)
MM A (g)
2. Bridge
___ mol CO\(_2\)
___ mol C\(_3\)H\(_8\)
3. Grams B
MM B (g)
1 mol CO\(_2\)
=
Result
____ g
Scenario 2: Decomposition
"How many grams of Oxygen are produced from 122.5g of KClO\(_3\)?"
Build your Road Map Set below:
=
Phase 4: Yield & Limits
5. Identifying the Limiting Reactant
Equation: \( \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \)
Stoichiometry Quest Recording Sheet Quest Recording Sheet
Stoichiometry Scavenger Hunt
Name:
Date:
How to Play: Start at any station. Solve the problem at the bottom of the card in the corresponding box below. Show all your work (dimensional analysis, molar mass calculations, etc.). Once you have an answer, "hunt" for the card with that answer at the top!
Station #
Show Work:
Answer:
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Stoichiometry Practice Exam Unit 8 DCA: Stoichiometry Mock Exam
Standardized Practice Assessment
Student Name:
Class Period:
Exam Strategy Guide
Conservation
Total Mass IN must equal Total Mass OUT. Atoms are rearranged, not lost.
Molar Mass
Sum of (Atomic Mass \(\times\) Subscript). Watch for parentheses!
Mole Ratio
Use Coefficients. Top = "Where you're going", Bottom = "Where you're from".
Percent Yield
(Actual Recovered / Theoretical Math) \(\times\) 100.
(C.9A) CONSERVATION PRINCIPLES
All chemical equations adhere to the law of conservation of mass. According to this law, the total mass of the reactants is the same as is different than the total mass of the products. This means that the number of atoms of each element remains constant may change throughout the reaction. However, the total number of moles will always be the same may vary since atoms rearrange into new products.
(C.8A) MOLAR MASS DETERMINATION
What is the molar mass of \( \text{Al}_2(\text{SO}_4)_3 \)?
Select the numerical value and the correct unit from the choices below:
mol
g/mol
123.0
342.1
284.5
g
Value
Unit
(C.9C) MOLE RATIO IDENTIFICATION
Given the balanced equation for the synthesis of ammonia:
N\(_2\) + 3 H\(_2\) → 2 NH\(_3\)
Identify the pieces needed to create the correct molar ratio of Hydrogen to Ammonia .
1 mol N\(_2\)
3 mol H\(_2\)
2 mol NH\(_3\)
17 g NH\(_3\)
Numerator
Denominator
(C.9C) MASS-TO-MASS CALCULATION
2 Mg + O\(_2\) → 2 MgO
A student reacts 4.86 grams of Magnesium (24.3 g/mol) with excess Oxygen. Calculate the mass of Magnesium Oxide (40.3 g/mol) produced.
A 4.03 grams
B 8.06 grams
C 12.1 grams
D 16.1 grams
(C.9C) THEORETICAL VS ACTUAL YIELD
"In a chemical reaction, the theoretical yield of products is 58.35 g. If 44.34 g of product is actually collected from the lab..."
A) Calculate the Percent Yield:
%
Stoichiometry Quest Key Quest Answer Key
Stoichiometry Scavenger Hunt Activity
Teacher Resource
The Quest Loop
1
→
2
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3
→
4
→
5
→
6
→
7
→
8
→
9
→
10
→
11
→
12
→
1
The hunt is a continuous loop. Students can start at any number and will return to their starting point after 12 stations.
Grading Tip
Verify that students have filled in the "Station #" boxes in the correct order. If they jump stations, they likely just found the answers without solving the problems. The dimensional analysis work in the boxes should reflect the specific mole ratios and molar masses provided on the cards.
Station Correct Answer Topic & Solution Path 1 74.10 g/mol Molar Mass: Ca(OH)2 40.08 + 2(16.00) + 2(1.01) = 74.10 2 10.0 mol Mole-to-Mole Ratio 5.0 mol O2 × (2 mol H2O / 1 mol O2) = 10.0 mol H2O 3 2.00 mol Mass-to-Mole 88.0 g CO2 / 44.01 g/mol = 1.9995 → 2.00 mol 4 29.22 g Mole-to-Mass 0.50 mol NaCl × 58.44 g/mol = 29.22 g 5 34.08 g Mass-to-Mass (Multi-step) 28.02 g N2 → 1 mol N2 → 2 mol NH3 → 34.08 g NH3 6 O2 Limiting Reactant 16.05g CH4 is 1 mol. 32g O2 is 1 mol. Reaction needs 2 mol O2 for 1 mol CH4. 7 18.02 g Theoretical Yield 1 mol O2 × (2 mol H2O / 2 mol O2) × 18.02 g/mol = 18.02 g 8 86.02% Percent Yield (15.50 g / 18.02 g) × 100 = 86.015... → 86.02% 9 6.0 mol Mole-to-Mole Ratio 4.0 mol Al × (3 mol Cl2 / 2 mol Al) = 6.0 mol Cl2 10 100 g Conservation of Mass Mass in = Mass out. No calculation needed, just conceptual understanding. 11 342.14 g/mol Complex Molar Mass 2(26.98) + 3(32.06) + 12(16.00) = 53.96 + 96.18 + 192.00 = 342.14
Stoichiometry Skill Builder Scaffolded Worksheet Stoichiometry Lab Practice
Unit 8: Quantitative Chemistry Review (Scaffolded Version)
Name:
Date:
Phase 1: Foundations
MODEL: Conservation of Mass
If 10g of A reacts with 5g of B to form C, how much C is made?
Answer: 10g + 5g = 15g. Mass is never lost!
1. PRACTICE: \( \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \)
Reactants
Mass of \( \text{CH}_4 \): 16.0 g
Mass of \( \text{O}_2 \): 64.0 g
Total Reactant Mass: ________
Products
Mass of \( \text{CO}_2 \): 44.0 g
Mass of \( \text{H}_2\text{O} \): ________
(Must equal reactant total)
MODEL: Molar Mass of \( \text{H}_2\text{SO}_4 \)
H: 2 \(\times\) 1.01 | S: 1 \(\times\) 32.07 | O: 4 \(\times\) 16.00 → Total = 98.09 g/mol
2. YOUR TURN: Calculate the molar mass of \( \text{Mg(NO}_3)_2 \)
Mg: ___ \(\times\) ___ = ___
N: ___ \(\times\) ___ = ___
O: ___ \(\times\) ___ = ___
Phase 2: The Bridge
MODEL: Mole Ratios
Reaction: \( 2\text{Fe} + 3\text{Cl}_2 \rightarrow 2\text{FeCl}_3 \)
Ratio of \( \text{Cl}_2 \) to Fe: 3 to 2 (\(\frac{3}{2}\))
Reaction: \( 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \)
Use the coefficients to write the following mole ratios:
H\(_2\) to O\(_2\)
______ : ______
O\(_2\) to H\(_2\)O
______ : ______
H\(_2\) to H\(_2\)O
______ : ______
3. Mole-to-Mole Walkthrough
Given: \( \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \). How many moles of \( \text{NH}_3 \) can be produced from 4.5 moles of \( \text{H}_2 \)?
4.5 mol H\(_2\) ×
2 mol NH\(_3\)
3 mol H\(_2\)
= _____ mol NH\(_3\)
Model: The units on the bottom MUST cancel the units on the left!
Phase 3: The Roadmap
STRATEGY: Mass-to-Mass
Follow the path: Grams A → Moles A → Moles B → Grams B
Grams A
Moles A
Moles B
Grams B
4. YOUR TURN: Mass-to-Mass Challenge
Reaction: \( \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \)
How many grams of \( \text{CO}_2 \) are produced when 25.0 grams of \( \text{C}_3\text{H}_8 \) (Molar Mass = 44.1 g/mol) is burned?
Stoichiometry Quest Cards Slice Space Fixed Stoichiometry Quest
Scavenger Hunt: Stations 1–12
Print: Pages 2–7 (Double-station Challenge Cards).
Prep: Cut cards along the dashed lines. Tape cards randomly around the room.
Play: Solve the problem at the bottom of any card, then find that answer at the top of another card!
STATION 1
Previous Answer:
8.00 g
Current Challenge:
Calculate the molar mass of calcium hydroxide, Ca(OH)2.
Ca: 40.08 | O: 16.00 | H: 1.01
Cut Along Line
STATION 2
Previous Answer:
74.10 g/mol
Current Challenge:
Given the reaction:
STATION 3
Previous Answer:
10.0 mol
Current Challenge:
How many moles are in 88.0 grams of carbon dioxide (CO2)?
Molar Mass of CO2 = 44.01 g/mol
Cut Along Line
STATION 4
Previous Answer:
2.00 mol
Current Challenge:
What is the mass in grams of 0.50 moles of sodium chloride (NaCl)?
Molar Mass of NaCl = 58.44 g/mol
STATION 5
Previous Answer:
29.22 g
Current Challenge:
Given: N2 + 3H2 → 2NH3
N2: 28.02 g/mol | NH3: 17.04 g/mol
Cut Along Line
STATION 6
Previous Answer:
34.08 g
Current Challenge:
Reaction: CH4 + 2O2 → CO2 + 2H2O
CH4: 16.05 g/mol | O2: 32.00 g/mol
STATION 7
Previous Answer:
O2
Current Challenge:
Based on the limiting reactant from Card 6, what is the theoretical yield of H2O in grams?
Ratio 2 O2 : 2 H2O | H2O = 18.02 g/mol
Cut Along Line
STATION 8
Previous Answer:
18.02 g
Current Challenge:
If you collect 15.50 g H2O in the lab, what is the percent yield?
Yield = (Actual / Theoretical) × 100
STATION 9
Previous Answer:
86.02%
Current Challenge:
Stoichiometry Quest Strategy Guide Quest Strategy Guide
Scavenger Hunt Cheat Sheet: Stations 1–12
Student Reference
#1
Molar Mass Basics
For Ca(OH)2, the "2" outside the parentheses distributes to everything inside!
• 1 × Ca mass
• 2 × O mass
• 2 × H mass
#2
Mole-to-Mole
Use the coefficients from the balanced equation.
5.0 mol O2 × ratio = ?
Ratio: (Wanted / Given)
#3
Grams to Moles
To go from Grams → Moles, you must divide by the molar mass.
Grams
÷
MM
#4
Moles to Grams
To go from Moles → Grams, you must multiply by the molar mass.
Moles
×
MM
#5
The 3-Step Marathon (Mass to Mass)
Grams A
→
Moles A
→
Moles B
→
Grams B
#6
Limiting Reactant (LR)
Find out how many moles of product each reactant can make. The reactant that makes LESS product is your LR.
Hint: If the ratio is 1:2, you need twice as much of the second reactant as the first.
#7
Theoretical Yield
Start your calculation with the Limiting Reactant from Station 6 (O2). Follow the same 3 steps as Station 5 to find grams of water.
#8
Percent Yield
Actual
Theoretical
× 100
Actual = Lab result | Theoretical = Math result
#9
Equation Ratios
Look at the coefficients for:
Al : Cl2
Example: If it's 2:3, then for every 4 moles of Al, you need 6 moles of Cl2.
#10
Mass Conservation
100g In
=
?g Out
#11
Big Formula MM
Al2(SO4)3
• 2 × Al
• 3 × S
• 12 × O
(4 × 3 = 12 oxygen atoms!)
#12
The Final Sprint
Simple Mole → Gram conversion for oxygen gas (O2).
0.25 mol × (32.00 g / 1 mol) = _____ g
Remember to round your final answers to the correct number of significant figures!
Stoichiometry Worksheet Answer Key Answer Key
Stoichiometry Skill Builder Worksheet (Updated)
Teacher Resource
Phase 1: Foundations
1. Conservation
A) 16.0g + 64.0g = 80.0g → 80.0g - 44.0g = 36.0g H\(_2\)O
B) 40.3g MgO - 16.0g O\(_2\) = 24.3g Mg
2. Molar Mass (g/mol)
a) Na\(_3\)PO\(_4\): 163.94
b) Ca(OH)\(_2\): 74.09
c) Al\(_2\)(SO\(_4\))\(_3\): 342.17
d) C\(_6\)H\(_12\)O\(_6\): 180.18
Phase 2: Mole Bridge
Mole Ratios (N\(_2\) + 3H\(_2\) → 2NH\(_3\))
a) H\(_2\) to NH\(_3\): 3 : 2
b) N\(_2\) to H\(_2\): 1 : 3
3. Mole-to-Mole
A) 8.4 mol O\(_2\) \(\times\) (2/1 ratio) = 16.8 mol H\(_2\)O
B) 15 mol O\(_2\) \(\times\) (2/3 ratio) = 10.0 mol KClO\(_3\)
Phase 3: The Roadmap
4. Mass-to-Mass
Scenario 1 (Propane):
50.0g / 44.1 \(\times\) (3/1 ratio) \(\times\) 44.01 = 149.7 grams CO\(_2\)
Scenario 2 (Magnesium):
10.0g / 24.31 \(\times\) (1/1 ratio) \(\times\) 2.016 = 0.83 grams H\(_2\)
Phase 4: Yield & Limits
5. Limiting Reactant Answer
Moles N\(_2\) = 25.0 / 28.02 = 0.89 mol
6. Percent Yield
A) (74.2 / 85.0) \(\times\) 100 = 87.3%
B) (2.1 / 4.5) \(\times\) 100 = 46.7%
Confidential Teacher Resource • Unit 8 Stoichiometry Mastery
Reaction Race Board Game Reaction Race
The Ultimate Stoichiometry Board Game
Unit 8 Review
Objective
Be the first chemist to navigate the "Stoichiometry Pipeline" and reach the Theoretical Yield Finish Line . Solve challenges, avoid spills, and use catalysts to speed ahead!
Players
2–4 Chemists (Players or Teams)
How to Play
Roll a die to move your token.
Land on a space and draw a card of that category.
If you solve it correctly, you stay. If not, go back to your previous space.
Special Spaces: Follow the instructions on the board immediately.First to reach the end exactly wins!
Start Lab Bench
100% Yield!
Space 1 Molar Mass
Space 2 Molar Mass
Glassware Break! Go Back 2
Space 4 Mole Ratio
Space 5 Mole Ratio
Catalyst! Advance 3
Space 7 Mass-to-Mass
Space 8 Mass-to-Mass
Acid Spill! Skip A Turn
Space 10 Limiting Reactant
Space 11 Limiting Reactant
Space 12 Percent Yield
Space 13 Percent Yield
Space 14 Mixed Review
Space 15 Mixed Review
Challenge Key
Molar Mass
Mole Ratio
Mass-to-Mass
Limiting Reactant
Percent Yield
Game Tokens
Cut these out to use as your player pieces! Fold the bottom tab to stand them up.
Erlenmeyer
Test Tube
Bunsen
Micro
No Dice? No Problem!
If you don't have a 6-sided die, use these "Energy Levels" cards. Cut them out, face down, and draw one each turn.
1
2
3
4
5
6
Stoichiometry Practice Exam Key v2 Mock Exam Master Key
Unit 8 DCA: Stoichiometry Practice Exam (11 Questions)
Answer Key
Conservation Sentences
1. is the same as
Molar Mass: Fe(NO\(_3\))\(_3\)
Value: 242 (or 241.88) | Unit: g/mol
Mole Ratio: Oxygen to Propane
Top: 5 mol O\(_2\) | Bottom: 1 mol C\(_3\)H\(_8\)
MC Calculation (CaO)
Correct Answer: C. 3.9 grams
Work: 2.2g CaO / 56.077 \(\times\) (1/1) \(\times\) 100.1 = 3.92g
Error Identification (KCl)
Correct Answer: C. Step 3—the mole ratio is wrong.
Explanation: The equation is 2K + Cl\(_2\) → 2KCl. The ratio should be 2 KCl / 1 Cl\(_2\). The student wrote 1/2.
Moles Needed (Mg + HCl)
Correct Answer: A. 5 moles Mg and 10 moles HCl
Yield discrepancy
Correct Answer: C. Limiting reagent controls theoretical yield... side reactions cause actual to vary.
Limiting Reactant (Equal Masses)
Correct Answer: A. Oxygen
Work: Need 3x more O\(_2\) by moles, but they have nearly equal moles by mass. O\(_2\) runs out first.
Setup Validation (Select TWO)
Correct Locations: Location 2 and Location 3
Yield Calculation (BaSO\(_4\))
75.99%
Math: (44.34 / 58.35) \(\times\) 100 = 75.989...
Particle Representation
Correct Image: C (12 circles)
Ratio 4K : 1O\(_2\). If flask has 3 molecules of O\(_2\), need 12 atoms of K.
Confidential Teacher Resource • Practice DCA Master Key • Standardized Mimic
Reaction Race Game Cards Plus Key Reaction Race Cards
Game Challenge Cards (1–12) & Answer Key
Game Component
Molar Mass
What is the molar mass of:
Mg(OH)2
Mg: 24.31 | O: 16.00 | H: 1.01
Molar Mass
What is the molar mass of:
K2SO4
K: 39.10 | S: 32.06 | O: 16.00
Mole Ratio
N2 + 3H2 → 2NH3
How many moles of H2 are needed to react with 2.0 moles of N2?
Mole Ratio
2H2 + O2 → 2H2O
How many moles of H2O are produced from 4.5 moles of O2?
Mass to Mass
2Mg + O2 → 2MgO
How many grams of MgO are produced from 24.31g of Mg?
Mg: 24.31 | MgO: 40.31
Mass to Mass
CH4 + 2O2 → CO2 + 2H2O
How many grams of CO2 are produced from 16.05g of CH4?
CH4: 16.05 | CO2: 44.01
Limiting
2H2 + O2 → 2H2O
If you start with 2 mol H2 and 2 mol O2, which is the Limiting Reactant?
Limiting
N2 + 3H2 → 2NH3
If you have 1 mol N2 and 2 mol H2, which is the Limiting Reactant?
Yield
Theoretical yield is 50.0g. You collect 40.0g in the lab.
What is the Percent Yield ?
Yield
A reaction should produce 10.0g, but you only get 9.5g.
What is the Percent Yield ?
Mixed
Mass to Mole:
How many moles are in 44.01g of CO2?
MM of CO2 = 44.01 g/mol
Mixed
Law of Conservation:
If 15g of Reactant A reacts with 10g of Reactant B, what is the total mass of the products?
Game Card Answer Key
Card 1 58.33 g/mol
Card 7 H2
Card 2 174.26 g/mol
Card 8 H2
Card 3 6.0 moles
Card 9 80.0%
Card 4 9.0 moles
Card 10 95.0%
Card 5 40.31 g MgO
Card 11 1.00 mole
Card 6 44.01 g CO2
Card 12 25 g
Teacher Note:
For larger groups, print two sets of cards or have students draw from different categories based on their space color. You can also allow students to "earn" a die roll by correctly solving a card while others are still playing!
Stoichiometry Master Slides Plus Practice v2 Stoichiometry Master
Unit 8 Review & Practice Blitz
The Golden Rule: Conservation
Key Concept
Matter cannot be created or destroyed. In a chemical reaction:
Total Atoms (Reactants) = Total Atoms (Products)
Total Mass (Reactants) = Total Mass (Products)
Watch Out!
Misconception: "The number of moles must be equal."
The Truth: Moles can change because atoms rearrange! Only mass and atom counts are strictly conserved.
PRACTICE: Conservation Round
Challenge 1
A reaction starts with 15g of reactant A and 25g of reactant B. If 30g of product C is formed, how much product D was also produced?
____ grams
Challenge 2
In the reaction \( 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \), there are 4 hydrogen atoms and 2 oxygen atoms on the reactant side. How many of each are on the product side?
H: ____ | O: ____
Molar Mass Calculations
The Logic
"Formula = Ingredients Count.
Distribute Those Subscripts!
Al\(_2\)(SO\(_4\))\(_3\)
Al: 2 \(\times\) 26.98
S: 3 \(\times\) 32.07
O: 12 \(\times\) 16.00
PRACTICE: Speed Round
Level 1
Na\(_2\)O
Na=22.99 | O=16.00
Level 2
Mg(NO\(_3\))\(_2\)
Mg=24.31 | N=14.01 | O=16.00
Boss Level
Al\(_2\)(CO\(_3\))\(_3\)
Al=26.98 | C=12.01 | O=16.00
The Mole Ratio Bridge
Coefficients are your EXCHANGE RATE .
1
mol N\(_2\)
↔
3
mol H\(_2\)
↔
2
mol NH\(_3\)
PRACTICE: Mole Ratio Blitz
CH\(_4\) + 2 O\(_2\) → CO\(_2\) + 2 H\(_2\)O
Challenge 1
Moles of O\(_2\) to CO\(_2\)
____ : ____
Challenge 2
Moles of CH\(_4\) to H\(_2\)O
____ : ____
Challenge 3
Moles of H\(_2\)O to O\(_2\)
____ : ____
Stoichiometry Review Task Cards Stoichiometry Review Task Cards
Phase-Based Review Stations
1
Phase 1: Molar Mass
Calculate the molar mass of Potassium Permanganate (\( \text{KMnO}_4 \)).
K=39.10 | Mn=54.94 | O=16.00
Unit 8 Stoichiometry
2
Phase 1: Conservation
A student reacts 24g of Magnesium with 16g of Oxygen. After the reaction, how many grams of Magnesium Oxide are produced?
Show the sum of reactants.
Unit 8 Stoichiometry
3
Phase 2: Mole Ratio
2 H\(_2\) + O\(_2\) → 2 H\(_2\)O
How many moles of Water are produced if 5.0 moles of Oxygen react completely?
Unit 8 Stoichiometry
4
Phase 2: Mole Ratio
N\(_2\) + 3 H\(_2\) → 2 NH\(_3\)
How many moles of Hydrogen are needed to produce 10 moles of Ammonia ?
Unit 8 Stoichiometry
5
Phase 3: Roadmap
How many grams of Sodium Chloride (MM=58.5) are produced from 10.0g of Sodium (MM=23.0)?
2 Na + Cl\(_2\) → 2 NaCl
Unit 8 Stoichiometry
6
Phase 3: Roadmap
Calculate the mass of Oxygen (MM=32.0) released when 50.0g of \( \text{KClO}_3 \) (MM=122.5) decomposes.
2 KClO\(_3\) → 2 KCl + 3 O\(_2\)
Unit 8 Stoichiometry
7
Phase 4: Limits
You have 4 moles of Iron and 4 moles of Oxygen . Which is the limiting reactant?
4 Fe + 3 O\(_2\) → 2 Fe\(_2\)O\(_3\)
Unit 8 Stoichiometry
8
Phase 4: Yield
Theoretical Yield = 150g
Calculate the Percent Yield.
Unit 8 Stoichiometry
Reaction Race Cheat Sheet Race Strategy Map
Mini Infographic Cheat Sheets for Game Cards 1–12
Student Guide
#1
The Parentheses Rule
Mg(OH)2
O × 2
H × 2
#2
Complex Molar Mass
K (39.10) × 2
S (32.06) × 1
O (16.00) × 4
Add them all up!
#3
The Ratio Factor
1
:
3
Need 3x more H2 than N2!
#4
Scaling Up
Input
1 O2
Output
2 H2O
#5
One-Mole Magic
Notice:
24.31g Mg
is exactly
1.0 Mole
#6
Mass Pathway
Mass A
Mole B
Mass B
If MM matches exactly, mol = 1!
#7
The "Recipe" Test
To use 2 mol O2, you would need 4 mol H2.
Have 2 < Need 4
#8
Yield Limits
1 mol N2
2 mol H2
Lower number = Limiting!
#9
Yield Efficiency
40
50
× 100 = ?%
#10
High Precision
95%
Think: 9.5 out of 10 is almost perfect...
#11
Mole Definition
Standard CO2
44.01 g = 1 mol
If you have exactly 44.01g, you have exactly 1 mole!
#12
Mass Balance
In 15g + 10g
=
Out ?? g
The Race Formula Bank
The Conversion Bridge
Grams ÷ MM Moles × Ratio Goal Moles
Efficiency Check
Yield % = (Actual / Theoretical) × 100
Actual is from the lab. Theoretical is from your math.
Atomic Mass Quick-Ref
Mg: 24.31
K: 39.10
S: 32.06
O: 16.00
H: 1.01
N: 14.01
C: 12.01
Cl: 35.45
Stoichiometry Task Cards Key Answer Key
Stoichiometry Task Cards Review
Teacher Resource
1
Molar Mass: KMnO\(_4\)
39.10 (K) + 54.94 (Mn) + [4 \(\times\) 16.00 (O)] = 158.04 g/mol
2
Conservation: Magnesium Oxide
24g (Mg) + 16g (O\(_2\)) = 40g MgO
3
Mole Ratio: 5.0 mol O\(_2\)
5.0 mol O\(_2\) \(\times\) (2 mol H\(_2\)O / 1 mol O\(_2\)) = 10.0 mol H\(_2\)O
4
Mole Ratio: 10 mol NH\(_3\)
10 mol NH\(_3\) \(\times\) (3 mol H\(_2\) / 2 mol NH\(_3\)) = 15 mol H\(_2\)
5
Roadmap: 10.0g Na to NaCl
(10.0g Na / 23.0) \(\times\) (2/2 ratio) \(\times\) 58.5 = 25.4 g NaCl
6
Roadmap: 50.0g KClO\(_3\) to O\(_2\)
(50.0g / 122.5) \(\times\) (3/2 ratio) \(\times\) 32.0 = 19.6 g O\(_2\)
7
Limits: 4 mol Fe & 4 mol O\(_2\)
4 mol Fe needs 3 mol O\(_2\) (ratio 4:3). We have 4 mol O\(_2\).
8
Yield: 120g Actual / 150g Theoretical
(120 / 150) \(\times\) 100 = 80% Yield
Unit 8 Stoichiometry Mastery • Task Card Solutions • Teacher Answer Key
Vocabulary Blitz Slides Vocabulary Blitz
The Fly Swatter Challenge
Speed Round
Mole Ratio
Stoichiometry
Actual Yield
Limiting Reactant
Molar Mass
Dimensional Analysis
Theoretical Yield
Excess Reactant
Percent Yield
Coefficients
Reactants
Products
Conservation of Mass
1
Two Students. Two Swatters.
2
Listen to the definition.
3
First to swat the word wins!
Teacher Call List
Call this:
"The reactant that runs out first and stops the reaction."
→ Limiting Reactant
Call this:
"The mass of one mole of a pure substance."
→ Molar Mass
Call this:
"The amount of product actually produced in the laboratory."
→ Actual Yield
Call this:
"The maximum amount of product that can be produced."
→ Theoretical Yield
Call this:
"A conversion factor that relates the amounts in moles of any two substances."
→ Mole Ratio
Call this:
"The efficiency of a reaction, calculated as a percentage."
→ Percent Yield
Call this:
"Matter cannot be created or destroyed in a chemical reaction."
→ Conservation of Mass
Call this:
"The mathematical method used for unit conversions."
→ Dimensional Analysis
Vocabulary Blitz Slides Refined Vocabulary Blitz
The Fly Swatter Challenge
Speed Round
Mole Ratio
Stoichiometry
Actual Yield
Limiting Reactant
Molar Mass
Dimensional Analysis
Theoretical Yield
Excess Reactant
Percent Yield
Coefficients
Reactants
Products
Conservation of Mass
1
Two Students.
2
Listen for definition.
3
Swat to Win!
Teacher Call List
Definition 1:
"The reactant that runs out first and stops the reaction."
→ Limiting Reactant
Definition 2:
"The mass of one mole of a pure substance."
→ Molar Mass
Definition 3:
"The amount of product actually produced in lab."
→ Actual Yield
Definition 4:
"A factor relating moles of any two substances."
→ Mole Ratio
Definition 5:
"The efficiency of a reaction as a percentage."
→ Percent Yield
Definition 6:
"The math method used for unit conversions."
→ Dimensional Analysis
Stoichiometry Mock Exam DCA Mimic v3 Chemistry Assessment Portfolio
IISD Unit 8 DCA: Stoichiometry Mock Exam
Name:
Date:
Exam Reference: Balance equation first. Molar mass = Atomic mass sum (distribute subscripts). Ratio = Bridge coefficients. % Yield = (Actual/Theoretical) × 100.
(C.9A) Choose the words to finish the sentences.
All chemical equations adhere to the law of conservation of mass. According to this law, the number of atoms on the reactant side is the same as is different than the number of atoms on the product side.
This means that the total mass of reactants is the same as is different than the total mass of products.
The total amount of moles in the reactants will always be the same may vary since atoms rearrange to form new products.
(C.8A) Calculate the molar mass of Fe(NO\(_3\))\(_3\).
Provide the numerical value and correct unit (g/mol, g, mol).
Numerical Value
___________
Correct Unit
______
(C.9C) Consider the balanced equation for propane combustion:
C\(_3\)H\(_8\) + 5 O\(_2\) → 3 CO\(_2\) + 4 H\(_2\)O
Construct the correct stoichiometric ratio of Oxygen molecules to Propane molecules .
Top
Bottom
(C.9C) Law of conservation of mass reaction:
CaO(s) + CO\(_2\)(g) → CaCO\(_3\)(s)
How much CaCO\(_3\) (100.1 g/mol) will be formed if 2.2 grams of CaO (56.077 g/mol) are used? Assume excess CO\(_2\). Round to 2 significant figures.
A 1.1 grams
B 2.2 grams
C 3.9 grams
D 4.3 grams
(C.9C) PROCESS EVALUATION
Producing KCl from 4.0g of Cl\(_2\) gas.
4.0 g Cl\(_2\) × 1 mol Cl\(_2\) / 70.9 g Cl\(_2\) × 1 mol KCl / 2 mol Cl\(_2\)
Step 1 Step 2 Step 3
Which step in the student's setup is incorrect ?
A Step 1—the starting mass is wrong.
B Step 2—the molar mass is wrong.
C Step 3—the mole ratio is wrong.
D Not enough information is provided.
(C.9C) BALANCED REACTION: Mg(s) + 2 HCl(aq) → MgCl\(_2\)(aq) + H\(_2\)(g)
In order to produce 5 moles of H\(_2\)(g) , how many moles of Mg and HCl are required by the reaction?
Vocabulary Blitz Final Slides Vocabulary Blitz
The Fly Swatter Challenge
Speed Round
Mole Ratio
Stoichiometry
Actual Yield
Limiting Reactant
Molar Mass
Dimensional Analysis
Theoretical Yield
Excess Reactant
Percent Yield
Coefficients
Reactants
Products
Conservation of Mass
1
Two Students.
2
Listen for definition.
3
Swat to Win!
The Call List (Key)
"Study of quantitative relationships in chemical reactions."
→ Stoichiometry
"The reactant that runs out first and stops the reaction."
→ Limiting Reactant
"The mass of one mole of a pure substance."
→ Molar Mass
"Amount of product actually produced in the laboratory."
→ Actual Yield
"The calculated maximum amount of product possible."
→ Theoretical Yield
"The mathematical method used for unit conversions."
→ Dimensional Analysis
"Matter is neither created nor destroyed in a reaction."
→ Conservation of Mass
"Conversion factor relating the moles of two substances."
→ Mole Ratio
"Efficiency of a reaction, calculated as a percentage."
→ Percent Yield
"Reactant that is not used up completely in a reaction."
→ Excess Reactant
"Numbers in front of formulas in a balanced equation."
→ Coefficients
"Starting substances in a chemical reaction."
→ Reactants
"Substances formed during a chemical reaction."
→ Products
Stoichiometry Skill Builder Roadmap Refined Stoichiometry Skill Builder
Unit 8: Quantitative Chemistry Review & Practice
Name:
Date:
Phase 1: Foundations
1. Conservation of Mass: Complete the missing values.
Reaction A: \( \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \)
Reactants: 16.0g \( \text{CH}_4 \) + 64.0g \( \text{O}_2 \)
Products: 44.0g \( \text{CO}_2 \) + ________ g \( \text{H}_2\text{O} \)
Reaction B: \( 2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO} \)
Reactants: ________ g Mg + 16.0g \( \text{O}_2 \)
Products: 40.3g MgO
2. Molar Mass Mission: Calculate the molar mass (g/mol) for each compound.
a) Sodium Phosphate: \( \text{Na}_3\text{PO}_4 \)
b) Calcium Hydroxide: \( \text{Ca(OH)}_2 \)
Phase 2: The Mole Bridge
Reaction: \( \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \)
Ratio of \( \text{H}_2 \) to \( \text{NH}_3 \):
Ratio of \( \text{N}_2 \) to \( \text{H}_2 \):
3. Mole-to-Mole Walkthrough
How many moles of \( \text{NH}_3 \) can be produced from 4.5 moles of \( \text{H}_2 \)?
4.5 mol H\(_2\) ×
___ mol NH\(_3\)
___ mol H\(_2\)
= ___ mol NH\(_3\)
Phase 3: The Roadmap
4. Stoichiometry Roadmap Set
Phase 3 Master Strategy
!
Scenario: Propane Combustion Challenge
"A student burns 50.0 grams of Propane (\( \text{C}_3\text{H}_8 \)) in excess oxygen.Carbon Dioxide (\( \text{CO}_2 \)) produced?"
C\(_3\)H\(_8\) + 5 O\(_2\) → 3 CO\(_2\) + 4 H\(_2\)O
Complete the 3-step conversion grid below
50.0g C\(_3\)H\(_8\)
Given Mass
×
1 mol C\(_3\)H\(_8\)
44.1 g
1. Moles of A
×
3 mol CO\(_2\)
1 mol C\(_3\)H\(_8\)
2. Mole Ratio
×
44.0 g
1 mol CO\(_2\)
3. Grams of B
=
____ g
Final Mass
Standard Road Map Set
Apply Strategy Solo:
"Scenario: 10.0g of Magnesium reacts with excess HCl. Calculate the mass of Hydrogen gas produced."
[ Setup your conversion set here ]
Phase 4: Yield & Limits
Vocabulary Blitz Teacher Script Fly Swatter Game Script
Teacher Call List & Gameplay Guide
Teacher Resource
How to Facilitate
The Setup: Project the "Vocabulary Blitz" slide. Divide the class into two teams. One representative from each team comes to the board, each holding a fly swatter.
The Action: Read one definition from the list below. The first student to "swat" the correct word on the screen wins a point for their team.
Lenny's Pro-Tip:
"Don't just read the words—read the 'Sneaky Scenarios' in Round 2 to see if they truly understand the concept rather than just memorizing a sentence!"
Round 1: Rapid-Fire Definitions
Teacher "Call" Statement Target Word "The study of quantitative relationships in a chemical reaction." Stoichiometry "The numbers in front of chemical formulas that show relative amounts." Coefficients "A conversion factor that relates moles of any two substances." Mole Ratio "The reactant that runs out first and determines how much product forms." Limiting Reactant "The reactant that remains after a reaction has stopped." Excess Reactant "The calculated maximum amount of product that can be made." Theoretical Yield "The amount of product measured by a scientist in the laboratory." Actual Yield "The mass in grams of exactly 6.02 × 1023 particles." Molar Mass
Round 2: Sneaky Scenarios
"You have 4 hamburger patties and 10 buns. The patties are the..."
Limiting Reactant
"The math says you should make 100g, but you spilled some and only got 80g. The 100g is the..."
Theoretical Yield
"This ensures the mass of your beakers at the start equals the mass at the finish."
Conservation of Mass
"I use the periodic table to find this number so I can convert grams to moles."
Molar Mass
"The 'Goal' of your math setup, used to describe the reaction's efficiency."
Percent Yield
End of Script • Unit 8 Stoichiometry Mastery
Stoichiometry Quest Mission Briefing Mission Briefing
Reaction Race & Stoichiometry Quest
Mastery
Essential Question
"How do chemical recipes allow us to predict the quantitative outcome of a reaction, and why is the result rarely perfect?"
Learning Objectives
01
Quantitative Analysis
Analyze balanced chemical equations to determine precise mole ratios between any two substances in a reaction.
02
Unit Transformation
Execute multi-step conversions using dimensional analysis to transition between grams and moles for reactants and products.
03
Reaction Evaluation
Identify limiting reactants based on available quantities and calculate the percent yield to evaluate the efficiency of a chemical process.
Success Criteria
I can calculate molar mass for complex formulas.
I can use coefficients as a bridge between chemicals.
I can convert mass to mass in 3 distinct steps.
I can prove mass is conserved in any reaction.
I can explain why the limiting reactant stops a reaction.
I can calculate percent yield from lab data.
Unit 8: Stoichiometry Mastery
Competitive Review Interactive Study
Reaction Race Lab Journal Reaction Race Journal
Game Documentation & Calculation Log
Name:
Date:
Scientist's Instructions: लैंडिंग परLand on a challenge space? Draw a card! Record the Card Category and Number below. Show your full setup (Molar Mass math or Dimensional Analysis) in the box to "earn" your stay on that space.
Card Type:
Final Answer:
Card Type:
Final Answer:
Card Type:
Final Answer:
Card Type:
Final Answer:
Card Type:
Final Answer:
Card Type:
Final Answer:
Card Type:
Final Answer:
Card Type:
Final Answer:
Victory Log
Reaction Efficiency:
Limiting Concept:
Summary: Explain one connection you made between the game's obstacles (spills, breaks) and real-life reasons why percent yield is rarely 100%.
Reaction Race Lab Journal Refined Reaction Race Journal
Game Documentation & Calculation Log
Name:
Date:
Scientist's Instructions: If you land on a challenge space, draw a card and record the Card Category and Card Number below. Show your full calculation setup (Molar Mass or Dimensional Analysis) in the workspace provided to confirm your move.
Card Category:
Card #
Final Answer:
Card Category:
Card #
Final Answer:
Card Category:
Card #
Final Answer:
Card Category:
Card #
Final Answer:
Card Category:
Card #
Final Answer:
Card Category:
Card #
Final Answer:
Card Category:
Card #
Final Answer:
Card Category:
Card #
Final Answer:
Victory Log
Reaction Efficiency:
Limiting Concept:
Summary & Reflection
Explain one connection you made between the game's obstacles (spills, breaks) and real-world reasons why percent yield is rarely 100%.
Reaction Race Reflection Key Reflection Scoring Guide
Suggested Answers for Reaction Race Journal
Teacher Key
Prompt 1: Reaction Efficiency
Expected Concept:
Students should define "Percent Yield" or the comparison of actual vs. theoretical results.
Ideal Student Answer:
"Reaction efficiency is measured by Percent Yield . It tells us what percentage of the mathematical maximum (theoretical yield) we actually produced in the real world."
Prompt 2: Limiting Concept
Expected Concept:
Students should explain that one reactant determines the "stop point" of the reaction.
Ideal Student Answer:
"The Limiting Reactant is the ingredient that runs out first. Even if you have plenty of the other reactant (the excess), the whole process has to stop as soon as that one substance is gone."
Prompt 3: The Big Connection
"Explain one connection between the game's obstacles and real-life reasons why yield is rarely 100%."
A
The "Spill" Connection: Just like the "Acid Spill" space in the game forced me to lose progress, real lab technicians might spill a tiny amount of liquid during a transfer between beakers, which reduces the mass of product they can collect.
B
The "Glassware" Connection: Landing on "Glassware Break" represents equipment failure. In a real lab, if a flask cracks or is dirty, the chemical reaction might not finish correctly, or impurities might get mixed in, making the actual yield lower than the math said it would be.
C
The "Catalyst" Connection: The "Catalyst" space allowed me to move faster. In real chemistry, a catalyst doesn't change the theoretical yield, but it lowers the energy needed so the reaction happens more efficiently and reaches the product stage faster.
Unit 8 • Reaction Race Scoring Guide
