Roadmap Mastery Answer Key ROADMAP MASTERY
Teacher Answer Key
Reference: Stoichiometry Fundamentals
1
Constructing the Master Blueprint
Pit Stop MASS (A)
(Grams)
Action
Molar Mass
Pit Stop MOLES (A)
\(n\)
The Bridge
Mole Ratio
Pit Stop MOLES (B)
\(n\)
Action
Molar Mass
Pit Stop MASS (B)
(Grams)
Note: The "Bridge Actions" use conversion factors from the Periodic Table (Molar Mass) or Balanced Equation (Mole Ratio).
2
Planning Your Route
Scenario Description Starting Pit Stop Destination Total Steps You have 5.0 grams of Oxygen and need to find the grams of Water produced. Mass A Mass B 3 You are given 2.5 moles of Magnesium and want to find how many moles of MgO form. Moles A Moles B 1 You have 10.0 grams of Iron and need to find the moles of Iron(III) Oxide produced. Mass A Moles B 2
3
The Critical Junction
Answer Guide:
Mass is not a direct way to compare substances because different atoms/molecules have different weights (molar masses). You cannot say "1 gram reacts with 1 gram" in chemistry because the number of particles will be different. The mole acts as the "universal currency" or counting unit. The balanced chemical equation only gives ratios in terms of particles (moles), never in terms of mass. Therefore, we must convert to moles to use the coefficients from the equation.
4
Navigating the Full Journey
The Reaction:
2 H2 (g) + O2 (g) → 2 H2O (l)
MISSION: If you start with 12.0 grams of H2, how many grams of H2O can be produced?
Step 1: Map your route
Mass H2
Moles H2
Moles H2O
Mass H2O
Step 2: Execute the Calculation (Dimensional Analysis)
<table class="text-xl font-roboto-slab text-blue-800"><tbody><tr><td class="px-4 py-2 text-center border-r-2 border-blue-300">12.0 g H<sub>2</sub></td><td class="px-4 py-2 text-center border-r-2 border-blue-300">1 mol H<sub>2</sub></td><td class="px-4 py-2 text-center border-r-2 border-blue-300">2 mol H<sub>2</sub>O</td><td class="px-4 py-2 text-center">18.02 g H<sub>2</sub>O</td></tr><tr class="border-t-4 border-blue-300"><td class="px-4 py-2 text-center border-r-2 border-blue-300"></td><td class="px-4 py-2 text-center border-r-2 border-blue-300">2.02 g H<sub>2</sub></td><td class="px-4 py-2 text-center border-r-2 border-blue-300">2 mol H<sub>2</sub></td><td class="px-4 py-2 text-center">1 mol H<sub>2</sub>O</td></tr></tbody></table>
Final Destination:
107.05 g H2O
5
Post-Trip Reflection
Common "Road Hazards":
Forgetting to use the mole ratio from the coefficients.
Inverting molar mass (e.g., putting g on top instead of bottom).
Rounding too early in the steps.
Identifying A vs B:
Substance "A" is the "Given"—it's the value with a known number. Substance "B" is the "Goal" or "Target"—it's what the question is asking you to find.
Reaction Roadmap Slides Reaction Roadmap
NAVIGATING THE STOICHIOMETRY FACTORY
What is Stoichiometry?
It is the "recipe" of chemistry. It allows us to predict how much product we can make from a given amount of reactant.
"If I have 50g of Iron, how many grams of Iron(III) oxide will form?"
To answer this, we need a map to convert between mass and moles across different substances.
Conservation of Mass
Matter is neither created nor destroyed, only rearranged into new substances.
The Stoichiometry Blueprint
Substance A MASS (g)
/ Molar Mass
Substance A MOLES
Mole Ratio (from Equation)
Substance B MOLES
* Molar Mass
Substance B MASS (g)
The Mole Ratio is the only way to "cross the bridge" from one substance to another.
Limiting Reactants
The Show-Stopper
The Limiting Reactant is the reactant that is completely consumed first. It dictates exactly how much product you can make.
The Excess Reactant
The reactant that is "left over" after the reaction stops.
The Sandwich Analogy
Recipe: 2 Slices Bread + 1 Slice Cheese = 1 Sandwich
If you have 10 slices of bread and 2 slices of cheese...
Cheese is Limiting.
Efficiency: Percent Yield
Theoretical Yield
The maximum amount of product predicted by stoichiometry (the "Perfect World" amount).
Actual Yield
The amount you actually measured in the lab (the "Real World" amount).
Calculated Percentage
Actual Yield
Theoretical Yield
× 100%
Stoichiometry Circuit
Rotate through 5 stations to master the roadmap logic.
1
Roadmap & Vocab
Build the blueprint and define key terms.
2
Pathfinder Scenarios
Map the path for mass and mole conversions.
3
Logic & Yield Lab
Predict limiting factors and analyze errors.
4
Molar Mass Match
Calculate masses for H₂O, CO₂, and NaCl.
5
Yield Efficiency
Calculate percentages from lab data.
Record all findings on your Station Recording Sheet
Roadmap Mastery Think Sheet ROADMAP MASTERY
Stoichiometry Think Sheet
NAME:
DATE:
1
Constructing the Master Blueprint
Before we can navigate a reaction, we need the map. Fill in the missing "Pit Stops" (units) and "Bridge Actions" (what tool/value you use to convert) in the diagram below.
Pit Stop MASS (A)
Action
Pit Stop MOLES (A)
The Bridge
Pit Stop MOLES (B)
Action
Pit Stop MASS (B)
2
Planning Your Route
Read each scenario below. Identify the STARTING POINT , the DESTINATION , and the NUMBER OF STEPS (conversions) needed to get there.
Scenario Description Starting Pit Stop Destination Total Steps You have 5.0 grams of Oxygen and need to find the grams of Water produced. You are given 2.5 moles of Magnesium and want to find how many moles of MgO form. You have 10.0 grams of Iron and need to find the moles of Iron(III) Oxide produced.
3
The Critical Junction
Concept Check: Why can we NEVER cross directly from "Mass of Substance A" to "Mass of Substance B" without going through "Moles" first? What is special about the mole that makes it the only way to compare two different substances?
4
Navigating the Full Journey
The Reaction:
2 H2 (g) + O2 (g) → 2 H2O (l)
"Two moles of hydrogen gas react with one mole of oxygen gas to produce two moles of water."
MISSION: If you start with 12.0 grams of H2, how many grams of H2O can be produced?
Step 1: Map your route
Mass H2
Pit Stop
Pit Stop
Mass H2O
Step 2: Execute the Calculation (Dimensional Analysis)
Start Value
Final Destination:
Ans:
5
Post-Trip Reflection
What is one "Road Hazard" (common mistake) that usually happens when crossing the mole bridge?
How do you know which substance is "A" and which is "B" when reading a word problem?
How confident are you navigating the Roadmap?
Stoichiometry Sort Cards Station 1: Roadmap & Vocab
SET 1
Mass (Grams)
Moles (A)
Moles (B)
Mass (Grams) B
Divide by Molar Mass A
Mole Ratio (B/A)
Multiply by Molar Mass B
Limiting Reactant
The reactant that runs out first and stops the reaction.
Theoretical Yield
The maximum amount of product predicted by calculation.
Actual Yield
The amount of product actually measured in the lab.
Station 2: Pathfinder
SET 2
"How many moles of O₂ are in 32.0 grams?"
START: Mass A → (/MM A) → END: Moles A
"What is the mass of 2.0 moles of H₂O?"
START: Moles A → (*MM A) → END: Mass A
"Calculate grams of product from grams of reactant."
Mass A → Moles A → Moles B → Mass B
"Find moles of B given moles of A."
START: Moles A → (Ratio) → END: Moles B
"How many moles of product from 50g of reactant?"
Mass A → Moles A → Moles B
"How many grams of A needed for 3 moles of B?"
Moles B → Moles A → Mass A
Station 3: Reactant & Yield Lab
SET 3
10 mol A, 10 mol B. Ratio 2A:1B.
A is Limiting
5 mol A, 5 mol B. Ratio 1A:1B. All B is gone.
B is Limiting
Student spills product before weighing.
Lower Actual Yield
Product is weighed while still damp.
Higher Apparent Mass
Impure reactants (sand mixed in).
Reduced Theoretical Yield
Station 4: Molar Mass Match
SET 4
H₂O
2(1.01) + 16.00
18.02 g/mol
CO₂
12.01 + 2(16.00)
44.01 g/mol
NaCl
22.99 + 35.45
58.44 g/mol
Station 5: Percent Yield Math
SET 5
Actual: 80.0g Theoretical: 100.0g
80.0%
Actual: 25.0g Theoretical: 50.0g
50.0%
Actual: 45.0g Theoretical: 50.0g
90.0%
Stoic Expedition Assessment Challenge STOIC EXPEDITION
Mastery Stoichiometry Assessment
NAME:
DATE:
1
Navigation Check
1. Which tool allows you to convert from substance A to substance B?
Molar Mass
Mole Ratio
2
Route Length
2. How many conversion steps are needed for Grams → Grams?
1 Step
3 Steps
MISSION 1: [Mole → Mole]
C3H8 + 5 O2 → 3 CO2 + 4 H2O . If you burn 2.5 moles of C3H8 , how many moles of O2 are needed?
Ans: ________ mol O2
MISSION 2: [Mass → Mass]
N2 + 3 H2 → 2 NH3 . If 56.0g of N2 reacts, how many grams of NH3 are produced?
Masses: N2=28.02 | NH3=17.03
Ans: ________ g NH3
MISSION 3: [Mass → Mole]
2 Mg + O2 → 2 MgO . React 12.15g of Mg . Moles of MgO formed?
Mass: Mg=24.30
Ans: ________ mol MgO
Mission 4: [Mole → Mass]
2 Na + Cl2 → 2 NaCl . Starting with 0.50 moles of Cl2 , how many grams of NaCl form?
NaCl = 58.44 g/mol
Ans: ________ g NaCl
Mission 5: The Carbon Cycle
6 CO2 + 6 H2O → C6H12O6 + 6 O2 . Consume 88.0g of CO2 . What is the theoretical yield of Glucose (C6H12O6) ?
CO2 = 44.01 | C6H12O6 = 180.16
Theoretical: ________ g Glucose
Mission 6: Actual Lab Yield
If the plant in Mission 5 actually produced only 50.0g of glucose, calculate the Percent Yield below.
%
Mission 7: Airbag Deployment [Mass → Volume]
2 NaN3 → 2 Na + 3 N2 . Use 130.0g of NaN3 . How many liters of N2 gas are produced at STP?
NaN3 = 65.01 | 1 mol gas at STP = 22.4 L
Volume: ________ L N2
3
Hazardous Territory
Mission 8: Resource Scarcity (LR Analogy)
"10 bread slices + 2 cheese slices (2:1 ratio)" Max Output: ____ Sandwiches LR: __________
Mission 9: 2 Al + 3 Cl2 → 2 AlCl3
React
Stoic Expedition Tool Kit Challenge Stoic Expedition
Reaction Base Mat
UNIVERSAL WORKSPACE
START
1
STEP 1
STEP 1
BRIDGE
MOLE RATIO
RATIO
MOLE RATIO
STEP 3
STEP 3
=
FINAL DESTINATION
Yield Calculator
x 100 =
%
Limiting Logic Check
Path 1 Max Product
Path 2 Max Product
Smallest Result Wins
Expedition Tiles: Set A
M1: Propane | M2: Haber | M3: Mg | M4: Salt
M1: Propane [Mole-Mole]
2.5 mol C3H8
5 mol O21 mol C3H8
12.5 mol
M3: Magnesium [Mass-Mole]
12.15g Mg
1 mol Mg24.30g Mg
2 mol MgO2 mol Mg
0.50 mol
M2: Haber Process [Mass-Mass Journey]
56.0g N2
1 mol N2
28.02g N2
2 mol NH3
1 mol N2
17.03g NH3
1 mol NH3
68.1g
Expedition Tiles: Set B
M5: Carbon | M7: Airbags | M9: AlCl3 | M10: Scrubber
M5: CO2 Theoretical Yield
88.0g CO2
1 mol C6H12O66 mol CO2
60.1g
83.3%
M7: Airbag Volume (STP)
130.0g NaN3
22.4 L N21 mol N2
3 mol N22 mol NaN3
67.2 L
M10: The Ultimate Space Scrubber Challenge
Path 1 (LiOH → H2O)
48.0g LiOH
1 mol H2O
2 mol LiOH
18.06g H2O
Path 2 (CO2 → H2O)
44.0g CO2
1 mol H2O
1 mol CO2
18.02g H2O
LIMITING CO2
MAX YIELD 18.02g
Teacher Dispatch
This Tool Kit is specifically balanced for the **Stoic Expedition Assessment**. Set B tiles for M10 are intentionally tricky—students must compare the results of both paths to find the smallest value.
Stoichiometry Sort Teacher Guide Stoichiometry Stations
Teacher Setup & Answer Key
Full Lesson
Station Management
This activity is designed for a 50-60 minute period. Divide students into groups of 3-4. Each group starts at a station and rotates every 8-10 minutes.
Station 1:
Station 2:
Station 3:
Station 4:
Station 5:
Station Answer Keys
Station 1: Roadmap
1. Divide by Molar Mass A
2. Mole Ratio (B/A)
3. Multiply by Molar Mass B
Station 2: Paths
Mass to Moles: Mass A → Moles A
Mass to Mass: Mass A → Mol A → Mol B → Mass B
Station 3: Logic Lab
10 mol A, 10 mol B (2A:1B) → A is Limiting
5 mol A, 5 mol B (1A:1B) → B is Limiting
Damp product → Higher Apparent Mass
Impure reactants → Lower Theoretical Yield
Station 4: Molar Mass
H₂O: 18.02 | CO₂: 44.01 | NaCl: 58.44
Station 5: Percent Yield
80/100: 80% | 25/50: 50% | 45/50: 90%
Mastery Practice Key
1. Mole-Mole (NH₃ from 9 mol H₂):
9.0 mol H₂ × (2 mol NH₃ / 3 mol H₂) = 6.0 moles NH₃
2. Mass-Mole (CO₂ from 32g CH₄):
32.0g CH₄ / 16.04g/mol = 1.995 mol CH₄ × (1/1 ratio) = ~2.00 moles CO₂
3. Mole-Mass (O₂ from 4 mol KClO₃):
4.0 mol KClO₃ × (3/2 ratio) = 6.0 mol O₂ × 32.00g/mol = 192.0 grams O₂
4. Limiting (Fe + S):
Fe: 27.9g/55.85 = 0.50 mol
5. Theo. Yield:
0.50 mol Fe × (1/1 ratio) × 87.91g/mol = 43.95 grams FeS
Yield Concept Checklist
Low Yield Factors (<100%)
Transfer loss (spilling, left in beaker)
Incomplete reaction (insufficient time)
Side reactions (forming byproducts)
Impure reactants (lower active mass)
High Yield Errors (>100%)
Damp/Wet product (water adds mass)
Contamination from atmosphere
Scale calibration errors
Incomplete washing of precipitate
Station Recording Worksheet Station Recording Sheet
Stoichiometry Sort Activity
Name:
Date:
Period:
Station 1: Roadmap & Vocab
List the 3 "Bridges" in order:
1. ____________________________
Define "Limiting Reactant":
Station 2: Pathfinder
Sketch the path for the following two scenarios from the cards:
Scenario 1 (Mass to Moles):
Start: __________ → __________ → End: __________
Scenario 2 (Mass to Mass):
Start: __________ → __________ → __________ → __________
Station 3: Logic Lab
Identify the limiting reactant if you have 10 mol A and 10 mol B (Ratio 2A:1B):
Answer: ____________________
Explain the yield impact of "damp/wet product":
Impact: ____________________
Station 4: Molar Mass
H₂O:
CO₂:
NaCl:
Station 5: Yield Math
Actual 80g / Theo 100g:
Actual 25g / Theo 50g:
Actual 45g / Theo 50g:
Mixed Mastery Practice
TYPE: MOLE-MOLE N₂ + 3 H₂ → 2 NH₃
1. How many moles of NH₃ can be produced from 9.0 moles of H₂ gas?
Answer: ______________ mol NH₃
TYPE: MASS-MOLE CH₄ + 2 O₂ → CO₂ + 2 H₂O
2. If you burn 32.0 grams of CH₄, how many moles of CO₂ will be produced? (MM: CH₄ = 16.04 g/mol)
Answer: ______________ mol CO₂
TYPE: MOLE-MASS 2 KClO₃ → 2 KCl + 3 O₂
3. How many grams of O₂ gas are produced by the decomposition of 4.0 moles of KClO₃? (MM: O₂ = 32.00 g/mol)
Answer: ______________ g O₂
TYPE: LIMITING & THEORETICAL YIELD Fe + S → FeS
Molar Masses: Fe = 55.85 g/mol | S = 32.06 g/mol | FeS = 87.91 g/mol
A chemist reacts 27.9 grams of Fe with 32.0 grams of S .
4. Determine the limiting reactant:
Limiting Reactant: ______________
5. Calculate the theoretical yield of FeS (in grams):
Theo. Yield: ______________ g FeS
Stoic Expedition Answer Key Challenge EXPEDITION KEY
Stoic Expedition Master Solutions
Assessment v1.0
Q1: The Bridge
Mole Ratio (Coefficients)
Q2: Route Length
3 Steps (G→M→M→G)
MISSION 1: [Mole-Mole]
\(2.5 \text{ mol C}_3\text{H}_8 \times \frac{5 \text{ mol O}_2}{1 \text{ mol C}_3\text{H}_8} = \mathbf{12.5 \text{ mol O}_2}\)
MISSION 2: [Mass-Mass]
\(56.0 \text{ g N}_2 \times \frac{1 \text{ mol N}_2}{28.02 \text{ g}} \times \frac{2 \text{ mol NH}_3}{1 \text{ mol N}_2} \times \frac{17.03 \text{ g}}{1 \text{ mol}} = \mathbf{68.1 \text{ g NH}_3}\)
MISSION 3: [Mass-Mole]
\(12.15 \text{ g Mg} \times \frac{1 \text{ mol Mg}}{24.30 \text{ g}} \times \frac{2 \text{ mol MgO}}{2 \text{ mol Mg}} = \mathbf{0.50 \text{ mol MgO}}\)
MISSION 4: [Mole-Mass]
\(0.50 \text{ mol Cl}_2 \times \frac{2 \text{ mol NaCl}}{1 \text{ mol Cl}_2} \times \frac{58.44 \text{ g}}{1 \text{ mol}} = \mathbf{58.44 \text{ g NaCl}}\)
Mission 5: Photo Theory
\(88.0 \text{g CO}_2 = 2.0 \text{ mol}\)
\(2.0 \times (1/6) \times 180.16 = \mathbf{60.05 \text{ g}}\)
Mission 6: % Yield
(50.0 / 60.05) x 100 = 83.3%
Mission 7: Airbag Volume [Mass-Volume]
130.0g NaN3 / 65.01 = 1.999 mol NaN3
Mission 8: Analogy Key
Max Output: 2 Sandwiches | LR: Cheese
Mission 9: AlCl3 LR Key [Mole-Mole]
Path 1 (Al): 4.0 mol Al → 4.0 mol AlCl3
Path 2 (Cl2): 3.0 mol Cl2 → 2.0 mol AlCl3
LR Verdict: Chlorine (Cl2)
Ultimate Mission 10 Solution
Path 1: LiOH Journey
48.0g / 23.95 = 2.004 mol LiOH
Path 2: CO2 Journey
44.0g / 44.01 = 0.999 mol CO2
Mission Verdict:
CO2 is Limiting
Max Yield:
18.0
Grams H2O
M1-M4 (Basic)
10 pts each
M5-M9 (Inter)
10 pts each
M10 (Master)
20 pts
Stoic Expedition Tool Kit Challenge Stoic Expedition
Reaction Base Mat
UNIVERSAL WORKSPACE
START
1
STEP 1
STEP 1
BRIDGE
MOLE RATIO
RATIO
MOLE RATIO
STEP 3
STEP 3
=
FINAL DESTINATION
Yield Calculator
x 100 =
%
Limiting Logic Check
Path 1 Result
Path 2 Result
Smallest Result Wins
Expedition Tiles: Set A
M1: Propane | M2: Haber | M3: Mg | M4: Salt | M5: Photo
M1: Propane
2.5 mol C3H8
5 mol O2 / 1 mol C3H8
12.5 mol
M3: Magnesium
12.15g Mg
1 mol / 24.30g
2 mol MgO / 2 mol Mg
M4: Salt
0.50 mol Cl2
2 mol NaCl / 1 mol Cl2
58.44g / 1 mol
M2: Haber Journey
56.0g N2
1 mol N228.02g N2
2 mol NH31 mol N2
17.03g NH31 mol NH3
M5: Photosynthesis Theory
88.0g CO2
1 mol CO244.01g CO2
1 mol Glu6 mol CO2
180.16g Glu1 mol Glu
Expedition Tiles: Set B
M7: Airbags | M9: AlCl3 | M10: Scrubber | Logic Tiles
M7: Airbag Deployment [Mass-Volume]
130.0g NaN3
1 mol NaN3
65.01g NaN3
3 mol N2
2 mol NaN3
22.4 L N2
1 mol N2
M9: AlCl3 Mole-Mole LR
4.0 mol Al
3.0 mol Cl2
2 mol AlCl32 mol Al
2 mol AlCl33 mol Cl2
M10: Space Scrubber LR
Path 1 (LiOH): 18.06g
Path 2 (CO2): 18.02g
Status Verdict & Yield Cards
LIMITINGCHEESE
LIMITINGCl2
LIMITINGCO2
MAX YIELD18.02g
MAX YIELD2.0 mol
50.0g
60.1g
83.3%
67.2 L
58.44g
Teacher Dispatch
This Tool Kit is specifically balanced for the **Stoic Expedition Assessment**. Page 3 includes the specialized cards for **Airbag Volume (M7)** and the **Space Scrubber LR Challenge (M10)**. Ensure students compare Path 1 and Path 2 on the base mat to find the true Limit.
Mole Mission Quiz Challenge Expanded MOLE MISSION
Ultimate Mastery Assessment
NAME:
DATE:
1
Roadmap Logic
1. Which "bridge" connects substance A to substance B?
Molar Mass
Mole Ratio
2
Route Planning
2. How many steps for Mass (A) → Mass (B)?
1 Step
3 Steps
MISSION 1: [Mole → Mole]
Burn 2.5 moles of Propane (C3H8) in C3H8 + 5 O2 → 3 CO2 + 4 H2O . Moles of O2 consumed?
Ans: ________ mol O2
MISSION 2: [Mass → Mass]
N2 + 3 H2 → 2 NH3 . If 56.0g of N2 react, how many grams of NH3 are produced?
Masses: N2=28.02 | NH3=17.03
Ans: ________ g NH3
MISSION 3: [Mass → Mole]
2 Mg + O2 → 2 MgO . Burn 12.15g of Mg . Moles of MgO formed?
Mass: Mg=24.30
Ans: ________ mol MgO
Mission 4: [Mole → Mass]
2 Na + Cl2 → 2 NaCl . From 0.50 moles of Cl2 , how many grams of Salt (NaCl) form?
Molar Mass: NaCl = 58.44 g/mol
Ans: ________ g NaCl
Mission 5: Photosynthesis Yield [Mass → Mass]
6 CO2 + 6 H2O → C6H12O6 + 6 O2
A plant consumes 88.0 grams of CO2 . What is the theoretical yield of Glucose (C6H12O6) in grams?
Masses: CO2 = 44.01 | C6H12O6 = 180.16
Theoretical: ________ g Glucose
MISSION 6
6. Actual Lab Results
If the plant in Mission 5 actually produced only 50.0 grams of glucose , calculate the Percent Yield based on your answer above.
Percent Yield = (Actual / Theoretical) x 100
%
3
Limiting Hazards
Mission 7: The "Universal Bridge" Analogy
You have 10 slices of bread and 2 slices of cheese . Each sandwich requires 2 bread + 1 cheese .
Limiting
Excess
Max Output
Master Mission 8
Aluminum Chloride Synthesis
2 Al + 3 Cl2 → 2 AlCl3
Mole Mission Challenge Answer Key MOLE MISSION KEY
Ultimate Mastery Assessment v3.0
Answer Key & Solutions
1
Roadmap Logic
1. Answer: Mole Ratio
Determined by the coefficients in the balanced equation.
2
Route Planning
2. Answer: 3 Steps
(Mass A → Moles A → Moles B → Mass B)
MISSION 1: [Mole → Mole]
\(2.5 \text{ mol C}_3\text{H}_8 \times \frac{5 \text{ mol O}_2}{1 \text{ mol C}_3\text{H}_8} = \mathbf{12.5 \text{ mol O}_2}\)
MISSION 2: [Mass → Mass]
\(56.0 \text{ g N}_2 \times \frac{1 \text{ mol N}_2}{28.02 \text{ g}} = 1.999 \text{ mol N}_2\)
\(1.999 \text{ mol N}_2 \times \frac{2 \text{ mol NH}_3}{1 \text{ mol N}_2} = 3.997 \text{ mol NH}_3\)
\(3.997 \text{ mol NH}_3 \times \frac{17.03 \text{ g}}{1 \text{ mol}} = \mathbf{68.1 \text{ g NH}_3}\)
MISSION 3: [Mass → Mole]
\(12.15 \text{ g Mg} \times \frac{1 \text{ mol Mg}}{24.30 \text{ g}} \times \frac{2 \text{ mol MgO}}{2 \text{ mol Mg}} = \mathbf{0.50 \text{ mol MgO}}\)
Mission 4: [Mole → Mass]
\(0.50 \text{ mol Cl}_2 \times \frac{2 \text{ mol NaCl}}{1 \text{ mol Cl}_2} \times \frac{58.44 \text{ g NaCl}}{1 \text{ mol NaCl}} = \mathbf{58.44 \text{ g NaCl}}\)
Mission 5: Photosynthesis Yield
\(88.0 \text{ g CO}_2 \times \frac{1 \text{ mol CO}_2}{44.01 \text{ g}} = 2.00 \text{ mol CO}_2\)
\(2.00 \text{ mol CO}_2 \times \frac{1 \text{ mol C}_6\text{H}_{12}\text{O}_6}{6 \text{ mol CO}_2} = 0.333 \text{ mol Glucose}\)
\(0.333 \text{ mol Glucose} \times \frac{180.16 \text{ g}}{1 \text{ mol}} = \mathbf{60.05 \text{ g Glucose}}\) (Accept 60.0 - 60.1 depending on rounding)
Mission 6: Percent Yield
Actual: 50.0g
83.3%
3
Limiting Hazards
Mission 7: Bread/Cheese Analogy Key
Limiting
Cheese
Excess
Bread
Max Output
2 Sandwiches
Note: 10 bread could make 5 sandwiches, but 2 cheese only makes 2. Cheese runs out first.
Master Mission 8 Solutions
Route 1: From 4.0 mol Al
\(4.0 \text{ mol Al} \times \frac{2 \text{ mol AlCl}_3}{2 \text{ mol Al}} = \mathbf{4.0 \text{ mol AlCl}_3}\)
Route 2: From 3.0 mol Cl2
\(3.0 \text{ mol Cl}_2 \times \frac{2 \text{ mol AlCl}_3}{3 \text{ mol Cl}_2} = \mathbf{2.0 \text{ mol AlCl}_3}\)
Verdict:
Chlorine (Cl2) is LR
Mole Mission Tool Kit Challenge Expanded Mole Mission
Reaction Place Mat
CHALLENGE LEVEL
START
1
STEP 1 TOP
STEP 1 BTM
BRIDGE
MOLE RATIO
RATIO BTM
MOLE RATIO
STEP 3 TOP
STEP 3 BTM
=
Final Answer
Percent Yield Calculator
x 100 =
%
Limiting Reactant Logic
Path 1 Product
Path 2 Product
THE SMALLEST IS THE LIMIT!
Mission Tiles: Set A
Basic Navigation & Conversions
M1: Propane Tiles
2.5 mol C3H8
5 mol O21 mol C3H8
12.5 mol O2
M2: Haber Tiles
56.0g N2
1 mol N228.02g N2
2 mol NH31 mol N2
17.03g NH31 mol NH3
M3: Magnesium Tiles
12.15g Mg
1 mol Mg24.30g Mg
2 mol MgO2 mol Mg
M4: Salt Tiles
0.50 mol Cl2
2 mol NaCl1 mol Cl2
58.44g NaCl1 mol NaCl
Mission Tiles: Set B
Theoretical Yield & Master Limiting Reactant
Mission 5: Photosynthesis [Mass → Mass]
88.0g CO2
1 mol CO2
44.01g CO2
1 mol C6H12O6
6 mol CO2
180.16g C6H12O6
1 mol C6H12O6
60.1g Glucose
Mission 8: AlCl3 Synthesis [Master LR]
PATH 1: Starting with 4.0 mol Al
4.0 mol Al
2 mol AlCl3
2 mol Al
4.0 mol
PATH 2: Starting with 3.0 mol Cl2
3.0 mol Cl2
2 mol AlCl3
3 mol Cl2
2.0 mol
Section 3 Tiles: Analogies & Yields
LIMITING CHEESE
EXCESS BREAD
OUTPUT 2 Total
LR VERDICT Cl2
50.0g
60.1g
83.3%
Instructor Guide
This Tool Kit covers all **8 Missions** in the Expanded Quiz. Set B is specifically designed for the **Photosynthesis Mass-to-Mass Yield (M5)** and the **AlCl3 Mastery Challenge (M8)**. For M8, have students complete both paths to identify the true limit.
