Atomic Ledger Test
Atomic Ledger
Chemistry Unit 03
Quantitative Analysis Test
Student Name
Date
Period
Standard Reference Data
Avogadro's Number: \(6.022 \times 10^{23}\) mol\(^{-1}\)
STP: \(0^\circ\)C, 1.00 atm
Molar Masses (g/mol):
H: 1.01 | C: 12.01 | O: 16.00 | N: 14.01
Na: 22.99 | Cl: 35.45 | S: 32.06 | Fe: 55.85
Cu: 63.55 | Ag: 107.87 | Pb: 207.2 | Al: 26.98
PART I
Multiple Choice
10 QUESTIONS • 2 PTS EACH
1. Which of the following contains exactly one mole of atoms?
A 12.01 grams of Carbon-12
B 6.022 grams of Hydrogen
C 22.4 liters of Liquid Water
D 1.00 gram of Oxygen gas (O\(_2\))
2. What is the molar mass of Calcium Nitrate, Ca(NO\(_3\))\(_2\)?
A 102.09 g/mol
B 164.10 g/mol
C 150.09 g/mol
D 116.10 g/mol
3. How many moles are in 50.0 grams of NaOH? (Molar Mass = 40.00 g/mol)
A 0.80 moles
B 2.00 moles
C 1.25 moles
D 2000 moles
4. How many moles are present in \(1.204 \times 10^{24}\) molecules of CO\(_2\)?
A 0.50 moles
B 2.00 moles
C 1.00 moles
D 6.02 moles
5. In the reaction N\(_2\) + 3H\(_2\) \(\rightarrow\) 2NH\(_3\), what is the mole ratio of H\(_2\) to NH\(_3\)?
A 1 : 2
B 3 : 2
C 2 : 3
D 1 : 3
6. What is the percent composition of Oxygen in water (H\(_2\)O)?
A 11.1%
B 50.0%
C 88.9%
D 16.0%
7. If you have \(3.011 \times 10^{23}\) atoms of Silver (Ag), how many moles do you have?
A 0.50 moles
B 2.00 moles
C 1.00 moles
D 6.02 moles
8. The reactant that is completely consumed in a chemical reaction is called the:
A Excess reactant
B Theoretical yield
C Limiting reactant
D Reaction Catalyst
9. What is the percent yield if the theoretical yield is 25.0g and the actual yield is 20.0g?
A 125%
B 80%
C 45%
D 5%
10. The maximum amount of product that can be produced from a given amount of reactant is the:
A Actual yield
B Theoretical yield
C Percent yield
D Mole ratio
PART II
Short Answer
5 QUESTIONS • 6 PTS EACH
Show all work, include units, and round to the correct number of significant figures.
11. Calculate the molar mass of Magnesium Phosphate, Mg\(_3\)(PO\(_4\))\(_2\).
12. How many molecules are in 15.5 grams of Carbon Dioxide (CO\(_2\))?
13. Determine the percentage of Iron (Fe) in Iron(III) Oxide (Fe\(_2\)O\(_3\)).
14. Given the balanced equation: 2KClO\(_3\) \(\rightarrow\) 2KCl + 3O\(_2\). How many moles of Oxygen gas are produced from 6.0 moles of KClO\(_3\)?
15. A sample of a hydrate (CuSO\(_4 \cdot n\)H\(_2\)O) weighs 5.00g. After heating, the dry residue weighs 3.20g. What is the mass percentage of water in the hydrate?
PART III
The Stoichiometric Challenge
1 QUESTION • 15 PTS TOTAL
Scenario: The Thermite Reaction (Mass-to-Mass Analysis)
The thermite reaction is used to produce molten iron for welding railroad tracks. \[2\text{Al}(s) + \text{Fe}_2\text{O}_3(s) \rightarrow \text{Al}_2\text{O}_3(s) + 2\text{Fe}(l)\]
A. If 100.0 grams of Aluminum (Al) is reacted with 400.0 grams of Iron(III) oxide (Fe\(_2\)O\(_3\)), identify the limiting reactant. Support your answer with a full calculation.
B. Based on your answer in part A, calculate the theoretical yield of molten Iron (Fe) in grams. (Perform a complete mass-to-mass conversion).
C. If the process actually produces 185.0g of Iron, what is the percent yield? Provide one chemical reason why the yield might be lower than 100%.
Atomic Ledger Key
Atomic Ledger Key
Teacher Resource
Answer Key & Grading Rubric
Part I: Multiple Choice
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A (12.01g C = 1 mole)
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B (164.10 g/mol)
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C (50.0 / 40.0 = 1.25 mol)
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B (1.204/0.6022 = 2.00 mol)
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B (3 : 2 ratio)
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C (16 / 18.02 = 88.9%)
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A (0.50 moles)
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C (Limiting reactant)
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B (20 / 25 = 80%)
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B (Theoretical yield)
Part II: Short Answer
11. Mg3(PO4)2 Molar Mass
3(24.31) + 2(30.97) + 8(16.00) = 72.93 + 61.94 + 128.00 = 262.87 g/mol
12. Molecules in 15.5g CO2
Moles = 15.5 g / 44.01 g/mol = 0.3522 mol
Molecules = 0.3522 mol × (6.022 × 1023) = 2.12 × 1023 molecules
13. % Fe in Fe2O3
Molar Mass: 2(55.85) + 3(16.00) = 159.70 g/mol
% Fe = (111.70 / 159.70) × 100 = 69.94%
14. 2KClO3 → 2KCl + 3O2 Stoichiometry
6.0 mol KClO3 × (3 mol O2 / 2 mol KClO3) = 9.0 moles O2
15. Hydrate Percent Water
Mass water = 5.00g - 3.20g = 1.80g
% Water = (1.80 / 5.00) × 100 = 36.0%
Part III: Stoic Challenge Solution
A. Limiting Reactant Analysis
Moles Al: 100.0 g / 26.98 g/mol = 3.706 mol Al
Moles Fe2O3: 400.0 g / 159.70 g/mol = 2.505 mol Fe2O3
Required Al = 2.505 mol Fe2O3 × (2 mol Al / 1 mol Fe2O3) = 5.01 mol Al.
We only have 3.706 mol Al. Al is the Limiting Reactant.
B. Theoretical Yield Calculation (g → g)
Step 1: 100.0 g Al / 26.98 g/mol = 3.706 mol Al
Step 2: 3.706 mol Al × (2 mol Fe / 2 mol Al) = 3.706 mol Fe
Step 3: 3.706 mol Fe × 55.85 g/mol = 206.9 g Fe
C. Percent Yield & Reasoning
% Yield = (185.0 g / 206.9 g) × 100 = 89.4%
Chemical reason: Material loss during molten transfer or reactant impurities.
