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CHEM 103 Module 2 Comprehensive Exam – General Chemistry I with Lab | Stoichiometry, Chemical Reactions & Redox Chemistry | Comprehensive Practice Questions with Answers & Rationales

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CHEM 103 Module 2 Comprehensive Exam – General Chemistry I with Lab | Stoichiometry, Chemical Reactions & Redox Chemistry | Comprehensive Practice Questions with Answers & Rationales

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1|Page


CHEM 103 Module 2 Comprehensive Exam – General
Chemistry I with Lab | Stoichiometry, Chemical
Reactions & Redox Chemistry | Comprehensive
Practice Questions with Answers & Rationales



Description: This comprehensive exam is designed for students enrolled
in CHEM103 General Chemistry I, specifically covering Module 2
content. The assessment focuses on chemical stoichiometry, including
molecular weight calculations, mole conversions, percent composition,
empirical and molecular formula determination, balancing chemical
equations, reaction classification, and redox chemistry. Questions range
from foundational theory to applied problem-solving, featuring realistic
clinical and laboratory scenarios. Based on actual Portage Learning,
UNLV, UW-Madison, and Bryn Mawr College CHEM103 curricula, this
150-question bank mirrors official exam patterns and tests both recall
and critical reasoning expected of a fully competent chemistry student.


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SECTION 1: QUESTIONS 1-50


Question 1
Show the calculation of the molecular weight for (NH4)2CrO4,
reporting your answer to 2 places after the decimal.

,2|Page


A) 150.07
B) 152.08
C) 154.12
D) 148.05


Answer: B


Rationale: The molecular weight is calculated by summing the atomic
weights of each atom in the compound. (NH4)2CrO4 contains 2 N
atoms (14.01 x 2 = 28.02), 8 H atoms (1.008 x 8 = 8.064), 1 Cr atom
(52.00), and 4 O atoms (16.00 x 4 = 64.00). Total = 28.02 + 8.064 +
52.00 + 64.00 = 152.08 g/mol.


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Question 2
Show the calculation of the molecular weight for C8H8NOI, reporting
your answer to 2 places after the decimal.


A) 245.05
B) 255.12
C) 261.05
D) 265.08


Answer: C

,3|Page




Rationale: The molecular weight of C8H8NOI is calculated as: 8 C
atoms (12.01 x 8 = 96.08), 8 H atoms (1.008 x 8 = 8.064), 1 N atom
(14.01), 1 O atom (16.00), and 1 I atom (126.90). Total = 96.08 + 8.064
+ 14.01 + 16.00 + 126.90 = 261.05 g/mol.


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Question 3
How many moles are in 12.0 grams of (NH4)2CrO4? Report your
answer to 3 significant figures.


A) 0.0789 mol
B) 0.0795 mol
C) 0.0812 mol
D) 0.0750 mol


Answer: A


Rationale: Moles = grams divided by molecular weight = 12.0 g divided
by 152.08 g/mol = 0.0789 mol (3 significant figures). This is a direct
application of the mole conversion formula.


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, 4|Page


Question 4
How many grams are in 0.0575 moles of C8H8NOI? Report your
answer to 1 place after the decimal.


A) 12.5 g
B) 15.0 g
C) 18.0 g
D) 20.0 g


Answer: B


Rationale: Grams = moles x molecular weight = 0.0575 mol x 261.05
g/mol = 15.0 g. This demonstrates the inverse of the mole calculation
above.


---


Question 5
Calculate the percent composition of oxygen in (NH4)2CrO4. Report
your answer to 1 decimal place.


A) 38.8%
B) 40.2%
C) 42.1%
D) 36.5%

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