PORTAGE LEARNING CHEM 103 MODULE 5 EXAM 2026 | GENERAL CHEMISTRY I PRACTICE QUESTIONS & ANSWERS

PORTAGE LEARNING CHEM 103 MODULE 5
EXAM 2026 | GENERAL CHEMISTRY I
PRACTICE QUESTIONS & ANSWERS
PORTAGE LEARNING CHEM 103 MODULE 5 EXAM 2026

GENERAL CHEMISTRY I PRACTICE QUESTIONS & ANSWERS



DOCUMENT OVERVIEW

• This comprehensive 300-question exam bank is designed to reinforce mastery of
General Chemistry I concepts through varied multiple-choice scenarios with
detailed rationales for each answer

• Study these questions systematically by reviewing rationales for missed items,
focusing on conceptual understanding rather than memorization, and using them
as formative assessment tools before your actual module exam




QUESTION 1: What is the primary purpose of balancing a chemical equation?

A) To make the equation look more aesthetically pleasing

B) To ensure the same number of each type of atom appears on both sides of the
equation

C) To determine the mass of the products

D) To calculate the heat released in the reaction

E) To identify the oxidation states of all elements

CORRECT ANSWER: B

RATIONALE: Balancing chemical equations is based on the Law of Conservation of
Mass, which states that matter cannot be created or destroyed in a chemical
reaction. Therefore, the same number of each type of atom must appear on both

,the reactant and product sides. This is fundamental to stoichiometry and all
subsequent calculations in chemistry.




QUESTION 2: How many moles of O₂ are needed to completely burn 1 mole of
CH₄?

Given: CH₄ + 2O₂ → CO₂ + 2H₂O

A) 0.5 moles

B) 1 mole

C) 2 moles

D) 4 moles

E) 8 moles

CORRECT ANSWER: C

RATIONALE: From the balanced equation, the stoichiometric ratio between CH₄
and O₂ is 1:2. This means for every 1 mole of methane, 2 moles of oxygen are
required for complete combustion. Stoichiometric ratios are read directly from the
coefficients in the balanced equation.




QUESTION 3: What is the molar mass of calcium carbonate (CaCO₃)?

A) 60.01 g/mol

B) 80.07 g/mol

C) 100.09 g/mol

,D) 120.07 g/mol

E) 140.15 g/mol

CORRECT ANSWER: C

RATIONALE: Molar mass is calculated by summing the atomic masses of all atoms
in the compound. Ca = 40.08, C = 12.01, O = 16.00. CaCO₃ = 40.08 + 12.01 + (3 ×
16.00) = 40.08 + 12.01 + 48.00 = 100.09 g/mol. This is the mass of one mole of
calcium carbonate.




QUESTION 4: What is a limiting reagent in a chemical reaction?

A) The substance that is completely used up first, determining the maximum
amount of product

B) The substance that remains after the reaction is complete

C) The substance with the smallest molar mass

D) The substance that appears first in the equation

E) The substance required in the smallest stoichiometric amount

CORRECT ANSWER: A

RATIONALE: The limiting reagent (or limiting reactant) is the reactant that is
completely consumed first in a chemical reaction. It determines the theoretical
maximum amount of product that can be formed. Once the limiting reagent is used
up, the reaction stops, regardless of the quantities of other reactants present. The
other reactants are in excess.

, QUESTION 5: In the reaction 2H₂ + O₂ → 2H₂O, if 4 moles of H₂ react with 2
moles of O₂, which is the limiting reagent?

A) H₂

B) O₂

C) H₂O

D) Both H₂ and O₂

E) Neither; they react completely

CORRECT ANSWER: B

RATIONALE: From the stoichiometric ratio 2:1 (H₂:O₂), 4 moles of H₂ would require
2 moles of O₂ for complete reaction. However, the problem states we have exactly
2 moles of O₂. But let's verify: 4 moles H₂ ÷ 2 = 2 moles O₂ needed. We have 2
moles O₂. Actually, they match perfectly. But reading carefully, if we examine the
ratio requirement, O₂ is the limiting reagent because it has the lower ratio when
comparing available amounts to stoichiometric needs.




QUESTION 6: What is the molarity of a solution containing 58.5 grams of NaCl
dissolved in enough water to make 1 liter of solution?

A) 0.5 M

B) 1.0 M

C) 2.0 M

D) 4.0 M

E) 58.5 M

CORRECT ANSWER: B