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Reactions

This document covers the basics of chemical reactions and stoichiometry, detailing types of reactions such as combination, decomposition, single and double replacement, and combustion. It explains the importance of balancing chemical equations and introduces stoichiometric calculations, including identifying limiting reactants and calculating theoretical versus actual yields. Key concepts include the Law of Conservation of Mass and the significance of mole ratios in chemical reactions.
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0% found this document useful (0 votes)
26 views9 pages

Reactions

This document covers the basics of chemical reactions and stoichiometry, detailing types of reactions such as combination, decomposition, single and double replacement, and combustion. It explains the importance of balancing chemical equations and introduces stoichiometric calculations, including identifying limiting reactants and calculating theoretical versus actual yields. Key concepts include the Law of Conservation of Mass and the significance of mole ratios in chemical reactions.
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as TXT, PDF, TXT or read online on Scribd
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Class Notes: Chemical Reactions and Stoichiometry (Chemistry 101)

Date: May 4, 2025


Topic: Chemical Reactions and Stoichiometry
Instructor: Dr. L. Evans

Introduction: What are Chemical Reactions?


A chemical reaction is a process in which one or more substances (reactants) are
converted into new substances (products) through the breaking and forming of
chemical bonds.

Key Concept: The total mass of the reactants must equal the total mass of the
products (this is the Law of Conservation of Mass).

Types of Chemical Reactions


Combination (Synthesis) Reactions:

Two or more reactants combine to form a single product.

General form: A + B → AB

Example:

𝐻
2

𝑂
+

2

𝐻
2

𝑂
2

2H
2

+O
2

→2H
2

Hydrogen gas reacts with oxygen gas to form water.

Decomposition Reactions:

A single reactant breaks down into two or more products.

General form: AB → A + B

Example:

𝐻
2

𝑂
2

2

2
𝐻

𝑂
2

𝑂
+

2
2H
2

O
2

→2H
2

O+O
2

Hydrogen peroxide decomposes into water and oxygen gas.

Single Replacement (Displacement) Reactions:

One element replaces another in a compound.

General form: A + BC → AC + B

𝑍
Example:

𝑛
+

𝐻
2

𝐶
𝑙

𝑍

𝑛
𝐶
𝑙
2

𝐻
+

2
Zn+2HCl→ZnCl
2

+H
2

Zinc reacts with hydrochloric acid to form zinc chloride and hydrogen gas.

Double Replacement (Displacement) Reactions:

The cations and anions of two ionic compounds switch places.


General form: AB + CD → AD + CB

𝐴
Example:

𝑔
𝑁
𝑂
3

𝑁
+

𝑎
𝐶
𝑙

𝐴

𝑔
𝐶
𝑙

𝑁
+

𝑎
𝑁
𝑂
3
AgNO
3

+NaCl→AgCl+NaNO
3

Silver nitrate reacts with sodium chloride to form silver chloride and sodium
nitrate.

Combustion Reactions:

A substance reacts with oxygen to produce energy, typically in the form of heat and
light, along with carbon dioxide and water.

General form: C_xH_y + O_2 → CO_2 + H_2O

𝐶
Example:

𝐻
4
+

𝑂
2

𝐶

𝑂
2
+

𝐻
2

𝑂
2
CH
4

+2O
2

→CO
2

+2H
2

Methane reacts with oxygen to form carbon dioxide and water.

Balancing Chemical Reactions


A balanced chemical equation has the same number of atoms of each element on both
sides of the equation.

Law of Conservation of Mass: Atoms cannot be created or destroyed in a chemical


reaction.

Steps to Balance a Reaction:

Write the unbalanced equation.

Balance atoms one element at a time, starting with elements that appear only once
on each side.

Balance hydrogen and oxygen atoms last.

Check that the number of atoms of each element is the same on both sides.

Example:
_{__}CH_4 + _{__}O_2 \rightarrow _{__}CO_2 + _{__}H_2O

Start with carbon: 1 C on both sides → balanced.

Balance hydrogen: 4 H on the left → 2 H₂O on the right.

Balance oxygen: 2 O in 2 H₂O → 2 O₂ molecules.

𝐶
Final balanced equation:

𝐻
4
+

𝑂
2

𝐶

𝑂
2
+

𝐻
2

2
𝑂
CH
4

+2O
2

→CO
2

+2H
2

Stoichiometry: The Study of Quantitative Relationships


Stoichiometry involves using the relationships between the quantities of reactants
and products in a chemical reaction.

Mole-to-mole ratio: From the balanced equation, we can determine the ratios of
moles of reactants and products.

Steps for Stoichiometric Calculations:

Write the balanced equation.

Convert known quantities (masses, volumes, or moles) into moles using molar mass or
molar volume.

Use the mole ratio from the balanced equation to find the moles of the unknown.

Convert moles back to desired quantity (grams, liters, etc.).

Example Problem:

Given: How many grams of water are produced when 16 grams of methane (CH₄) reacts
with excess oxygen?

𝐶
Step 1: Write the balanced equation:

𝐻
4
+

𝑂
2

𝐶

𝑂
2
+

𝐻
2

𝑂
2

CH
4

+2O
2

→CO
2

+2H
2

Step 2: Convert grams of CH₄ to moles:


Molar mass of CH₄ = 16.04 g/mol, so
16
g
×
1
mol
16.04
g
=
0.998
mol CH
4
16 g×
16.04 g
1 mol

=0.998 mol CH
4

Step 3: Use the mole ratio:


From the balanced equation, 1 mole of CH₄ produces 2 moles of H₂O.
0.998
mol CH
4
×
2
mol H

𝑂
2

1
mol CH
4
=
1.996
mol H

𝑂
2

0.998 mol CH
4

×
1 mol CH
4

2 mol H
2

=1.996 mol H
2

Step 4: Convert moles of H₂O to grams:


Molar mass of H₂O = 18.02 g/mol, so
1.996
mol
×
18.02
g/mol
=
35.96
g H

𝑂
2

1.996 mol×18.02 g/mol=35.96 g H


2

Answer: 35.96 grams of water are produced.

Limiting Reactants and Excess Reactants


The limiting reactant is the reactant that is completely consumed in a chemical
reaction, limiting the amount of product that can be formed.

The excess reactant is the reactant that is not completely used up in the reaction.

How to Identify the Limiting Reactant:

Convert all reactant quantities to moles.

Use the mole ratio from the balanced equation to determine how much of each
reactant is required.

The reactant that runs out first is the limiting reactant.

Example:

Given: 4 moles of CH₄ and 6 moles of O₂. How many moles of CO₂ are produced?

Step 1: From the balanced equation, 1 mole of CH₄ reacts with 2 moles of O₂.

For CH₄: 4 moles of CH₄ would need 8 moles of O₂.

But you only have 6 moles of O₂, so O₂ is the limiting reactant.

Step 2: Use the mole ratio to find the moles of CO₂ produced:
From the balanced equation, 2 moles of O₂ produce 1 mole of CO₂.
6
mol O
2
×
1
mol CO
2
2
mol O
2
=
3
mol CO
2
6 mol O
2

×
2 mol O
2

1 mol CO
2

=3 mol CO
2

Answer: 3 moles of CO₂ are produced.

Theoretical Yield vs. Actual Yield


Theoretical Yield: The maximum amount of product that can be formed from a given
amount of reactants, calculated based on the stoichiometry of the reaction.

Actual Yield: The amount of product actually obtained from the reaction in the
laboratory.

Percent Yield:

Percent Yield
=
Actual Yield
Theoretical Yield
×
100
Percent Yield=
Theoretical Yield
Actual Yield

×100
Example:
If the theoretical yield of CO₂ is 3.00 grams, and the actual yield is 2.50 grams,
the percent yield is:

2.50
g
3.00
g
×
100
=
83.33
%
3.00 g
2.50 g

×100=83.33%
Discussion Points:
What is the significance of the limiting reactant in a reaction?

The limiting reactant determines how much product can be formed, as it is the
reactant that is completely consumed first.

Why is it important to balance chemical equations before performing stoichiometric


calculations?

Balancing ensures that the conservation of mass is followed and that the correct
mole ratios are used in calculations.

What factors might cause the actual yield to be less than the theoretical yield?

Incomplete reactions, loss of product during recovery, side reactions, or


measurement errors can all reduce the actual yield.

Quick Recap:
Chemical reactions involve the transformation of reactants into products

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