Stoichiometry Tutorial

Applewood Heights Seconadary School, Science Department

Written by: D. Bilic

 

Table of Contents

Introduction

Mole to Mole Stoichiometry

Mass to Mass Stoichiometry

Mass to Molecule Stoichiometry

Volume to Volume Stoichiometry

Mass to Volume Stoichiometry



 Introduction

Stoichiometry is the prediction of how much of one substance will react or be produced in a chemical reaction relative to the amount of another substance in the reaction. Stocihiometry is a review, it assumes a knowledge of:

- nomenclature (check out the tutorial!)

- writing balanced chemical equations

- predicting products for chemical reactions

- converting mass to # of moles

- converting # of moles to mass

In many ways then, stoichiometry will serve as a review of the units studied in this course- refer back to previously studied sections when necessary.

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Mole to Mole Stoichiometry

Every problem in stoichiometry is based on a chemical equation and its interpretation of moles.

The equation :

2 C + O2 -------> 2 CO

- may be interpreted as saying that 2 moles of carbon combine with 1 mole of oxygen to form 2 moles of carbon monoxide. The coefficients in the equation express the relative reacting quantities in terms of moles.

From this it follows that if the reaction quantitiy of carbon is 4 moles (twice as much as is indicated by the equation) the reaction quantities of oxygen is 2 moles and 4 moles of carbon monoxide:

4 C + 2 O2 -------> 4 CO

Likewise if the reacting quantity of carbon is half as much as indicated by the original equation, what would the number of moles be for O2 and CO?



Example 1:

Balance the following equation and fill in the table:

N2

  +

H2  

  ----->

NH3
  6 mol   mol   mol
  mol    6 mol     mol  
  mol   mol     6 mol



Example 2:

Methane burns in oxygen to produce carbon dioxide and water. If 8.2 moles of methane are burned in the reaction, how many moles of water will be produced?

Solution

Step 1

Write the balanced equation for the reaction and the mole ratio for the given and required substances.

 CH4  +  2O2  ------->  CO2  +  2H2O
  1 mol            2 mol 

Step 2

Write the given and the unknown beneath the appropriate substance.

 CH4  +  2O2  ------->  CO2  +  2H2O
  1 mol            2 mol
 8.2 mol            X mol

 

Step 3

Set up a ratio:

1 mol 

 =
  2 mol

8.2 mol

    X mol 

Step 4

Solve for "X". Use cross multiplication.

1 mol x X mol = 8.2 mol x 2 mol

X = 16.4 mol

Therefore the number of moles of H2O formed from 8.2 mol of methane is 16.4 mol

 

Example 3: You try this one!

Propane burns in oxygen to produce carbon dioxide and water. If 0.2 moles of propane are burned in the reaction, how many moles of oxygen is required?

Solution

Step 1

Write the balanced equation for the reaction and the mole ratio of the given and required substances.

C3H3   + O2 --> CO2  + H2O
  mol     mol         

Step 2

Write the given and the unknown beneath the appropriate substance.

C3H3   + O2 --> CO2  + H2O
  mol     mol        
  mol   mol         

Step 3
 X

Set up a ratio:

1 mol 

 =
  5 mol

0.2 mol

    X mol 

Step 4

Solve for "X". Use cross multiplication.

1 mol x X mol = 0.2 mol x 5 mol

X = 1 mol

Therefore, 1 mole of oxygen is required when 0.2 mol of propane are burned.

For some more problems click here.

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E-mail me at: dbilic@sympatico.ca