Stoichiometry is the quantitative study of reactants and products in chemical reactions, focusing on determining how much product can be formed from given reactants. Key steps include balancing equations, converting mass to moles, and using mole ratios to find the unknown quantities. The document provides examples and practice problems related to mass-mass and mass-volume stoichiometry calculations.

1. STOICHIOMETRY

2. What is stoichiometry?
Stoichiometry is the
quantitative study of
reactants and products in
a chemical reaction.

3. What You Should Expect
 Given : Amount of reactants
 Question: how much of products can be
formed.
 Example
2 A + 2B 3C
 Given 20.0 grams of A and sufficient B,
how many grams of C can be produced?

4. What do you need?
You will need to use
i. molar ratios,
ii. molar masses,
iii. balancing and interpreting equations, and
iv. conversions between grams and moles.
Note: This type of problem is often called "mass-mass."

5. Steps Involved in Solving Mass-Mass
Stoichiometry Problems
 Balance the chemical equation correctly
 Using the molar mass of the given substance,
convert the mass given to moles.
 Construct a molar proportion (two molar
ratios set equal to each other)
 Using the molar mass of the unknown
substance, convert the moles just calculated
to mass.

6. Mole Ratios
A mole ratio converts moles
of one compound in a
balanced chemical equation
into moles of another
compound.

7. Example
Reaction between magnesium and oxygen
to form magnesium oxide. ( fireworks)
2 Mg(s) + O2(g) 2 MgO(s)
Mole Ratios:
2 : 1 : 2

8. 1) N2 + 3 H2 ---> 2 NH3
Write the mole ratios for N2 to H2 and
NH3 to H2.
2) A can of butane lighter fluid
contains 1.20 moles of butane
(C4H10). Calculate the number of
moles of carbon dioxide given off
when this butane is burned.
Practice Problems

9. Mole-Mole Problems
Using the practice question 2) above:
Equation of reaction
2C4H10 + 13O2 8CO2 + 10H2O
Mole ratio
C4H10 CO2
1 : 4 [ bases]
1.2 : X [ problem]
By cross-multiplication, X = 4.8 mols of CO2 given off

10. Mole-Mass Problems
Problem 1: 1.50 mol of
KClO3 decomposes. How
many grams of O2 will be
produced? [k = 39, Cl =
35.5, O = 16]
2 KClO3 2 KCl + 3 O2

11. Three steps…Get Your
Correct Answer
Use mole ratio
Get the answer in moles and then
Convert to Mass. [Simple Arithmetic]
Hello!
If you are given a mass in the problem,
you will need to convert this to moles
first. Ok?

12. Let’s go!
2 KClO3 2 KCl + 3 O2
2 : 3
1.50 : X
X = 2.25mol
Convert to mass
2.25 mol x 32.0 g/mol = 72.0 grams

13. Try This:
 We want to produce 2.75 mol of KCl. How many
grams of KClO3 would be required?
Soln
KClO3 : KCl
2 : 2
X : 2.75
X = 2.75mol
In mass: 2.75mol X 122.55 g/mol
= 337 grams zooo zimple!

14. Mass-MassProblems
There are four steps involved in solving
these problems:
 Make sure you are working with a properly
balanced equation.
 Convert grams of the substance given in the
problem to moles.
 Construct two ratios - one from the
problem and one from the equation and set
them equal. Solve for "x," which is usually
found in the ratio from the problem.
 Convert moles of the substance just solved
for into grams.

15. Just follow mass-
mass problem to
the penultimate
level
Mass-Volume Problems

16. Like this:
There are four steps involved in solving these
problems:
 Make sure you are working with a properly
balanced equation.
 Convert grams of the substance given in
the problem to moles.
 Construct two ratios - one from the
problem and one from the equation and set
them equal. Solve for "x," which is usually
found in the ratio from the problem.
 Convert moles of the substance just solved
for into Volume.

17. Conversion of mole to
volume
No of moles = Volume
Molar
volume
Can you remember a similar equation?

18. Molar volume
The molar volume is the
volume occupied by one
mole of ideal gas at STP.
Its value is: 22.4dm3

19. Practice Problems
Calculate the volume of carbon dioxide formed at
STP in ‘dm3
' by the complete thermal
decomposition of 3.125 g of pure calcium
carbonate (Relative atomic mass of Ca=40,
C=12, O=16)
Solution:
Convert the mass to mole:
Molar mass of CaCO3
=
40 + 12 + (16 x 3) = 100gmol-1
Mole = mass/molar mass
3.125/100 = 0.03125mol

20. Practice Problems
As per the equation,
Mole ratio 1 : 1
problem 0.03125mol X
X = 0.03125mol of CO2
Convert mole to volume [slide 17]
Volume = (0.03125 x 22.4)dm3

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