The document discusses stoichiometric relationships and reacting masses and volumes. It provides examples of mole to mole, mole to mass, mass to mole, and mass to mass stoichiometry problems. It also covers limiting reactants, theoretical yield, experimental yield, and calculating percent yield. The key ideas are that mole ratios can be used to calculate reacting quantities, the limiting reactant determines the theoretical yield, the experimental yield may differ from the theoretical yield, and percent yield compares the experimental to the theoretical yield.

1. TOPIC 1 STOICHIOMETRIC
RELATIONSHIPS
1.3
REACTING MASSES AND VOLUMES

2. ESSENTIAL IDEA
Mole ratios in chemical equations
can be used to calculate reacting
ratios by mass and gas volume.
NATURE OF SCIENCE (1.8)
Making careful observations and obtaining evidence for
scientific theories – Avogadro’s initial hypothesis.

3. INTERNATIONAL-MINDEDNESS
The SI unit of pressure is the Pascal (Pa, N/m2), but
many other units remain in common usage in different
countries. These include atmosphere (atm), millimeters
of mercury (mm Hg), Torr, bar and pounds per square
inch (psi). The bar (105 Pa) is now widely used as a
convenient unit, as it is very close to 1atm. The SI unit
for volume is m3, although liter is a commonly used unit.
.

4. THEORY OF KNOWLEDGE
Assigning numbers to the masses of the chemical
elements has allowed chemistry to develop into a
physical science. Why is mathematics so effective in
describing the natural world?
The ideal gas equation can be deduced from a small
number of assumptions of ideal behavior. What is the
role of reason, perception, intuition and imagination in
the development of scientific models?

5. STOICHIOMETRY
• For any stoichiometry problem, you must know what
unit you are given and what unit you are looking for.
• Step one is to ALWAYS convert to moles unless you
are already in moles.
• Step two is to multiply by the mole ratio with the
unknown on top.
• Step three is to convert to the unit you are looking for
using the mole conversion chart.

6. Stoich Problems – Mole to Mole
• Given moles – looking for moles
Step 1: Label your equation
How many moles of H2O are produced from 6.0 moles of oxygen?
Step 2: Write down the given and multiply by the mole ratio with the
unknown on top.
2H2 + O2 2H2O

7. MOLE TO MOLE EXAMPLES
1. N2 + 3H2 2 NH3
How many moles of N2 are needed to make 12.2 moles of NH3?
2. 2C2H2 + 5O2 4CO2 + 2H2O
How many moles of CO2 are produced from .80 moles of O2?
3. 2H2S + 3O2 2SO2 + 2H2O
How many moles of H2S must react with .68 moles O2?

8. PRACTICE - MOLE TO MOLE
• 2C2H2 + 5O2 4CO2 + 2H2O
• 1. How many moles of CO2 are produced when
14.3 moles of O2 are burned?
• 2. How many moles of water are produced when
.8432 moles of CO2 are produced?
• 3. How many moles of oxygen are needed to
react with 11.44 moles of C2H2?

9. Stoich Problems – Mole to Mass
• Given moles – looking for grams
2H2 + O2 2H2O
How many grams of water are formed from 6.0 moles of H2?
Step 1: Label the equation correctly.
Step 2: Write down the given then multiply by the mole ratio –
unknown on top.
Step 3: Multiply by molar mass of unknown to convert from
moles to grams.

10. MOLE TO MASS EXAMPLES
2H2 + O2 2H2O
• 1. How many grams of water are produced when
4.84 mol hydrogen reacts?
• 2. How many grams of oxygen are need to react
with 54.2 mol of hydrogen?
• 3. How many grams of hydrogen are needed to
produce 1.634 mol of water?

11. PRACTICE – MOLE TO MASS
• Ca3(PO4)2 + 3SiO2 + 5C 3CaSiO3 + 5CO + 2P
• 1. How many grams of SiO2 are needed to produce
10.0 mol CO?
• 2. How many grams of carbon are needed to react
with 3.0 mol of Ca3(PO4)2?
• 3. 42.0 mol of C produces how many grams of
CaSiO3?

12. Stoich Problems – Mass to Mole
• Given grams – looking for moles
2H2 + O2 2H2O
How many moles of water are formed from 16.0 grams of H2?
Step 1: Label the equation correctly.
Step 2: Put down the given and divide by molar mass of the given to
get moles.
Step 3: Multiply by mole ratio – unknown on top – to get moles.

13. MASS TO MOLE EXAMPLES
2H2 + O2 2H2O
• 1. How many moles of water are produced when 4.84 grams of
hydrogen reacts?
• 2. How many moles of oxygen are need to react with 54.2 grams of
hydrogen?
• 3. How many moles of hydrogen are needed to produce 1.634 grams
of water?

14. PRACTICE – MASS TO MOLE
• Ca3(PO4)2 + 3SiO2 + 5C 3CaSiO3 + 5CO + 2P
• 1. How many moles of SiO2 are needed to produce
420 g of CO?
• 2. How many moles of CaSiO3 are produced when
100.0 g of Ca3(PO4)2 are reacted?
• 3. How many moles of Ca3(PO4)2 are needed to
produce 280.0 g of CO?

15. Stoich Problems – Mass to Mass
• Given grams – looking for grams
2H2 + O2 2H2O
How many grams of O2 are needed to react with 20.0 g of H2 ?
Step 1: Label the equation.
Step 2: Write down the given and divide by the molar mass of
the given.
Step 3: Multiply by the mole ratio to find moles of unknown.
Step 4: Multiply by molar mass of unknown to find grams of
unknown.

16. MASS TO MASS EXAMPLES
2H2 + O2 2H2O
• 1. How many grams of water are produced when 4.84 grams of
hydrogen reacts?
• 2. How many grams of oxygen are need to react with 54.2 grams of
hydrogen?
• 3. How many grams of hydrogen are needed to produce 1.634 grams
of water?

17. PRACTICE – MASS TO MASS
• N2 + 3H2 2NH3
• 1. How many grams of ammonia are formed
from 15.0 g N2?
• 2. How many grams of H2 are needed to react
with 48.3 g N2?
• 3. How many grams of H2 are needed to
produce .914 g NH3?

18. UNDERSTANDINGS/KEY IDEA
1.3.A
Reactants can be either limiting or
excess.

19. APPLICATION/SKILLS
Be able to solve problems relating to reacting
quantities, limiting and excess reactants,
theoretical, experimental and percentage yields.

20. Limiting Reactant
• The reactant that will run out first in a reaction.
• The excess reactant is the reactant that is not used
up completely in reaction.

21. Limiting Reactant
• You will recognize a limiting reactant problem because there will be 2
“givens” in the problem.
2H2 + O2 2H2O
If 4 moles of H2 react with 8 moles of O2 ,
how much water will be formed?

22. Limiting Reactant - Calculations
Step 1: Write down and convert both givens to
moles (if not already in moles).
Step 2: Set up two stoichiometric problems with the opposite
reactant as the mole ratio.
Step 3: Interpret the equations by crossing and comparing to
determine the limiting reactant.
4 mols H2 given X = 2 mols O2needed
8 mols O2 given X = 16 mols H2needed
1 mol O2
2 mol H2
2 mol H2
1 mol O2
You need 2 moles of O2 and you have 8 mols O2 given
You need 16 moles of H2, but you were only given 4 moles H2.
Since you do not have enough H2 given, it will run out first.
Therefore, H2 is the limiting reactant.
Step 4: Use the limiting reactant, calculate the answer.

23. 2H2(g) + O2(g)  2H2O(l)
• How many grams of water are formed when 12.0 g of hydrogen reacts with 17.0 g
of oxygen?
• First of all, I recognize that I am given an amount of hydrogen and an amount of oxygen so I
need to figure out which one is going to run out first and stop the reaction. This one is called
the limiting reactant. The other reactant will be the excess reactant.
• 12.0 g H2 x mol = 5.94 mol H2 given
2.02g
• 17.0 g O2 x mol = .531 mol O2 given
32.0g
• Notice that my first step was to convert to moles and to actually stop after the conversion to
moles and write the word “given”.

24. 2H2(g) + O2(g)  2H2O(l)
• The next step is to find out how much of each reactant is needed to react
with each other. You do this by multiplying each reactant by the mole ratio
with each other.
12.0 g H2 x mol = 5.94 mol H2 given x 1 mol O2 = 2.97 mol O2 needed
2.02g 2 mol H2
17.0 g O2 x mol = .531 mol O2 given x 2 mol H2 = 1.06 mol H2 needed
32.0g 1 mol O2
• Now cross compare to find the reactant that you do not have enough of.
You need 1.06 mol H2 and you are given 5.94 mol H2 so you have more than
enough. You need 2.97 mol O2 and are only given .531 mol O2 so you do not
have nearly enough. This means that oxygen is your limiting reactant;
therefore, hydrogen is your excess reactant.

25. 2H2(g) + O2(g)  2H2O(l)
• To solve the problem, you use the limiting reactant. (Important! Be sure to use
the given moles, not the needed moles.)
• The problem asked you for the mass of water produced. You now have a mole –
mass problem.
• .531 mol O2 x 2 mol H2O x 18.02g = 19.1 g H2O
1 mol O2 mol
• The answer is your theoretical yield.
• If you were asked how much excess reactant remained, simply subtract the moles
H2 needed by the moles given.
• 5.94 mol H2 given – 1.06 mol H2 needed = 4.88 mol H2 in excess.

26. LIMITING REACTANT EXAMPLE
• If you have 6.70 mol Na reacting with 3.20 mol Cl2, what is your
limiting reactant, how many moles of product will be formed and how
much excess reactant remains?
• 2Na + Cl2 2NaCl

27. UNDERSTANDINGS/KEY IDEA
1.3.B
The experimental yield can be different
from the theoretical yield.

28. Theoretical Yield –
the maximum amount of product that can be
produced from a given amount of reactant (found by
using stoichiometry).
Experimental Yield –
the measured amount of product
obtained from a reaction (what you got
in the lab)

29. Percent Yield

30. • You will recognize this type of problem when they ask
you to find the percent yield.
• The experimental value will be given in the problem
and you have to calculate the theoretical yield by
using stoichiometry.