Mg + O2  MgO
 If
you have 3.2 g of magnesium to start
with (and plenty of oxygen) how many
grams of MgO will be produced in this
reaction?
Unit 8 Stoichiometry
LIMITING REACTANT
Back to our Cake Analogy

1 cake mix + 3 eggs + 1 cup water  1 cake

If you have 2 mixes, 3 eggs and 5 cups of water, how many cakes can
you make?

In this case we would call the eggs the
limiting reactant. We can only make one
cake because we don’t have enough eggs
to match the other materials that are
available. In this case we would make 1
cake and have 1 mix and 4 cups water left
over.
Limiting reactant in a chemical
Rxn
2 H2 + O2  2 H2O
If you have 2 moles H2 and 3 moles O2, which is the
limiting reactant?
2 mol H2 X 1 mol O2 = 1 mol O2
2 mol H2
You only need one mole of O2 to react with the
2 moles of H2 (you have an excess of 2 mol O2
[3-1=2]).
LR- H2
ER- O2

Calculating Limiting reactant
Always in moles!
2 H2 + O2  2 H2O
If you have 7.6 moles H2 and 3.5 moles O2,
Needed to
which is the limiting reactant?
react with

7.6 mol H2 X 1 mol O2 =
3.8 mol O2
7.6 mol H2
2 mol H2
The 3.5 mol O2 you have is smaller than the 3.8 mol O2 that you would need
to react with all the H2 so the O2 is the limiting reactant.

Remember to balance the equation first!
Calculating Limiting reactant
Always in moles!
2 2O
SiO2 + 4 HF  SiF4 + H
If you have 2.0 mol HF and 4.5 mol SiO2,
which is the limiting reactant?

2.0 mol HF X 1 mol SiO2 =
0.50 mol SiO2
4 mol HF
The 4.5 moles SiO2 you are given are MORE than enough to react with the
2.0 mol HF so HF is our limiting reactant
Calculating Limiting reactant
Always goes through moles!
2 HCl + Ca(OH)2  CaCl2 + 2 H2O
If you have 15 g HCl and 12 g Ca(OH)2,
which is the limiting reactant?
mol Ca(OH)
15 g HCl X 1 mol HCl = 0.41 mol HClX 21 mol
HCl
2
= 0.205 mol Ca(OH)2
36.5g HCl
12 g Ca(OH)2 X 1 mol Ca(OH)2 = 0.16 mol Ca(OH)2
74 g Ca(OH)2
The Ca(OH)2 is the limiting reactant because we do not have
enough to react with all of the HCl we are given.
Using limiting reactant for stoichiometry

Zn +2 HCl 
ZnCl2 + H2
If you are given 70 g Zn and 71 g HCl, what will be the mass of the salt
produced?
71 g HCl X 1 mol HCl = 1.95 mol HCl
36.5g HCl
70 g Zn X 1 mol Zn = 1.07 mol Zn X 2 mol HCl = 2.14 mol HCl
1 mol Zn
65.4 g Zn
The HCl is the limiting reactant because we do not have enough
HCl to react with all of the Zn we are given.
Using limiting reactant

Zn + 2HCl 
ZnCl2 + H2
If you are given 70 g Zn and 71 g HCl, what
will be the mass of the salt produced?
HCl is our limiting reactant
71 g HCl X 1 mol HCl X 1 mol ZnCl2 X 136.3 g ZnCl2 = 132.57 g
ZnCl2
36.5g HCl
2 mol HCl
1 mol ZnCl2
132.57 g is the theoretical yield for this reaction.
% yield

% yield = experimental yield x 100%
theoretical yield
% expressing what you u collected vs what you were
supposed to collect.
% Yield

If you did the reaction on the last two slides and you only collected 110 g
of ZnCl2, what is your % yield?
% Yield =
% Yield =
mass collected
mass expected
110 g
X
132.57 g
X
100%
100% = 82.97%

Mg +
HCl
-- >
MgCl2 + H2
If you have 25.4 g Mg and combine it with 50.0 mL of 1.25 M HCl, what mass of
MgCl2 will you collect?
Do Now 9/22 – Do this then discuss
questions from your reading on gasses

Zn + 2 HCl 
ZnCl2 + H2
If you are given 70.0 g Zn and 71.0 g HCl,
what will be the mass of the salt
produced?
If you did the reaction and you only
collected 110 g of ZnCl2, what is your %
yield?
Do Now 9/24 – Do this then discuss
questions from your reading on gasses
A sample of a compound contains 0.99 g Na,
1.365 g S and 1.021 g O. Determine its
empirical formula.
ATP is an important molecule in living cells. A
sample with a mass of 0.8138 g was
analyzed and found to contain 0.1927 g C,
0.02590 g H, 0.1124 g N, and 0.1491 g P. The
remainder was O. Determine the empirical
formula of ATP. Its formula mass was found
to be 507 g mol-1. Determine the molecular
formula.
Do Now 9/25
A sample of a compound contains 0.99 g Na,
1.365 g S and 1.021 g O. Determine its
empirical formula.
ATP is an important molecule in living cells. A
sample with a mass of 0.8138 g was
analyzed and found to contain 0.1927 g C,
0.02590 g H, 0.1124 g N, and 0.1491 g P. The
remainder was O. Determine the empirical
formula of ATP. Its formula mass was found
to be 507 g mol-1. Determine the molecular
formula.