Chemical Stoichiometry
The Mathematics
of
Chemical Reactions
Quantitative Relationships between the reactants
and products in a chemical reaction
Chemical Stoichiometry
Understanding chemical reactions helps us to
predict what will be produced by a reaction.
We also need to know how much will be produced
by a reaction.
We also can determine how much reactant is
required to produce a desired amount of product.
Stoichiometry – Solving the problems:
A simple Process Step #1: Balance the Equation.
Step #2: Problem on the Top.
Vertical
put the measurements or quantities that you were given
over the reactant or product to which that value pertains.
Step #3: Equation on the Bottom.
A
put the measurements or quantities from the equation
greement
under the reactant or product to which that value pertains.
Step #4:
Solve the Problem.
Cross multiply and divide.
Mass – Mole Stoichiometry
Given 100.0 g of potassium chlorate, determine the moles
of oxygen gas produced.
100.0 g
X mol
2 KClO3
==> 2 KCl
245.2 g
•
•
(122.6) g
K = 1 x 39.1 = 39.1
Cl = 1 x 35.5 = 35.5
O = 3 x 16.0 = 48.0
122.6g
+ 3 O2
3 mol
300 g • mol
245.2 g
=
1.22 mol
Mass – Mass Stoichiometry
Determine the mass of iron which could be produced by the reaction
of 100.0 grams of iron(III) oxide with enough carbon to use it all.
100.0 g
X g
2 Fe2O3 + 3C
(159.6)
==> 4 Fe
(55.8)
•
•
319.2 g
Fe = 2 x 55.8 = 111.6
O = 3 x 16.0 = 48.0
159.6
+ 3CO2
223.2 g
69.9 g
Volume in Stoichiometry
Determine the volume of carbon dioxide produced by the reaction
of 0.750 L of ethane with enough oxygen to use it all.
0.750 L
X L
2 C2H6 + 7 O2
(22.4)
22.4 L
Remember :
Volume of a Gas
@ STP
1 mole = 22.4 L
==> 6 H2O
•
•
+ 4 CO2
(22.4)
89.6 L
3.00 L
% Yield Problems
What is the percent yield of salt in the following reaction when 50.0 g of
sodium bicarbonate reacts with HCl, and your lab partner measured the
salt produced in the evaporating dish at 28.6 g.
50.0 g
NaHCO3
84.0 g
Na =
H=
C =
O =
1 x 23.0
1 x 1.0
1 x 12.0
3 x 16.0
+
=
=
=
=
HCl
•
•
23.0
1.0
12.0
48.0
84.0
xg
==> NaCl +
58.5 g
+
CO2
Na = 1 x 23.0 = 23.0
Cl = 1 x 35.5 = 35.5
58.5
34.8 g
28.6
34.8
H 2O
X
100 % = 82.2 %
Limiting Reactant Problems
NH3 Remains
(1.50g – 0.788g = 0. 712g )
The conversion of ammonia to nitrogen monoxide is
Y = 0.788g 1.50
Y g
1.85 g
Xg
4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g)
X = 2.65g
68 g
160 g
120 g
X = 1.39g
a) How many grams of NO form when 1.50 g of
NH3 reacts with 1.85 g O2? 1.39 g NO
b) Which reactant is the limiting reactant and
which is the excess reactant?
c) How much of the excess reactant remains after
the limiting reactant is completely consumed?
0.788g NH3 consumed and 0. 712g NH3 remains