3.11 & 4.2
Introduction
to Reaction
Stoichiometry
Writing and Balancing Chemical
Equations
How do you know if a chemical reaction has occurred?
One or more substances have to have been converted into different
substances.
Let’s write the unbalanced chemical equation for the combustion on propane:
C3H8(g) + O2(g)  CO2(g) + H20(g)
Does this equation obey the law of conservation of mass? How do you know?
No, this equation does not obey the law of mass conservation because there is
a different number of atoms, of each element, on both the reactant side and
the product side of the chemical equation.
How to Balance a Chemical Equation
• Since atoms don’t just vanish or appear, we must
balance the equation by listing coefficient which
represents the mole ratio of each of the substances
in the equation.
Let’s try to balance the reaction on the previous slide:
C3H8(g) + O2(g)  CO2(g) + H20(g)
C3H8(g) + 5O2(g)  3CO2(g) + 4H20(g)
Let’s Try a Sample Problem
Write the balanced equation for the reaction between
aqueous lead(II) nitrate and aqueous potassium chloride
to form solid lead(II) chloride and aqueous potassium
nitrate.
First let’s write the unbalanced equation:
Pb(NO3)2(aq) + KCl(aq)  PbCl2(s) + KNO3(aq)
Now let’s balance it!
Pb(NO3)2(aq) + 2KCl(aq)  PbCl2(s) + 2KNO3(aq)
Reaction Stoichiometry: How Much
CO2?
Balance the combustion of octane:
C8H18(l) + O2(g)  CO2(g) + H2O(g)
2C8H18(l) + 25O2(g)  16CO2(g) + 18H2O(g)
Reaction stoichiometry is the numerical relationship
between chemical amounts in a balanced chemical
equation, and make predictions based on these
amounts.
A Chemical Recipe
• We can compare reaction stoichiometry to a recipe.
For example: (and we’ll keep it simple)
Mr. Stevenson’s egg sandwich:
2 eggs + 4 bacon strips + 1 slice of cheese + 1 English muffin + butter 
Mr. Stevenson’s egg sandwich
How many sandwiches could you make if you had 24 strips of bacon?
1 sandwich
24 strips of bacon X ------------------------- = 6 sandwiches
4 strips of bacon
Mole-to-Mole Conversions
Let’s use our balanced chemical equation for the combustion of octane to
determine how many moles of CO2 enter our atmosphere when we combust
36.0 moles of octane.
2C8H18(l) + 25O2(g)  16CO2(g) + 18H2O(g)
Step 1: List mol-to-mol ratio of octane to carbon dioxide.
2 mol octane: 16 mol carbon dioxide (1:8 ratio)
Step 2: Use this ratio to covert moles of reactant to moles of product.
16 mol CO2
36 mol C8H18 X --------------------- = 288 mol CO2
2 mol C8H18
Mass-to-Mass Conversion
Step 1: Convert the mass of what’s given to grams, if
it’s not already in grams.
Step 2: Convert grams of given to moles of given
Step 3: Convert moles of given to moles of what you
want. (Use mole-to-mole ratio)
Step 3: Convert moles of what you want to mass in the
units asked to find.
Let’s Try a Sample Problem
A component of acid rain is nitric acid, which forms NO2,
also a pollutant, reacts with oxygen and water according to
the simplified equation:
4NO2(g) + O2(g) + 2H2O(l)  4HNO3(aq)
The generation of the electricity used by a n
medium sized home produces about 16 kg of NO2 per
year. Assuming that there is aqequate O2 and H2O,
what mass of HNO3, in kg, can form from this amount of
NO2 pollutant.
Sample Problem 2 Worked Out
I am going to complete this whole problem in one
step using dimensional analysis.
1000 g NO2
1 mol NO2
4 mol HNO3
16 kg NO2 X ---------------- X ----------------- X ---------------1 kg NO2
46.01 g NO2 4 mol NO2
63.02 g HNO3
1 kg HNO3
----------------- X -------------------- = 21.9 kg HNO3(aq)
1 mol HNO3
1000 g HNO3
Chapter 4 pg. 186 #’s 26, 28, 32, 36 (Just the a’s)
Read 4.3 pgs. 145-151