Have out for today
• Your periodic table
• Your scientific calculator
9.1 Mole Ratios
• Now that we know what a balanced
equation means, we can use an
equation to predict the moles of
products that a given number of
moles of reactants will yield.
• For example:
2H20(l)  2H2(g) + O2(g)
2 mol H2O yields 2 mol H2 and 1 mol O2
So what if…
I start with…
2H20(l)  2H2(g) + O2(g)
Then instead of 2 mol,
I have 4 mol of H2O…how many mol of
each product do I get?
4 2(g) + __O
2 2(g)
4H20(l)  __H
Now suppose…
Given the same equation
2H20(l)  2H2(g) + O2(g)
If we decompose 5.8 mol of water, how
many moles of products are formed? Start
with O2.
Let’s use a mole ratio – ratio of moles of
one substance to moles of another
substance in a balanced chemical
equation
2H20(l)  2H2(g) + O2(g)
2 mol H2O = 1 mol O2
forms 2 conversions factors:
1 mol O2
2 mol H2O
and
2 mol H2O
1 mol O2
Start with the number given in the problem
5.8 mol H2O 1 mol O2  2.9 mol O

2
1
2 mol H2O
And since…
2H20(l)  2H2(g) + O2(g)
…the ratio of water to hydrogen is
2 mol H2
1 mol H2

2 mol H2O 1 mol H2O
the amount of H2 produced is 5.8 mol.
So try this.
• Calculate the number of moles of oxygen
required to react exactly with 4.3 mol of propane,
C3H8, in the reaction described by the following
balanced equation.
C3H8 (g) + 5O2(g) 3CO2(g) + 4H2O(g)
4.3 mol C3H8 ?? mol O2
4.3 mol C3H8 5 mol O2
 21.5 mol O2

1
1 mol C3H8
Last example
• Ammonia is used in huge quantities as a
fertilizer. It is manufactured by combining
nitrogen and hydrogen according to the following
equation:
N2(g) + 3H2(g) 2NH3(g)
Calculate the moles of ammonia that are produced
from 1.3 mol of hydrogen reacting with excess N2
1.3 mol H2 2 mol NH3
 0.867 mol NH3

1
3 mol H2
One Step
Stoichiometry
Worksheet
9.2 Mass calculations
Consider the reaction of powdered Al and finely
ground iodine to produce aluminum iodide.
Al(s) + I2(s)  AlI3(s)
What mass of iodine is needed to react with 35.0 g
solid aluminum?
FIRST: be sure the equation is balanced.
2Al(s) + 3I2(s)  2AlI3(s)
Because the balanced equation
deals with moles instead of grams
• SECOND – convert 35.0 grams of
aluminum to mol of aluminum
35.0 g Al 1 mol Al


1.30
mol
Al
26.8 g Al
1
THIRD – Use the coefficients to determine
the mole ratio required to go from mol of
Al to mol I2 and convert.
1.3 mol Al 3 mol I2

 1.95 mol I2
1
2 mol Al
Mole ratio
Last step!
FOURTH – Convert the number of
moles of I2 into grams.
1.95 mol l2 253.8 g l2

mol I2
1
 495 g l2
So…given
2Al(s) + 3I2(s)  2AlI3(s)
It takes 495g of I2 to react with 35.0g of Al
Now, given the same
reaction…
2Al(s) + 3I2(s)  2AlI3(s)
Find the mass of AlI3(s) formed by the
35g of Al(s) with 495g of I2(s).
35.0 g Al 1 mol Al 1 mol AlI3 407.68 g AlI3



26.8
g
Al
1 mol Al
1
1 mol AlI3
 530 g AlI3
Homework
• Section 9.2 review
– # 2,4 & 6 only!
• Due tomorrow!!! I will not accept this
late for credit!