(STOY-KEE-AHM-EH-TREE)
• Stoichiometry is the part of chemistry that studies
amounts of reactants and products that are
involved in reactions.
• Chemists use a balanced chemical equation as a
basis to calculate how much reactant is used or
product that is produced in a reaction.
• A balanced chemical equation can be interpreted
in terms of different quantities, including number
of atoms, molecules, or moles.
• In a balanced chemical equation, the coefficient
in an equation represents not only numbers of
individual molecules but also numbers of moles.
• The basis for stoichiometry is the law of
conservation of mass, which states the mass of
the reactants will equal the mass of the products.
2 Cr
+
1 Mm +
3 Ch

1 Cr2MmCh3
Balance the equation.
1
3
2
____N
+
____H

____NH
2
2
3
This equation can be looked at with
________
molecule of N2 reacts with
one
three molecules of H2 to produce
__________
molecules
two ________________
of NH3.
1 N2
+
3 H2  2 NH3
This equation can be looked at as one
mole of N2 reacts with three
moles
____________
of H2 to produce
two moles of NH3.
_________
conversion
A mole ratio is a _______________
factor
coefficients of a
derived from the ______________
balanced chemical equation interpreted in
moles
terms of ______________.
Find
Given
Balance the equation and write all six mole
ratios.
3
2
4
____Fe
+
____O
____Fe
2 
2O3
4 Fe
3 O2
4 Fe
2 Fe2O3
4 Fe
3 O2
2 Fe2O3
4 Fe
3 O2
2 Fe2O3
2 Fe2O3
3 O2
*
1.Balance the equation
2.Identify given and finish line
3.Use mole ratio and/or molar mass to
solve
4.Check sig. figs. and round if needed
*
1 C6H12O6 +
6 O2
2.00 mol C6H12O6
6 mol CO2
1 mol C6H12O6
6 CO2 + 6 H2O
= 12.0 mol CO2
3 H2 + 1
8.50 mol H2
N2
2 NH3
2 mol NH3
3 mol H2
= 5.67 mol NH3
*
2 Al + 6 HCl
6.000 mol Al
2 AlCl3 + 3 H2
3 mol H2
2.02 g H2
2 mol Al
1 mol H2
= 18.18 g H2
2 Al + 6 HCl
4.00 mol HCl
2 mol AlCl3
6 mol HCl
2 AlCl3 + 3 H2
133.33 g AlCl3
1 mol AlCl3
= 178 g AlCl3
2 C8H18 + 25 O2
325 g C8H18
16 CO2 + 18 H2O
1 mol C8H18
18 mol H2O
114.26 g C8H18
2 mol C8H18
= 25.6 mol H2O
2 C8H18 + 25 O2
145 g O2
1 mol O2
32.00 g O2
16 mol CO2
25 mol O2
16 CO2 + 18 H2O
= 2.90 mol CO2
*
1 C5H12 + 8 O2
100.0 g C5H12
5 CO2 + 6 H2O
1 mol C5H12
5 mol CO2
44.01 g CO2
72.17 g C5H12
1 mol C5H12
1 mol CO2
= 304.9 g CO2
4 Zn + 10 HNO3
8.75 g N2O
4 Zn(NO3)2 + 1 N2O + 5 H2O
1 mol N2O
10 mol HNO3
63.02 g HNO3
44.02 g N2O
1 mol N2O
1 mol HNO3
= 125 g HNO3
1 N2 + 3 H2
5.40 g H2
2 NH3
1 mol H2
2 mol NH3 17.04 g NH3
2.02 g H2
3 mol H2
1 mol NH3
= 30.4 g NH3