Guide to Stoich
Christopher Martinez period 1
Equation, Type, and how to balance
Aluminum and Iron(III) Oxide using permanganate catalyst
Al(s) + Fe2O3(s) → Al2O3(s) + Fe(s) You first give the chemicals their charges
2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s) Then you make the chemicals balance each other on both sides of the arrow.
This type of equation is know as a single replacement because only one chemical is switching with another
IUPAC name for each reactant and product With molar mass
26.982+71.844 → 42.981+ 55.845 The molar mass is the atomic number of the chemicals.
Mole to mole conversion
l 1mole Fe2O3 l
What you do to find the mole of Aluminum you multiply the amount of Iron Oxide by the amount of aluminum divided by the amount of Iron Oxide on the bottom.
12.1 Al l 2mole Fe2O3 l
l 2mole Al l
The amount of Fe2O3 is 12.1 because the 2's cancel out which just leaves a multiple of 12.1.
Mass to mass conversion
12.3 Al2O3 l 1mole Al2O3 l 2mole Al l 53.964 Al
l 101.961mole Fe2O3 l 1mole Al2O3 l 1mole Al
To find the mass of the aluminum you multiply grams by moles by the second chemical mole and molar mass divided by the molar mass of the first chemical by the mole of the first chemical and 1 mole of the second chemical.
Limiting and excess reactant
To get the amount of Al2O3 you multiply chemical one's grams by its mole and by the second chemical two's mole by molar mass and divide that by the first chemical molar mass and its mole and one mole of the second chemical.
Fe2O3=10.44 from equaling 100 from the percent yield
You find percent yield by dividing the mass to mass problem by the limiting factor number given at the begining and multiply the product by 100.