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

Aluminum + Iron Oxide → Aluminum Oxide + Iron

26.982+71.844 → 42.981+ 55.845 The molar mass is the atomic number of the chemicals.

Mole to mole conversion

11.3 Fe2O3 l 2mole Al l

l 1mole Fe2O3 l

11.3*2=22.2mole Al

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


12.1=12.1mole Fe2O3

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

12.3*1*2*53.964

101.961*1*1 =12.73gAl

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

5.25Al*1moleAl*1Al2O3*101.96Al2O3/26.981538Al*2mole Al*1mole Al2O3=9.92Al2O3

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.


12.18FeO3*1moleFeO3*2moleFe*55.845Fe/103.842FeO3*1moleFeO3*1moleFe=13.10g Fe

Theoretical yield

Al=12.73 from mass to mass

Fe2O3=10.44 from equaling 100 from the percent yield

Percent yield

Al=5.25/12.73=41.24%

Fe2O3=17.77/10.44=58.76%


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.

Background of reaction

The chemical reaction is used in the real world for production of incendiary devices in the welding industry.