guide to stoich

great way to solve stoichiometry

Al+Fe(2)O(3) This is an example reaction. This is a single replacement reaction


2Al+Fe(2)O(3)=====> Al(2)O(3)+2Fe is a balanced reaction this is balanced because there are the same number of all elements on reactant and reaction side


Aluminum+ Iron oxide ======> Aluminum oxide+ Iron is the IUPAC name for this reaction

Molar Mass

26.98+101.84====>99.96+55.845 g/mole are the molar mass of all the reactants and products you find this by adding all parts


11.11*1

______________== 5.5

----- 2


this is an example of a mole to mole problem in this problem you multiply across the top and divide by the bottom

12.1 (fe(2)O(3) )*1 mole*1*99.96

____________________________________

---- 101.84 1 1


this is an example of mass to mass conversion. In this problem you would multiply across the top and the divide the answer by the bottom you get the answer of 11.87 grams of Al(2)O(3)

limiting and excess

A limiting reactant is the reactant that gives the max number of the product you can have given by he smallest number in grams that you have of each reactant. in this reaction iron oxide is the limiting reactant with 12.07 grams. this leaves Aluminum as the excess with 22.78 grams. there are 10.71 grams of aluminum left.

this leaves a theoretical yield which is the max possible number that could be made which is 12.07 and the percent yield which is the actual yield (10.981) divided by the theoretical yield (12.07) multiplied by 100 giving the percent yield of 90.96%