3.17 g of iron 3 oxide mixed with excess Aluminum. reaction yields 1.54 g of Aluminum oxide. what is the theoretical yield alumium oxide?
m = 3.17 g of Fe2O3
the reaction
Fe2O3(s) + Al(s) = Al2O3 +Fe(s)
balanced reaction
Fe2O3(s) + 2Al(s) = Al2O3 + 2Fe(s)
m of Al2O3 = 1.54 g
NOTE
the theoretical yield of Al2O3 is INDEPENDENT on how much product was formed
then
3.17 g of Fe2O3 sohuld produce the stoichiometrical amount of Al2O3
change ot mol
MW of Fe2O3 = 159.69 g/mol
mol Fe2O3 = mass/MW = 3.17/159.69 = 0.0198509 mol of Fe2O3
then
since ratio is 1:1 according to equation
we cna produces
0.0198509 mol of Al2O3
mass = mol*MW
Mw of Al2O3 = 101.96 g/mol
then
mass = 101.96 *0.0198509 = 2.0239 g of Al2O3
probably you are wondering for the % yield so
% yield = real / theoretical * 100 = 1.54 /2.0239*100 = 76.0907 %
Get Answers For Free
Most questions answered within 1 hours.