Question

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?

Answer 1

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 %