answer and explanation
we are given the balanced reaction and so we begin by calculating the number of mols of each reactant
for Al
mols = mass/Molar mass
= 22.0g / 26.98g/mol
= 0.82 mols
for oxygen
mols = mass / Molar mass
= 26.0g / 16.00g/mol
= 1.6 mols
theerefore
a. the limiting reagent is Al
b. theoretical yield will be determine using the limiting reagent.
from the balanced reaction we see that the mol ratio between aluminum and aluminum oxide is 4:2
therefore
2/4 x0.82 = 0.41 mols of aluminum oxide will form
the mass will be
mass = n x M
= 0.41 mols x 101.96g/mol
= 41.8 grams
c. percentage yield is:
actual yield/ theorectical yield x 100
= 40.0/41.8 x 100%
= 95.7 %
