20.0g sample of al reacts to produce 32.7g of Al2O3. what is the percentage yield of the reaction?

Question

4Al(s) + 3O₂(g) ==> 2Al₂O₃(s) ... balanced equation

use the method

(Al) mole: mass Al: Al3O3 mole: mass (Al2O3)

Answer 1

4Al(s) + 3O₂(g) ==> 2Al₂O₃(s) ... balanced equation

1). Find the limiting reactant:

20.0 g Al x 1 mol Al / 27 g = 0.741 mols Al

Assuming there is excess O2, we can proceed as follows:

2). Find theoretical yield:

0.741 mols Al x 2 mol Al₂O₃ / 4 mols Al x 102 g Al₂O₃/ mol = 37.79 g Al₂O₃ formed

3). Find percent yield:

% yield = actual yield / theoretical yield (x100%)

32.7 g / 37.79 g (x100%) = 86.5% yield