Question

if you start with 580 g aluminum oxide and you isolate 250g of aluminum at the end what was your percent yield

Answer 1

The mass of starting material, i.e. aluminium oxide (Al₂O₃) = 580 g

The molar mass of Al₂O₃ = 101.96 g/mol

i.e. The no. of moles of Al₂O₃ = 580 g /101.96 g mol^-1, i.e. 5.6885 mol

The mass of productl, i.e.aluminium (Al) = 250 g

The molar mass of Al = 26.982 g/mol

i.e. The no. of moles of Al = 250 g /26.982 g mol^-1, i.e. 9.2654 mol

The balanced reaction for the reduction of Al₂O₃ to Al can be written as follows.

Al₂O₃ 2Al + (3/2)O₂

i.e. 1 mole of Al₂O₃ is converted into 2 moles of Al, which means 100% theoretical yield.

The yield in the given case = {(9.2654 mol / (2 * 5.6885 mol)} * 100, i.e. 81.44 %