17kg of aluminium was produced from 51kg of aluminium oxide (Al2O3) by electrolysis. What was the percentage yield?

Consider this reaction : 2Al2O3 → 4Al + 3O2

The number of moles of Al2O3 can be determined from its mass →

Number of moles of AL2O3 = mass/molar mass

=51 / (27x2+16x3)

= 0.5 mol

Now, calculate aluminium's number of moles from the mole ratio →

Mole ratio 4 (Al) : 2 (Al2O3)

Number of moles of Al = (4/2) x0.5mol = 1 mol

Mass of Al = number of moles x molar mass

= 1x27

= 27kg (theoretical yield)

Given, actual yield = 17kg

Therefore, percentage yield = (actual yield / theoretical yield) x 100

= (17/27) x100

= 62.96%