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%

where's the diagram?