What is the theoretical yield in grams of aluminum that can

What is the theoretical yield (in grams) of aluminum that can be produced by the reaction of 60.0 g of aluminum oxide with 30.0 g of carbon according to the following chemical reaction? Al_2O_3 + 3C rightarrow 2Al + 3CO

Solution

Al₂O₃ + 3 C = 2 Al + 3 CO

60 gm of Al2O3 = 60/ 101.96 = 0.5884 Moles

30 gm of carbon = 30 / 12.01 = 2.497 Moles

One molecule of Al2O3 needs 3 moelcule of carbon. Hence here carbon present more than 3 equivalent and Al2O3 is the limiting reagent

0.5584 Moles of Al2O3 can produce maximum (0.5584 x 2) 1.17691 Moles of aluminium

1.17691 Moles of aluminium = 1.17691 x 26.98 = 31.75 gm

Hence maximum of yield of aluminium is 31.75 gm