27 g of Al reacts with excess of oxygen to give 4.59 g of $A{l}_2{O}_{3,}$ The percentage  yield  of reaction is ___________. (Use molar mass: Al=27 g/mol, O=16 g/mol)

$4Al+3O_2\to 2A{l}_2{O}_3$ 
$4\times 27g2\times 102g$ 
27g                              ?
                    $=\frac{2\times 102}{4\times 27}\times 27=51g$    
$\%yield=\frac{Actualyield}{Theoriticalyield}\times 100=\frac{4.59}{51}\times 100=9$