How many grams of oxygen (o2), x, are needed to react with the 700 g of iron to produce 1000 g of iron (iii) oxide, fe2o3?

As we know that, molecular mass of ferric oxide, Fe2O3, is 159.69 grams.
Out of which, iron contributes 111.69 g (2 X 55.845 g) and oxygen contributes
48 g (3 X 16 g).   
 Each gram of iron (III) oxide contains 111.69/159.69 g of iron and 48/159.69
g of oxygen.   
 To produce 1000 g iron (III) oxide we need,    
 Iron = 111.69*1000/159.69 = 699.42 g   
 Oxygen = 48*1000/159.69 = 300.58 g

x - y = 2 is the equation in standard form.