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Respuesta :

Willchemistry
Molar mass 

O2 = 31.99 g/mol
Fe = 55.8 g/mol
Fe2O3 = 159.68 g/mol

Identifying excess reagent and the limiting of the reaction :

3 O2 + 4 Fe = 2 Fe2O3

3 x 31.99 g O2 ----------- 2 x 159.68 g Fe2O3
   ?? g O2 ----------------- 250 g Fe2O3

250 x 3 x 31.99 / 2 x 159.68 =

23992.5 / 319.36 => 75.12 g of O2   iron is  the  limiting reactant

4 x 55.8 g Fe ------------ 2 x 159.68 g Fe2O3
175 g Fe ----------------- ??

2 x 159.68 x 175 / 4 x 55.8 =

55888 / 223,2 => 250.394 g of Fe2O3 , Fe is the excess reagent 

therefore 75.12 g of O2 

hope this helps!

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