Question

En la reaccion 2Al + Fe2O3 produce Al2O3 + 2Fe. Si se tienen 800g de aluminio con 2 libras de oxido férrico. ¿Cuantas moles tiene el reactivo limitante? (Al= 27, Fe= 56, O=16)

Answer 1

                        2 Al + Fe2O3 → Al2O3 + 2Fe

Mm    Al = 27 g/mol
  Fe2O3 = 159.68 g/mol
  Al2O3 = 101.96 g/mol

1. convertir libras a gramos

g = libras x 453,59

g = 2 x 453.59 = 907.18 g de Al2O3

2. calcular moles

n Al = 800 g  x  1 mol
                        ``````````
                           27 g

n Al = 29.63 moles

n Fe2O3 = 907.18 g  x  1  mol
                                     ```````````
                                     159.68 g

n Fe2O3 = 5.68 moles

3. calcular moles de Al2O3  a partir de la Reacción

n Al2O3 = 29.63 moles Al  x   1 mol Al2O3
                                               ``````````````````
                                                   2 mol Al

n Al2O3 = 14.815

n Al2O3 = 5.68 moles Fe2O3 x 1 mol Al2O3
                                                   ``````````````````
                                                   1 mol Fe2O3

n Al2O3 = 5.68 moles

Reactivo Limitante: Fe2O3 produce 5.68 moles