Question

The reaction between aluminum and iron(III) oxide can generate temperatures approaching 3000 degrees C and is used in welding metals: 2 Al + Fe2O3 = Al2O3 + 2Fe In one process, 126 g of Al are reacted with 601 g of Fe2O3 Calculate the mass, in grams, of Al2O3 formed

Answer 1

the reaction is

2 Al + Fe203 ---> Al203 + 2 Fe

we know that

moles = mass / molar mass

so

moles of Al taken = 126 / 27 = 4.667

moles of Fe203 taken = 601 / 159.69 = 3.7635

now

consider the given reaction

2 Al + Fe203 ---> Al203 + 2 Fe

we can see that

moles of Al required = 2 x moles of Fe203

so

moles of Al required = 2 x 3.7635

moles of Al required = 7.527

but

only 4.667 moles of Al is present

so

Al is the limiting reagent

now

we can see that

moles of Al203 formed = 0.5 x moles of Al reacted

so

moles of Al203 formed = 0.5 x 4.6667 = 2.3335

now

mass = moles x molar mass

so

mass of Al203 = 2.335 x 102 = 238

so

238 grams of Al203 is formed