Question

In this case you have unlimited iron, but only 6.40 moles of oxygen (O2). How much Fe2O3 can you produce?

Answer 1

Answer: 682 grams

Step-by-step explanation:

The balanced chemical reaction is:

`4Fe+3O_2\rightarrow 2Fe_2O_3`

`O_2` is the limiting reagent as it limits the formation of product and
`Fe` is the excess reagent.

According to stoichiometry :

3 moles of
`O_2` produce = 2 moles of
`Fe_2O_3`

Thus 6.40 moles of
`O_2` will produce=
`(2)/(3)* 6.40=4.27moles` of
`Fe_2O_3`

Mass of
`Fe_2O_3=moles* {\text {Molar mass}}=4.27moles* 159.69g/mol=682g`

Thus 682 g of
`Fe_2O_3` will be produced from 6.40 moles of oxygen