Question

Question #967d1

Answer 1

155.5g will be formed.

`4Fe+3O_2rarr2Fe_2O_3`

So 3 mol `O_2` `rarr` 2 mol `Fe_2O_3`

Using approximate `A_r` values:

`3xx(2xx16)`g `O_2` `rarr2[(56xx2)+(3xx16)]` g `Fe_2O_3`

So:

`64grarr320g`

So:

`1grarr320/64g`

So:

`31.1grarr(320)/(64)xx31.1=155.5g`